Physics - Modern Physics Question with Solution | TestHub

When hydrogen gas is bombarded with electrons of energy 12.09 eV, the hydrogen atoms are excited to higher energy levels. The energy levels of the hydrogen atom are given by:

$$E_n=-\frac{13.6\mathrm{eV}}{n^2}$$

where $n$ is the principal quantum number.

The ground state energy is $E_1=-13.6\mathrm{eV}$.

The energy of the electron is 12.09 eV, so the total energy of the excited state is:

$E_n=-13.6\mathrm{eV}+12.09\mathrm{eV}=-1.51\mathrm{eV}$

Now, we find the value of $n$ for this energy:

$-1.51\mathrm{eV}=-\frac{13.6\mathrm{eV}}{n^2}$

$n^2=\frac{13.6}{1.51}\approx 9$

$n=3$

So, the electron is excited to the $n=3$ level. When the electron returns to lower energy levels, it can make transitions from $n=3$ to $n=2$, $n=3$ to $n=1$, and $n=2$ to $n=1$.

The energy of a photon emitted during a transition from $n_i$ to $n_f$ is given by:

$$E=13.6\mathrm{eV}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$$

The frequency of the emitted photon is given by $E=hf$, where $h=4.135\times 10^{-15}\mathrm{eV}\cdot \mathrm{s}$ is Planck's constant.

1. $n=3$ to $n=2$:

$E=13.6\mathrm{eV}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\mathrm{eV}\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\mathrm{eV}\left(\frac{5}{36}\right)\approx 1.89\mathrm{eV}$

$f=\frac{E}{h}=\frac{1.89\mathrm{eV}}{4.135\times 10^{-15}\mathrm{eV}\cdot \mathrm{s}}\approx 4.57\times 10^{14}\mathrm{Hz}=4.57\times 10^{14}{\mathrm{s}}^{-1}$

2. $n=3$ to $n=1$:

$E=13.6\mathrm{eV}\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=13.6\mathrm{eV}\left(1-\frac{1}{9}\right)=13.6\mathrm{eV}\left(\frac{8}{9}\right)\approx 12.09\mathrm{eV}$

$f=\frac{E}{h}=\frac{12.09\mathrm{eV}}{4.135\times 10^{-15}\mathrm{eV}\cdot \mathrm{s}}\approx 2.92\times 10^{15}\mathrm{Hz}=2.92\times 10^{15}{\mathrm{s}}^{-1}$

3. $n=2$ to $n=1$:

$E=13.6\mathrm{eV}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6\mathrm{eV}\left(1-\frac{1}{4}\right)=13.6\mathrm{eV}\left(\frac{3}{4}\right)\approx 10.2\mathrm{eV}$

$f=\frac{E}{h}=\frac{10.2\mathrm{eV}}{4.135\times 10^{-15}\mathrm{eV}\cdot \mathrm{s}}\approx 2.47\times 10^{15}\mathrm{Hz}=2.47\times 10^{15}{\mathrm{s}}^{-1}$

Therefore, the frequencies of the spectral lines emitted are approximately $2.92\times 10^{15}{\mathrm{s}}^{-1}$, $2.47\times 10^{15}{\mathrm{s}}^{-1}$, and $4.57\times 10^{14}{\mathrm{s}}^{-1}$.