Q. A hydrogen atom is initially at rest and free to move is in the second excited state. It comes to ground state by emitting a photon, then the momentum of hydrogen atom will he approximately: (in $kg-m/s$)

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A

$12.1 \times 0^{-27}$

13%

B

$6.45 \times 10^{-27}$

46%

C

$3 \times 10^{-27}$

26%

D

$1.5 \times 0^{27}$

15%

Solution:

Momentum of photon is given as
$p=\frac{h}{\lambda}=\frac{h v }{c}=\frac{E}{c}$
$\Rightarrow p=\frac{\left(13.6-\frac{13.6}{3^{2}}\right) \times 1.6 \times 10^{-19}}{3 \times 10^{8}}$
$\Rightarrow p=6.45 \times 10^{-27} kg\, m / s$

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