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The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = $2.18 \times 10^{-18}$ J atom$^{-1}$ and h = $6.625 \times 10^{-34}$ Js):
A
$1.03 \times 10^{15}$ s$^{-1}$
B
$3.08 \times 10^{15}$ s$^{-1}$
C
$2.00 \times 10^{15}$ s$^{-1}$
D
$1.54 \times 10^{15}$ s$^{-1}$
Detailed Solution
Energy of the nth level of hydrogen: $E_n = -\dfrac{2.18 \times 10^{-18}}{n^2}$ J (the ionisation energy is the energy needed to remove the electron from n = 1).
$\Delta E = E_4 - E_1 = 2.18 \times 10^{-18}\left(\dfrac{1}{1^2} - \dfrac{1}{4^2}\right)$
$= 2.18 \times 10^{-18} \times \dfrac{15}{16} = 2.044 \times 10^{-18}$ J
Frequency: $\nu = \dfrac{\Delta E}{h} = \dfrac{2.044 \times 10^{-18}}{6.625 \times 10^{-34}}$
$\nu = 3.08 \times 10^{15}$ s$^{-1}$
(This line belongs to the Lyman series in the ultraviolet region.)
$\Delta E = E_4 - E_1 = 2.18 \times 10^{-18}\left(\dfrac{1}{1^2} - \dfrac{1}{4^2}\right)$
$= 2.18 \times 10^{-18} \times \dfrac{15}{16} = 2.044 \times 10^{-18}$ J
Frequency: $\nu = \dfrac{\Delta E}{h} = \dfrac{2.044 \times 10^{-18}}{6.625 \times 10^{-34}}$
$\nu = 3.08 \times 10^{15}$ s$^{-1}$
(This line belongs to the Lyman series in the ultraviolet region.)