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A hydrogen atom moves with velocity \mathrm{u} and makes head on inelastic collision with another stationary \mathrm{H}-atom. Both atoms are in ground state before collision. The minimum value of \mathrm{u}, if one of the them is to be given a minimum excitation energy is
 

Option: 1

2.64 \times 10^4 \mathrm{~ms}^{-1}
 


Option: 2

6.24 \times 10^4 \mathrm{~ms}^{-1}
 


Option: 3

2.64 \times 10^8 \mathrm{~ms}^{-1}
 


Option: 4

6.24 \times 10^8 \mathrm{~ms}^{-1}


Answers (1)

best_answer

\mathrm{m v =2 m v \text { or } v=\frac{u}{2} }

\mathrm{\Delta E =\frac{1}{2} m u^2-\frac{1}{2}(2 m)\left(\frac{u}{2}\right)^2=\frac{m u^2}{4} }

\mathrm{\frac{1}{4} m u^2 =13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) }

\mathrm{\frac{1}{4}(1.0078)\left(1.66 \times 10^{-27}\right) u^2 =10.2 \times 1.6 \times 10^{-19} }

\mathrm{\Rightarrow \quad u =6.24 \times 10^4 \mathrm{~ms}^{-1}}

Hence option 2 is correct.




 

Posted by

shivangi.bhatnagar

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