Hydrogen atom in ground state is excited by a monochromatic radiation of $$\lambda = 975\,\mathop {\text{A}}\limits^ \circ .$$ Number of spectral lines in the resulting spectrum emitted will be
A.
3
B.
2
C.
6
D.
10
Answer :
6
Solution :
Energy provided to the ground state electron
$$\eqalign{
& = \frac{{hc}}{\lambda } = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{975 \times {{10}^{ - 10}}}} = \frac{{6.6 \times 3}}{{975}} \times {10^{ - 16}} \cr
& = 0.020 \times {10^{ - 16}} = 2 \times {10^{ - 18}}J \cr
& = \frac{{20 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}eV = \frac{{20}}{{1.6}}eV = 12.75\,eV \cr} $$
It means the electron jumps to $$n = 4$$ from $$n = 1.$$ When electron will fall back, number of spectral lines emitted $$ = \frac{{n\left( {n - 1} \right)}}{2} = \frac{{4\left( {4 - 1} \right)}}{2} = 6.$$
Releted MCQ Question on Modern Physics >> Atoms And Nuclei
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