The correctness of the statements,“The first excited state of an electron in a hydrogen atomis n = 4 and it takes more energy to ionize the electron from n = 4 than from ground state,”“The electron is farther from the nucleus in n = 4 than in the ground state,”“The wavelength of light when the electron drops from n = 4 to n = 1 is longer than that from n = 4 to n = 2 ,” and “The wavelength of absorbed light during transition from n = 1 to n = 4 and n = 4 to n = 1 is same,”isto be commented on. Concept introduction: When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy ( Δ E ) is determined as follows: Δ E = − 2.18 × 10 − 18 J ( 1 n f 2 − 1 n i 2 ) Here, n f is the excited state and n i is the ground state. Energy of the electron for the given value of n (energy state) ( E n ) is inversely proportional to the square of n . E n = − 2.18 × 10 − 18 J ( 1 n 2 ) Here, n f is the excited state and n i is the ground state. The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows: 1 λ = 2.18 × 10 − 18 J h c ( 1 n f 2 − 1 n i 2 ) Here, n f is the excited state and n i is the ground state.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 6, Problem 109AP

Interpretation Introduction

Interpretation:

The correctness of the statements,“The first excited state of an electron in a hydrogen atomis $n=4$ and it takes more energy to ionize the electron from $n=4$ than from ground state,”“The electron is farther from the nucleus in $n=4$ than in the ground state,”“The wavelength of light when the electron drops from $n=4 to n=1$ is longer than that from $n=4 to n=2$ ,” and “The wavelength of absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is same,”isto be commented on.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy $(ΔE)$ is determined as follows:

$ΔE=−2.18×10−18 J(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Energy of the electron for the given value of n (energy state) $(En)$ is inversely proportional to the square of n.

$En=−2.18×10−18 J(1n2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows:

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Expert Solution & Answer

Answer to Problem 109AP

Solution:

(a)False. $n=2$ is the first excited state.

(b)False. In the state, $n=4$ , electrons can easily be removed as they are away from the nucleus.

(c)True. The distance between the nucleus and the electron in $n=4$ is more as compared to the distance from the nucleus and the electron in ground state.

(d)False. Transition from $n=4 to n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1 to n=4$ , and from $n=4 to n=1$ is the same.

Explanation of Solution

a) $n=4$ is the first excited state.

First excited state is the energy state, which is immediate next to the ground state. Excited state can be counted as one less than the electron in the energy state. Therefore, $n=4$ is not the first excited state. It is the third excited state.

b)It takes more energy to ionize the electron from $n=4$ then from the ground state.

The energy of electron for the given value of $n$ (energy state), $(En)$ is inversely proportional to the square of $n$ . The equation is as follows:

$En=−2.18×10−18 J(1n2)$

The energy of electron in $n=4$ and in ground state, $n=1$ , can be compared as follows:

$E4E1=−2.18×10−18 J(142)−2.18×10−18 J(112)E1=16E4$

Therefore, $E4<E1$

Therefore, the energy required for removing an electron from $n=4$ is lesser than the energy required for removing electron from ground state.

c) The electron is farther the nucleus in $n=4$ than in ground state.

The distance between the electrons in energy states increases as the value of $(n)$ increases. Therefore, the distance between the nucleus and the electron in $n=4$ is more as compared to the distance between the nucleus and the electron in ground state, $n=1$ .

d) The wavelength of light emitted when the electron drops from $n=4$ to $n=1$ longer than that from $n=4 to n=2$ .

The wavelength of electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

For transition from $n=4 to n=1$ , the wavelength of emitted light is longer than the wavelength of light during the transition from $n=4 to n=2$ .

$(1λ4→1)(1λ4→2)=2.18×10−18 Jhc(112−142)2.18×10−18 Jhc(122−142)λ4→2λ4→1=51λ4→2=5×λ4→1$

Therefore, $λ4→2>λ4→1$ , which means that when the electron drops from $n=4 to n=1$ , shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4 to n=2$ .

e) The wavelength the atom absorbs in going from $n=1 to n=4$ is the same as that emitted as it goes from $n=4 to n=1$ .

