Question

Some hydrogen-like atom in ground state absorb $n$ photons having the same energy and on de-excitation, it emits exactly $n$ photons. Then the energy of absorbed photon may be:

• A. $91.8eV$
• B. $40.8eV$
• C. $48.4eV$
• D. $54.4eV$

Answer 1

Answer: (A), (B)

The correct options are
A $91.8eV$
B $40.8eV$

Solution:- (A) $91.8eV$ and (B) $40.8eV$

As we know that,

$E=13.6\times Z^2[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]$

Number of dark lines (in absorption) i.e excitation $=$ number of bright lines (in emission),i.e de-excitation

It is possible only when the $e^{-}$ is excited from $n=1$ to $n=2$.

Therefore,

$E=13.6\times Z^2[\frac{1}{1^2}-\frac{1}{2^2}]$

$\Rightarrow E=13.6\times Z^2[1-\frac{1}{4}]$

$\Rightarrow E=10.2Z^2$

Now,

For $Z=1$

$E=10.2\times {\left(1\right)}^2=10.2$

For $Z=2$

$E=10.2\times {\left(2\right)}^2=40.8$

For $Z=3$

$E=10.2\times {\left(3\right)}^2=91.8$

Hence the correct answer is $\left(A\right)91.8eV$ and $\left(B\right)40.8eV$.