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 $n=4\text{ to }n=1$ is longer than that from $n=4\text{ to }n=2$,” and “The wavelength of absorbed light during transition from $n=1\text{ to }n=4$ and $n=4\text{ to }n=1$ 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 $\left(\Delta E\right)$ is determined as follows:

$\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

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) $\left(E_n\right)$ is inversely proportional to the square of n.

$E_n=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_{}^2}\right)$

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:

$\frac{1}{\lambda }=\frac{2.18\times {10}^{-18}\text{ J}}{\text{h}c}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

Here, $n_f^{}$ is the excited state and $n_i^{}$ is the ground state.

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 $n=4\text{ to }n=1$ takes high energy, which corresponds to shorter wavelength.

(e) True. The wavelength of light absorbed for transition from $n=1\text{ to }n=4$, and from $n=4\text{ to }n=1$ 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), $\left(E_n\right)$ is inversely proportional to the square of $n$. The equation is as follows:

$E_n=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_{}^2}\right)$

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

$\begin{array}{c}\frac{E_4}{E_1}=\frac{-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{4_{}^2}\right)}{-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{1_{}^2}\right)}\\ E_1=16E_4\end{array}$

Therefore, $E_4<E_1$

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 $\left(n\right)$ 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 $n=4\text{ to }n=2$.

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

$\frac{1}{\lambda }=\frac{2.18\times {10}^{-18}\text{ J}}{\text{hc}}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

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

$\begin{array}{c}\frac{\left(\frac{1}{{\lambda }_{4\to 1}}\right)}{\left(\frac{1}{{\lambda }_{4\to 2}}\right)}=\frac{\frac{2.18\times {10}^{-18}\text{ J}}{\text{hc}}\left(\frac{1}{1^2}-\frac{1}{4^2}\right)}{\frac{2.18\times {10}^{-18}\text{ J}}{\text{hc}}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}\\ \frac{{\lambda }_{4\to 2}}{{\lambda }_{4\to 1}}=\frac{5}{1}\\ {\lambda }_{4\to 2}=5\times {\lambda }_{4\to 1}\end{array}$

Therefore, ${\lambda }_{4\to 2}>{\lambda }_{4\to 1}$, which means that when the electron drops from $n=4\text{ to }n=1$, shorter wavelength of light is emitted than the wavelength of light when the electron drops from $n=4\text{ to }n=2$.

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

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

$\frac{1}{\lambda }=\frac{2.18\times {10}^{-18}\text{ J}}{\text{hc}}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

So, $\lambda \propto |\frac{n_i^2{n}_f^2}{n_i^2-n_f^2}|$

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

$\frac{{\lambda }_{1\to 4}}{{\lambda }_{4\to 1}}=\frac{|\raisebox{1ex}{$\left(n_1^2{n}_4^2\right)$}\!\left/ \!\raisebox{-1ex}{$\left(n_1^2-n_4^2\right)$}\right.|}{|\raisebox{1ex}{$\left(n_1^2{n}_4^2\right)$}\!\left/ \!\raisebox{-1ex}{$\left(n_4^2-n_1^2\right)$}\right.|}$

On solving it further, we get

$\begin{array}{l}\frac{{\lambda }_{1\to 4}}{{\lambda }_{4\to 1}}=\frac{|-\left(n_1^2-n_4^2\right)|}{|\left(n_1^2-n_4^2\right)|}\\ \frac{{\lambda }_{1\to 4}}{{\lambda }_{4\to 1}}=\frac{|-1|}{|1|}\\ \frac{{\lambda }_{1\to 4}}{{\lambda }_{4\to 1}}=1\\ {\lambda }_{1\to 4}={\lambda }_{4\to 1}\end{array}$

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