The wavelength of an electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

So, $λ∝|ni2nf2ni2−nf2|$

The wavelength of radiation in both the cases can be compared as

$λ1→4λ4→1=|(n12n42)(n12−n42)||(n12n42)(n42−n12)|$

On solving it further, we get

$λ1→4λ4→1=|−(n12−n42)||(n12−n42)|λ1→4λ4→1=|−1||1|λ1→4λ4→1=1λ1→4=λ4→1$

Therefore, the wavelength of the absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is the same.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

esc Draw the Markovnikov product of the hydration of this alkene. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for this problem. Explanation Check BBB + X 0 1. Hg (OAc)2, H₂O 2. Na BH 5 Click and drag to start drawing a structure. © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility Bl P 豆 28 2 28 N 9 W E R T Y A S aps lock G H K L Z X C V B N M T central H command #e command

C A student proposes the transformation below in one step of an organic synthesis. There may be one or more products missing from the right-hand side, but there are no reagents missing from the left-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. . If the student's transformation is possible, then complete the reaction by adding any missing products to the right-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. (X) This transformation can't be done in one step. + T

く Predict the major products of this organic reaction. If there aren't any products, because nothing will happen, check the box under the drawing area instead. No reaction. Explanation Check OH + + ✓ 2 H₂SO 4 O xs H₂O 2 Click and drag to start drawing a structure. © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center

Answer 2

Question

Chapter 6, Problem 109AP

Interpretation Introduction

Interpretation:

The correctness of the statements,“The first excited state of an electron in a hydrogen atomis $n=4$ and it takes more energy to ionize the electron from $n=4$ than from ground state,”“The electron is farther from the nucleus in $n=4$ than in the ground state,”“The wavelength of light when the electron drops from $n=4 to n=1$ is longer than that from $n=4 to n=2$ ,” and “The wavelength of absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is same,”isto be commented on.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy $(ΔE)$ is determined as follows:

$ΔE=−2.18×10−18 J(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Energy of the electron for the given value of n (energy state) $(En)$ is inversely proportional to the square of n.

$En=−2.18×10−18 J(1n2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows:

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Expert Solution & Answer

Answer to Problem 109AP

Solution:

(a)False. $n=2$ is the first excited state.

(b)False. In the state, $n=4$ , electrons can easily be removed as they are away from the nucleus.

(c)True. The distance between the nucleus and the electron in $n=4$ is more as compared to the distance from the nucleus and the electron in ground state.

(d)False. Transition from $n=4 to n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1 to n=4$ , and from $n=4 to n=1$ is the same.

Explanation of Solution

a) $n=4$ is the first excited state.

First excited state is the energy state, which is immediate next to the ground state. Excited state can be counted as one less than the electron in the energy state. Therefore, $n=4$ is not the first excited state. It is the third excited state.

b)It takes more energy to ionize the electron from $n=4$ then from the ground state.

The energy of electron for the given value of $n$ (energy state), $(En)$ is inversely proportional to the square of $n$ . The equation is as follows:

$En=−2.18×10−18 J(1n2)$

The energy of electron in $n=4$ and in ground state, $n=1$ , can be compared as follows:

$E4E1=−2.18×10−18 J(142)−2.18×10−18 J(112)E1=16E4$

Therefore, $E4<E1$

Therefore, the energy required for removing an electron from $n=4$ is lesser than the energy required for removing electron from ground state.

c) The electron is farther the nucleus in $n=4$ than in ground state.

The distance between the electrons in energy states increases as the value of $(n)$ increases. Therefore, the distance between the nucleus and the electron in $n=4$ is more as compared to the distance between the nucleus and the electron in ground state, $n=1$ .

d) The wavelength of light emitted when the electron drops from $n=4$ to $n=1$ longer than that from $n=4 to n=2$ .

The wavelength of electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

For transition from $n=4 to n=1$ , the wavelength of emitted light is longer than the wavelength of light during the transition from $n=4 to n=2$ .

$(1λ4→1)(1λ4→2)=2.18×10−18 Jhc(112−142)2.18×10−18 Jhc(122−142)λ4→2λ4→1=51λ4→2=5×λ4→1$

Therefore, $λ4→2>λ4→1$ , which means that when the electron drops from $n=4 to n=1$ , shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4 to n=2$ .

e) The wavelength the atom absorbs in going from $n=1 to n=4$ is the same as that emitted as it goes from $n=4 to n=1$ .

