Question

Monochromatic radiation of specific wavelength is incident on H-atoms in the ground state. H-atoms absorb energy and emit subsequently radiations of six different wavelengths.
Find the wavelength of incident radiations:
(A) 9.75nm
(B) 50nm
(C) 85.8nm
(D) 97.25nm

Answer 1

Hint: First find out the excited state of the electron and then the energy associated with the excited state and ground state. The difference of these energies gives us the energy of the incident monochromatic radiation. Now we can find out the wavelength using the formula:
$E = \dfrac{{hc}}{\lambda }$

Formulas used:
-Number of spectral lines = $\dfrac{{n(n - 1)}}{2}$ (1)
Where, n = excited state
-Energy for H atom: ${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}$ (2)
-Energy for a wavelength: $E = \dfrac{{hc}}{\lambda }$ (3)
Where, h = $6.626 \times {10^{ - 34}}$ J.sec
              c = $3 \times {10^8}$ m/sec

Complete step by step answer:
-The question says that there are radiations of 6 different wavelengths emitted. So, the number of spectral lines = 6.
Now using the equation (1) we can find out the excited state (n).
Number of spectral lines = $\dfrac{{n(n - 1)}}{2}$
6 = $\dfrac{{n(n - 1)}}{2}$
12 = ${n^2} - n$
${n^2} - n - 12 = 0$
  By solving the above equation we get: n = 4
So, the excited state is 4.

-We will now find out the energy associated with the excited state and the ground state using equation (2).
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}$
So, energy for the excited state: n =4
${E_4} = \dfrac{{ - 13.6}}{{{4^2}}}$ = - 0.85 eV
Energy for excited state: n = 1
${E_1} = \dfrac{{ - 13.6}}{{{1^2}}}$ = - 13.6 eV

-The energy required for excitation of electron to n = 4 will be: $\Delta E = {E_4} - {E_1}$
 $\Delta E = ( - 0.85) - ( - 13.6)$
 = 12.75 eV = $12.75 \times 1.6 \times {10^{ - 19}}$ J
-The energy required by the monochromatic radiation for excitation of the H electron from n = 1 to n = 4 will also be = 12.75 Ev
And we know that: $E = \dfrac{{hc}}{\lambda }$ (3)

So, using this formula we can calculate the wavelength of the incident monochromatic light.
  $12.75 \times 1.6 \times {10^{ - 19}}$= $\dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda }$
  $\lambda = \dfrac{{19.878 \times {{10}^{ - 26}}}}{{20.416 \times {{10}^{ - 19}}}}$
= $0.9736 \times {10^{ - 7}}$ m
= 97.36 nm
Hence, the correct option is: (D) 97.25 nm
So, the correct answer is “Option D”.

Note: The energy associated with the incident monochromatic radiation is not the energy of the excited electron, but it is the energy required to excite the electron from the ground state to the excited state. It can be calculated by taking the difference of energy between the excited state and ground state.