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

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detailed solution

Correct option is C

Energy of the photon, E=hcλE=6.63×10-34×3×108975×10-10 J=6.63×10-34×3×108975×10-10×1.6×10-19eV=12.75eVAfter absorbing a photon of energy 1275 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. ∴ Number of spectral lines emitted=(n)(n-1)2=(4)(4-1)2=6

Similar Questions

In a sample of ions, a Li⁺⁺ ion electron in first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ ?

$(Given, h = 6.63 \times 10^{- 34} Js, c = 3 \times 10^{8} {ms}^{- 1})$