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← JEE Advanced 2016 · Paper 1
PhysicsInteger+3Question 16 of 54

JEE Advanced 2016 · Paper 1, Physics question 16

A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking hc/e=1.237×10−6hc/e=1.237\times10^{-6} eV m and the ground state energy of hydrogen atom as −13.6-13.6 eV, the number of lines present in the emission spectrum is

Answer and worked solution

Official answer: 6

Photon energy.
E=1.237×10−6970×10−10=12.75 eV.E=\frac{1.237\times10^{-6}}{970\times10^{-10}}=12.75\text{ eV}.

Level reached. −13.6+12.75=−0.85 eV=−13.616-13.6+12.75=-0.85\text{ eV}=-\dfrac{13.6}{16}, so the atom is excited to n=4n=4.

Emission lines. n(n−1)2=4×32=6\dfrac{n(n-1)}2=\dfrac{4\times3}2=6.

Answer: 6

Topics:Bohr model and hydrogen spectrum
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