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Physics Question on Atoms

Hydrogen atom in ground state is excited by a monochromatic radiation of λ\lambda = 975 A˚\mathring{A}. Number of spectral lines in the resulting spectrum emitted will be

A

3

B

2

C

6

D

10

Answer

6

Explanation

Solution

Energy of the photon, E=hcλE =\frac{hc}{\lambda}
E=6.63×10−34×3×108975×10−10JE=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10}} J
=6.63×10−34×3×108975×10−10×1.6×10−19eV=12.75eV=\frac{6.63\times10^{-34}\times3\times10^{8}}{975\times10^{-10}\times1.6\times10^{-19}} eV=12.75 eV
After absorbing a photon of energy 12.75 eV, the electron will reach to third excited state of energy -0.85 eV, since energy difference corresponding
to n = 1 and n = 4 is 12.75 eV.
∴\therefore Number of spectral lines emitted
=(n)(n−1)2=(4)(4−1)2=6=\frac{(n)(n-1)}{2}=\frac{(4)(4-1)}{2}=6