Chemistry [hardcover]
Chemistry [hardcover]
5th Edition
ISBN: 9780393264845
Author: Geoffery Davies
Publisher: NORTON
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Chapter 7, Problem 7.50QP

(a)

Interpretation Introduction

Interpretation: The given transitions in the hydrogen atom in order of increasing frequency of the electromagnetic radiation are to be ranked.

Concept introduction: Frequency of light is defined as the number of waves passes in space during a given interval of time of one second. The energy of a photon is defined as the energy of a single photon. The frequency of photon is inversely proportional to the wavelength of the electromagnetic wave.

To determine: The frequency of the given transition.

(a)

Expert Solution
Check Mark

Answer to Problem 7.50QP

Solution

The value of 1n121n22 for the given transition is 0.0347_ .

Explanation of Solution

Explanation

Given

The transition is from n=4ton=6 .

The relation between the wavelength of an electronic transition and the value of orbit in a hydrogen atom is given by Rydberg equation,

1λ=R(1n121n22) (1)

Where,

  • λ is the wavelength of photons.
  • R is the Rydberg constant (1.09×102(nm)1) .
  • n1 is the lower energy state.
  • n2 is the higher energy state.

The frequency of photons is given by the formula,

ν=cλ (2)

Where,

  • ν is the frequency of electromagnetic radiation.
  • c is the speed of light.
  • λ is the wavelength.

Therefore, from equation (1) and (2) it is clear that the frequency of photons is inversely proportional to the wavelength of photons.

να1λ

Hence,

να(1n121n22)

Therefore, the value of 1n121n22 for the given transition is calculated as,

1n121n22=1(4)21(6)2=116136=361616×36=20576

Simplify the above equation.

1n121n22=20576=0.0347_

Hence, the value of 1n121n22 for the given transition is 0.0347_ .

(b)

Interpretation Introduction

To determine: The frequency of the given transition.

(b)

Expert Solution
Check Mark

Answer to Problem 7.50QP

Solution

The value of 1n121n22 for the given transition is 0.0121_ .

Explanation of Solution

Explanation

Given

The transition is from n=6ton=8 .

The value of 1n121n22 for the given transition is calculated as,

1n121n22=1(6)21(8)2=136164=643636×64=282304

Simplify the above equation.

1n121n22=282304=1.0121_

Hence, the value of 1n121n22 for the given transition is 0.0121_ .

(c)

Interpretation Introduction

To determine: The frequency of the given transition.

(c)

Expert Solution
Check Mark

Answer to Problem 7.50QP

Solution

The value of 1n121n22 for the given transition is 0.0040_ .

Explanation of Solution

Explanation

Given

The transition is from n=9ton=11 .

The value of 1n121n22 for the given transition is calculated as,

1n121n22=1(9)21(11)2=1811121=1218181×121=409801

Simplify the above equation.

1n121n22=409801=0.0040_

Hence, the value of 1n121n22 for the given transition is 0.0040_ .

(d)

Interpretation Introduction

To determine: The frequency of the given transition.

(d)

Expert Solution
Check Mark

Answer to Problem 7.50QP

Solution

The value of 1n121n22 for the given transition is 0.0023_ .

Explanation of Solution

Explanation

Given

The transition is from n=11ton=13 .

The value of 1n121n22 for the given transition is calculated as,

1n121n22=1(11)21(13)2=11211169=169121121×169=4820449

Simplify the above equation.

1n121n22=4820449=0.0023_

Hence, the value of 1n121n22 for the given transition is 0.0023_ .

Conclusion

The transitions in the hydrogen atom in order of increasing frequency of the electromagnetic radiation are,

n=11ton=13<n=9ton=11<n=6ton=8<n=4ton=6 .

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Chapter 7 Solutions

Chemistry [hardcover]

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