Subjective Type

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

Solution

The electron in the ground state of H-atom jumps to the nth state after absorbing the radiation.
Wavelength of the radiation, $$\lambda = 970 A^o =970 \times 10^{-10} m$$
Energy gained by the electron, $$E' = \dfrac{hc}{e \lambda} eV =\dfrac{1.237 \times 10^{-6}}{970 \times 10^{-10}} = 12.75 eV$$

Thus the energy of the $$n^{th}$$ state, $$E_n = -13.6 + 12.75 =-0.85 eV$$
Using: $$E_n = \dfrac{-13.6}{n^2} eV$$

$$\therefore$$ $$-0.85 = \dfrac{-13.6}{n^2}$$

$$\implies n = 4$$

Number of (emission) spectral line, $$N = \dfrac{n (n-1)}{2} = \dfrac{4 (4-1)}{2} = 6 $$ lines

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