Question

Electrons are bombarded to excite hydrogen atoms and six spectral lines are observed. If \[{E_g}\] is the ground state energy of hydrogen, the minimum energy the bombarding electrons should possess is
A. \[\dfrac{{8{E_g}}}{9} \\ \]
B. \[\dfrac{{15{E_g}}}{{16}} \\ \]
C. \[\dfrac{{35{E_g}}}{{36}} \\ \]
D. \[\dfrac{{48{E_g}}}{{49}}\]

Answer 1

Hint: To solve this question, we first need to find the maximum ${n^{th}}$ excited state to which the electrons of the hydrogen atoms will get excited. Then we will find the energy difference between the ${1^{st}}$ and the ${n^{th}}$ excited state, that is ${\left( {n + 1} \right)^{th}}$ energy state to find the minimum energy that the bombarding electrons must possess.

Formula Used:
$n = \dfrac{{\vartriangle n\left( {\vartriangle n + 1} \right)}}{2}$
where $n$ is the number of spectral lines observed and $\vartriangle n$ denotes the ${n^{th}}$ excited state.
Energy of ${n^{th}}$ energy state,
${E_n} = \dfrac{{{E_g}}}{{{n^2}}}$

Complete step by step solution:
Given: Number of spectral lines observed when electrons are bombarded to excite hydrogen atoms is, $n = 6$. A hydrogen atom may enter a higher energy state as a result of being bombarded. Light is released as the excited electron reverts to its lower energy states. We know that,
$n = \dfrac{{\vartriangle n\left( {\vartriangle n + 1} \right)}}{2}$

Substituting the value of $n$ in the above expression, we get,
$6 = \dfrac{{\vartriangle n\left( {\vartriangle n + 1} \right)}}{2}$
Solving for $\vartriangle n$ , we get,
$\vartriangle n\left( {\vartriangle n + 1} \right) = 12$
Simplifying this,
$\vartriangle n\left( {\vartriangle n + 1} \right) = 3 \times 4$
Therefore,
$\vartriangle n = 3$
This implies that the electrons of the hydrogen atoms will excite till ${3^{rd}}$ excited state, that is, till ${4^{th}}$ energy state.

We know that, the energy of ${n^{th}}$ energy state, ${E_n} = \dfrac{{{E_g}}}{{{n^2}}}$
The required minimum energy is the energy difference between ${1^{st}}$ and ${3^{rd}}$ excited state.
That is, ${n_1} = 1$ and ${n_2} = 4$
Therefore,
${E_{\min }} = \dfrac{{{E_g}}}{{{n_1}^2}} - \dfrac{{{E_g}}}{{{n_2}^2}} \\ $
Substituting the values of ${n_1}\,\& \,{n_{_2}}$ , we get,
${E_{\min }} = \dfrac{{{E_g}}}{{{1^2}}} - \dfrac{{{E_g}}}{{{4^2}}} \\ $
Solving this, we get,
${E_{\min }} = \dfrac{{{E_g}}}{1} - \dfrac{{{E_g}}}{{16}} \\ $
$\therefore {E_{\min }} = \dfrac{{15{E_g}}}{{16}}$

Hence, option B is the answer.

Note: An important point to be noted here is the difference in the meaning of the terms ${n^{th}}$ excited energy state and ${n^{th}}$ energy state. This is because ${n^{th}}$ excited state means ${\left( {n + 1} \right)^{th}}$ energy state. While solving this question, do not take the value of $\vartriangle n$ as the maximum energy state $\left( {{n_2}} \right)$ to which the electrons of the atoms are excited. Rather take the value of $\vartriangle n + 1$ as the maximum state to which the electrons are excited.