The wavelength of an electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

So, $λ∝|ni2nf2ni2−nf2|$

The wavelength of radiation in both the cases can be compared as

$λ1→4λ4→1=|(n12n42)(n12−n42)||(n12n42)(n42−n12)|$

On solving it further, we get

$λ1→4λ4→1=|−(n12−n42)||(n12−n42)|λ1→4λ4→1=|−1||1|λ1→4λ4→1=1λ1→4=λ4→1$

Therefore, the wavelength of the absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is the same.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 6, Problem 109AP

Interpretation Introduction

Interpretation:

The correctness of the statements,“The first excited state of an electron in a hydrogen atomis $n=4$ and it takes more energy to ionize the electron from $n=4$ than from ground state,”“The electron is farther from the nucleus in $n=4$ than in the ground state,”“The wavelength of light when the electron drops from $n=4 to n=1$ is longer than that from $n=4 to n=2$ ,” and “The wavelength of absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is same,”isto be commented on.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy $(ΔE)$ is determined as follows:

$ΔE=−2.18×10−18 J(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Energy of the electron for the given value of n (energy state) $(En)$ is inversely proportional to the square of n.

$En=−2.18×10−18 J(1n2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows:

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Expert Solution & Answer

Answer to Problem 109AP

Solution:

(a)False. $n=2$ is the first excited state.

(b)False. In the state, $n=4$ , electrons can easily be removed as they are away from the nucleus.

(c)True. The distance between the nucleus and the electron in $n=4$ is more as compared to the distance from the nucleus and the electron in ground state.

(d)False. Transition from $n=4 to n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1 to n=4$ , and from $n=4 to n=1$ is the same.

Explanation of Solution

a) $n=4$ is the first excited state.

First excited state is the energy state, which is immediate next to the ground state. Excited state can be counted as one less than the electron in the energy state. Therefore, $n=4$ is not the first excited state. It is the third excited state.

b)It takes more energy to ionize the electron from $n=4$ then from the ground state.

The energy of electron for the given value of $n$ (energy state), $(En)$ is inversely proportional to the square of $n$ . The equation is as follows:

$En=−2.18×10−18 J(1n2)$

The energy of electron in $n=4$ and in ground state, $n=1$ , can be compared as follows:

$E4E1=−2.18×10−18 J(142)−2.18×10−18 J(112)E1=16E4$

Therefore, $E4<E1$

Therefore, the energy required for removing an electron from $n=4$ is lesser than the energy required for removing electron from ground state.

c) The electron is farther the nucleus in $n=4$ than in ground state.

The distance between the electrons in energy states increases as the value of $(n)$ increases. Therefore, the distance between the nucleus and the electron in $n=4$ is more as compared to the distance between the nucleus and the electron in ground state, $n=1$ .

d) The wavelength of light emitted when the electron drops from $n=4$ to $n=1$ longer than that from $n=4 to n=2$ .

The wavelength of electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

For transition from $n=4 to n=1$ , the wavelength of emitted light is longer than the wavelength of light during the transition from $n=4 to n=2$ .

$(1λ4→1)(1λ4→2)=2.18×10−18 Jhc(112−142)2.18×10−18 Jhc(122−142)λ4→2λ4→1=51λ4→2=5×λ4→1$

Therefore, $λ4→2>λ4→1$ , which means that when the electron drops from $n=4 to n=1$ , shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4 to n=2$ .

e) The wavelength the atom absorbs in going from $n=1 to n=4$ is the same as that emitted as it goes from $n=4 to n=1$ .

The wavelength of an electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

So, $λ∝|ni2nf2ni2−nf2|$

The wavelength of radiation in both the cases can be compared as

$λ1→4λ4→1=|(n12n42)(n12−n42)||(n12n42)(n42−n12)|$

On solving it further, we get

$λ1→4λ4→1=|−(n12−n42)||(n12−n42)|λ1→4λ4→1=|−1||1|λ1→4λ4→1=1λ1→4=λ4→1$

Therefore, the wavelength of the absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is the same.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 6, Problem 109AP

Interpretation Introduction

Interpretation:

The correctness of the statements,“The first excited state of an electron in a hydrogen atomis $n=4$ and it takes more energy to ionize the electron from $n=4$ than from ground state,”“The electron is farther from the nucleus in $n=4$ than in the ground state,”“The wavelength of light when the electron drops from $n=4 to n=1$ is longer than that from $n=4 to n=2$ ,” and “The wavelength of absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is same,”isto be commented on.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy $(ΔE)$ is determined as follows:

$ΔE=−2.18×10−18 J(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Energy of the electron for the given value of n (energy state) $(En)$ is inversely proportional to the square of n.

