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  • Minimum K.E. of a moving hydrogen atom which on its inelastic headon collision with another stationary hydrogen atom produce a photon emitted by one of the atoms? Both the atoms are supposed to be

Minimum K.E. of a moving hydrogen atom which on its inelastic headon collision with another stationary hydrogen atom produce a photon emitted by one of the atoms? Both the atoms are supposed to be in ground state prior to the collision. No heat is lost.

Option: 1

10.2 \mathrm{~eV}
 


Option: 2

20.4 \mathrm{~eV}


Option: 3

12.75 \mathrm{~eV}


Option: 4

25.5 \mathrm{~eV}


Answers (1)

best_answer

Let u be the velocity of the moving hydrogen atom before collision and just after
       collision its velocity becomes v1.
       Then from conservation of momentum\mathrm{mv}_1+\mathrm{mv}_2=\mathrm{mu} \Rightarrow \quad \mathrm{v}_1+\mathrm{v}_2=\mathrm{u}
where v_2  is the velocity of the another hydrogen just after collision also, \frac{v_2-v_1}{0-u}=-\mathrm{e}

\Rightarrow \quad \mathrm{v}_2-\mathrm{v}_1=\mathrm{eu} \quad \quad \quad \dots (ii)
from equation (I) and (ii)

\mathrm{V}_1=\frac{(1-e) u}{2} ; \quad \mathrm{V}_2=\frac{(1+e) u}{2} \\

 loss of kinetic energy  =\left\{\frac{1}{2} m u^2-\left(\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2\right)\right\} .
\left\{u^2-\left(\frac{\left(1-e^2\right) u^2}{4}+\frac{\left(1+e^2\right) u^2}{4}\right)\right\}=\frac{\left(1-e^2\right) m}{4} \mathrm{u}^2
Minimum energy required for emission of photon

=\frac{13.6}{1^2}-\frac{13.6}{2^{12}}=10.2 \mathrm{eV}
Therefore \frac{\left(1-\varepsilon^2\right) m u^2}{4} \geq 10.2 \mathrm{eV} \quad \Rightarrow \frac{1}{2} \mathrm{mu}^2 \geq \frac{20.4}{1-e^2}
Therefore minimum value of KE of hydrogen atom

=\frac{20.4}{1-0}=20.4 \mathrm{eV}

Posted by

Ritika Jonwal

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