Question

A hydrogen atom is excited from $n=1$ to $n=2$ energy state. What is the wavelength of the radiation emitted by the atom when it goes back to its ground state ? $(hc=12431\text{eV}\stackrel{\circ }{\text{A}})$

• A. $1024.5\stackrel{\circ }{\text{A}}$
• B. $1122.2\stackrel{\circ }{\text{A}}$
• C. $1218.7\stackrel{\circ }{\text{A}}$
• D. $1324.5\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is C $1218.7\stackrel{\circ }{\text{A}}$

The energy of the $n^{th}$ orbit in the hydrogen atom $(Z=1)$ is given by,

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

So, when the electron jumps from $n=1$ to $n=2$, then energy required is given as

$\Delta E=E_2-E_1=\frac{-13.6}{2^2}-[-\frac{-13.6}{1^2}]$

$=-3.4\text{eV}-(-13.6\text{eV})=10.2\text{eV}$

So, this same amount of energy is released when the electron jumps $n=2$ to $n=1$.

Therefore, the corresponding wavelength will be:-

$E=\frac{hc}{\lambda }$

$\Rightarrow \lambda =\frac{hc}{E}=\frac{12431\text{eV}\stackrel{\circ }{\text{A}}}{10.2\text{eV}}=1218.72\stackrel{\circ }{\text{A}}$

Hence, $\left(C\right)$ is the correct answer.

Why this Question? Useful in understanding of the "Electron transition" concept from one state to another state.