$En=−2.18×10−18 J(1n2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows:

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Expert Solution & Answer

Answer to Problem 109AP

Solution:

(a)False. $n=2$ is the first excited state.

(b)False. In the state, $n=4$ , electrons can easily be removed as they are away from the nucleus.

(c)True. The distance between the nucleus and the electron in $n=4$ is more as compared to the distance from the nucleus and the electron in ground state.

(d)False. Transition from $n=4 to n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1 to n=4$ , and from $n=4 to n=1$ is the same.

Explanation of Solution

a) $n=4$ is the first excited state.

First excited state is the energy state, which is immediate next to the ground state. Excited state can be counted as one less than the electron in the energy state. Therefore, $n=4$ is not the first excited state. It is the third excited state.

b)It takes more energy to ionize the electron from $n=4$ then from the ground state.

The energy of electron for the given value of $n$ (energy state), $(En)$ is inversely proportional to the square of $n$ . The equation is as follows:

$En=−2.18×10−18 J(1n2)$

The energy of electron in $n=4$ and in ground state, $n=1$ , can be compared as follows:

$E4E1=−2.18×10−18 J(142)−2.18×10−18 J(112)E1=16E4$

Therefore, $E4<E1$

Therefore, the energy required for removing an electron from $n=4$ is lesser than the energy required for removing electron from ground state.

c) The electron is farther the nucleus in $n=4$ than in ground state.

The distance between the electrons in energy states increases as the value of $(n)$ increases. Therefore, the distance between the nucleus and the electron in $n=4$ is more as compared to the distance between the nucleus and the electron in ground state, $n=1$ .

d) The wavelength of light emitted when the electron drops from $n=4$ to $n=1$ longer than that from $n=4 to n=2$ .

The wavelength of electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

For transition from $n=4 to n=1$ , the wavelength of emitted light is longer than the wavelength of light during the transition from $n=4 to n=2$ .

$(1λ4→1)(1λ4→2)=2.18×10−18 Jhc(112−142)2.18×10−18 Jhc(122−142)λ4→2λ4→1=51λ4→2=5×λ4→1$

Therefore, $λ4→2>λ4→1$ , which means that when the electron drops from $n=4 to n=1$ , shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4 to n=2$ .

e) The wavelength the atom absorbs in going from $n=1 to n=4$ is the same as that emitted as it goes from $n=4 to n=1$ .

The wavelength of an electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

So, $λ∝|ni2nf2ni2−nf2|$

The wavelength of radiation in both the cases can be compared as

$λ1→4λ4→1=|(n12n42)(n12−n42)||(n12n42)(n42−n12)|$

On solving it further, we get

$λ1→4λ4→1=|−(n12−n42)||(n12−n42)|λ1→4λ4→1=|−1||1|λ1→4λ4→1=1λ1→4=λ4→1$

Therefore, the wavelength of the absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is the same.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 6, Problem 109AP

Interpretation Introduction

Interpretation:

The correctness of the statements,“The first excited state of an electron in a hydrogen atomis $n=4$ and it takes more energy to ionize the electron from $n=4$ than from ground state,”“The electron is farther from the nucleus in $n=4$ than in the ground state,”“The wavelength of light when the electron drops from $n=4 to n=1$ is longer than that from $n=4 to n=2$ ,” and “The wavelength of absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is same,”isto be commented on.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy $(ΔE)$ is determined as follows:

$ΔE=−2.18×10−18 J(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Energy of the electron for the given value of n (energy state) $(En)$ is inversely proportional to the square of n.

$En=−2.18×10−18 J(1n2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows:

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Expert Solution & Answer

Answer to Problem 109AP

Solution:

(a)False. $n=2$ is the first excited state.

(b)False. In the state, $n=4$ , electrons can easily be removed as they are away from the nucleus.

(c)True. The distance between the nucleus and the electron in $n=4$ is more as compared to the distance from the nucleus and the electron in ground state.

(d)False. Transition from $n=4 to n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1 to n=4$ , and from $n=4 to n=1$ is the same.

Explanation of Solution

a) $n=4$ is the first excited state.

First excited state is the energy state, which is immediate next to the ground state. Excited state can be counted as one less than the electron in the energy state. Therefore, $n=4$ is not the first excited state. It is the third excited state.

b)It takes more energy to ionize the electron from $n=4$ then from the ground state.

