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Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda$ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be
A
3
B
2
C
6
D
10
Detailed Solution
Energy of the photon: $E = \frac{12400}{975}$ eV ≈ 12.75 eV
Energy of the excited state $= -13.6 + 12.75 = -0.85$ eV, which is the n = 4 level.
Number of spectral lines from n = 4: $\frac{n(n-1)}{2} = \frac{4\times3}{2} = 6$
Energy of the excited state $= -13.6 + 12.75 = -0.85$ eV, which is the n = 4 level.
Number of spectral lines from n = 4: $\frac{n(n-1)}{2} = \frac{4\times3}{2} = 6$