SATHEE: Modern Physics 1 Question 47

Modern Physics 1 Question 47

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

(2016 Adv.)

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Solution:

Energy of incident light (in $e V$ )

$E = \frac{12375}{970} = 12.7 e V$

After excitation, let the electron jumps to $n$ th state, then

$\frac{- 13.6}{n^{2}} = - 13.6 + 12.7$

Solving this equation, we get

$n = 4$

$∴$ Total number of lines in emission spectrum,

$= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$

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Modern Physics 1 Question 47

43. A hydrogen atom in its ground state is irradiated by light of wavelength 970\AA. Taking hc/e=1.237×10−6eVm and the ground state energy of hydrogen atom as −13.6eV, the number of lines present in the emission spectrum is

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