The energy of electron for the given value of $n$ (energy state), $(En)$ is inversely proportional to the square of $n$ . The equation is as follows:

$En=−2.18×10−18 J(1n2)$

The energy of electron in $n=4$ and in ground state, $n=1$ , can be compared as follows:

$E4E1=−2.18×10−18 J(142)−2.18×10−18 J(112)E1=16E4$

Therefore, $E4<E1$

Therefore, the energy required for removing an electron from $n=4$ is lesser than the energy required for removing electron from ground state.

c) The electron is farther the nucleus in $n=4$ than in ground state.

The distance between the electrons in energy states increases as the value of $(n)$ increases. Therefore, the distance between the nucleus and the electron in $n=4$ is more as compared to the distance between the nucleus and the electron in ground state, $n=1$ .

d) The wavelength of light emitted when the electron drops from $n=4$ to $n=1$ longer than that from $n=4 to n=2$ .

The wavelength of electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

For transition from $n=4 to n=1$ , the wavelength of emitted light is longer than the wavelength of light during the transition from $n=4 to n=2$ .

$(1λ4→1)(1λ4→2)=2.18×10−18 Jhc(112−142)2.18×10−18 Jhc(122−142)λ4→2λ4→1=51λ4→2=5×λ4→1$

Therefore, $λ4→2>λ4→1$ , which means that when the electron drops from $n=4 to n=1$ , shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4 to n=2$ .

e) The wavelength the atom absorbs in going from $n=1 to n=4$ is the same as that emitted as it goes from $n=4 to n=1$ .

The wavelength of an electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

So, $λ∝|ni2nf2ni2−nf2|$

The wavelength of radiation in both the cases can be compared as

$λ1→4λ4→1=|(n12n42)(n12−n42)||(n12n42)(n42−n12)|$

On solving it further, we get

$λ1→4λ4→1=|−(n12−n42)||(n12−n42)|λ1→4λ4→1=|−1||1|λ1→4λ4→1=1λ1→4=λ4→1$

Therefore, the wavelength of the absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is the same.

Answer 6

Chapter 6, Problem 109AP

Interpretation Introduction

Interpretation:

The correctness of the statements,“The first excited state of an electron in a hydrogen atomis $n=4$ and it takes more energy to ionize the electron from $n=4$ than from ground state,”“The electron is farther from the nucleus in $n=4$ than in the ground state,”“The wavelength of light when the electron drops from $n=4 to n=1$ is longer than that from $n=4 to n=2$ ,” and “The wavelength of absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is same,”isto be commented on.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy $(ΔE)$ is determined as follows:

$ΔE=−2.18×10−18 J(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Energy of the electron for the given value of n (energy state) $(En)$ is inversely proportional to the square of n.

$En=−2.18×10−18 J(1n2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows:

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Expert Solution & Answer

Answer to Problem 109AP

Solution:

(a)False. $n=2$ is the first excited state.

(b)False. In the state, $n=4$ , electrons can easily be removed as they are away from the nucleus.

(c)True. The distance between the nucleus and the electron in $n=4$ is more as compared to the distance from the nucleus and the electron in ground state.

(d)False. Transition from $n=4 to n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1 to n=4$ , and from $n=4 to n=1$ is the same.

Explanation of Solution

a) $n=4$ is the first excited state.

First excited state is the energy state, which is immediate next to the ground state. Excited state can be counted as one less than the electron in the energy state. Therefore, $n=4$ is not the first excited state. It is the third excited state.

b)It takes more energy to ionize the electron from $n=4$ then from the ground state.

The energy of electron for the given value of $n$ (energy state), $(En)$ is inversely proportional to the square of $n$ . The equation is as follows:

$En=−2.18×10−18 J(1n2)$

The energy of electron in $n=4$ and in ground state, $n=1$ , can be compared as follows:

$E4E1=−2.18×10−18 J(142)−2.18×10−18 J(112)E1=16E4$

Therefore, $E4<E1$

Therefore, the energy required for removing an electron from $n=4$ is lesser than the energy required for removing electron from ground state.

c) The electron is farther the nucleus in $n=4$ than in ground state.

The distance between the electrons in energy states increases as the value of $(n)$ increases. Therefore, the distance between the nucleus and the electron in $n=4$ is more as compared to the distance between the nucleus and the electron in ground state, $n=1$ .

d) The wavelength of light emitted when the electron drops from $n=4$ to $n=1$ longer than that from $n=4 to n=2$ .

The wavelength of electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

For transition from $n=4 to n=1$ , the wavelength of emitted light is longer than the wavelength of light during the transition from $n=4 to n=2$ .

