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The recoil speed of hydrogen atom after it emits a photon in going from \mathrm{n=5} state to \mathrm{n=1} state is \mathrm{ (Given, \mathrm{R}_{\infty}=1.097 \times 10^7 \mathrm{~m}^{-1}, \mathrm{~h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{M}_H=1.67 \times 10^{-27} \mathrm{~kg} ) }

 

Option: 1

2.2 \mathrm{~ms}^{-1}
 


Option: 2

4.18 \mathrm{~ms}^{-1}
 


Option: 3

6.2 \mathrm{~ms}^{-1}
 


Option: 4

1 \mathrm{~ms}^{-1}


Answers (1)

best_answer

\mathrm{\frac{1}{\lambda} =R_{\infty}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) }

\mathrm{ =\left(1.097 \times 10^7\right)\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=1.053 \times 10^7 \mathrm{~ms}^{-1} }

\mathrm{\lambda =0.950 \times 10^{-7} \mathrm{~m}}

Momentum of photon,

\mathrm{ p=\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{0.95 \times 10^{-7}}=6.98 \times 10^{-27} }

From conservation of linear momentum,

Momentum of recoiled \mathrm{H}-atom = Momentum of photon

\mathrm{m_{\mathrm{H}} \times v =\frac{\hbar}{\lambda} }

\mathrm{\therefore \quad\left(1.67 \times 10^{-27}\right) \times v =6.98 \times 10^{-27} }

\mathrm{ v=\frac{6.98}{1.67}=4.18 \mathrm{~ms}^{-1}}

Hence option 2 is correct.




 

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Sayak

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