$(1λ4→1)(1λ4→2)=2.18×10−18 Jhc(112−142)2.18×10−18 Jhc(122−142)λ4→2λ4→1=51λ4→2=5×λ4→1$

Therefore, $λ4→2>λ4→1$ , which means that when the electron drops from $n=4 to n=1$ , shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4 to n=2$ .

e) The wavelength the atom absorbs in going from $n=1 to n=4$ is the same as that emitted as it goes from $n=4 to n=1$ .

The wavelength of an electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

So, $λ∝|ni2nf2ni2−nf2|$

The wavelength of radiation in both the cases can be compared as

$λ1→4λ4→1=|(n12n42)(n12−n42)||(n12n42)(n42−n12)|$

On solving it further, we get

$λ1→4λ4→1=|−(n12−n42)||(n12−n42)|λ1→4λ4→1=|−1||1|λ1→4λ4→1=1λ1→4=λ4→1$

Therefore, the wavelength of the absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is the same.

Answer 7

Chapter 6, Problem 109AP

Interpretation Introduction

Interpretation:

The correctness of the statements,“The first excited state of an electron in a hydrogen atomis $n=4$ and it takes more energy to ionize the electron from $n=4$ than from ground state,”“The electron is farther from the nucleus in $n=4$ than in the ground state,”“The wavelength of light when the electron drops from $n=4 to n=1$ is longer than that from $n=4 to n=2$ ,” and “The wavelength of absorbed light during transition from $n=1 to n=4$ and $n=4 to n=1$ is same,”isto be commented on.

Concept introduction:

When an electron jumps from one energy state to another energy state, it is termed as transition of electron. Energy is either lost or absorbed during the transition of an electron. The transition energy $(ΔE)$ is determined as follows:

$ΔE=−2.18×10−18 J(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Energy of the electron for the given value of n (energy state) $(En)$ is inversely proportional to the square of n.

$En=−2.18×10−18 J(1n2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

The wavelength of an electron when it jumps from one energy state to another energy state is determined as follows:

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

Here, $nf$ is the excited state and $ni$ is the ground state.

Answer 8

Expert Solution & Answer

Answer to Problem 109AP

Solution:

(a)False. $n=2$ is the first excited state.

(b)False. In the state, $n=4$ , electrons can easily be removed as they are away from the nucleus.

(c)True. The distance between the nucleus and the electron in $n=4$ is more as compared to the distance from the nucleus and the electron in ground state.

(d)False. Transition from $n=4 to n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1 to n=4$ , and from $n=4 to n=1$ is the same.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

a) $n=4$ is the first excited state.

First excited state is the energy state, which is immediate next to the ground state. Excited state can be counted as one less than the electron in the energy state. Therefore, $n=4$ is not the first excited state. It is the third excited state.

b)It takes more energy to ionize the electron from $n=4$ then from the ground state.

The energy of electron for the given value of $n$ (energy state), $(En)$ is inversely proportional to the square of $n$ . The equation is as follows:

$En=−2.18×10−18 J(1n2)$

The energy of electron in $n=4$ and in ground state, $n=1$ , can be compared as follows:

$E4E1=−2.18×10−18 J(142)−2.18×10−18 J(112)E1=16E4$

Therefore, $E4<E1$

Therefore, the energy required for removing an electron from $n=4$ is lesser than the energy required for removing electron from ground state.

c) The electron is farther the nucleus in $n=4$ than in ground state.

The distance between the electrons in energy states increases as the value of $(n)$ increases. Therefore, the distance between the nucleus and the electron in $n=4$ is more as compared to the distance between the nucleus and the electron in ground state, $n=1$ .

d) The wavelength of light emitted when the electron drops from $n=4$ to $n=1$ longer than that from $n=4 to n=2$ .

The wavelength of electron when it jumps from one energy state to another energy state can be evaluated as

$1λ=2.18×10−18 Jhc(1nf2−1ni2)$

For transition from $n=4 to n=1$ , the wavelength of emitted light is longer than the wavelength of light during the transition from $n=4 to n=2$ .

$(1λ4→1)(1λ4→2)=2.18×10−18 Jhc(112−142)2.18×10−18 Jhc(122−142)λ4→2λ4→1=51λ4→2=5×λ4→1$

Therefore, $λ4→2>λ4→1$ , which means that when the electron drops from $n=4 to n=1$ , shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4 to n=2$ .