$(a)\;7.2 \times 10^{5} ms^{-1}\qquad(b)\;6.3 \times 10^{4} ms^{-1}\qquad(c)\;8.6 \times 10^{6} ms^{-1}\qquad(d)\;4.2 \times 10^{3} ms^{-1}$

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Answer : (b) $\;6.3 \times 10^{4} ms^{-1}$

Explanation :

Let u be velocity of hydrogen atom before collision and V the ve locity of two atoms moving together after collision By principal of conservation of momentum , we have

$M_{u}+M \times 0=2 MV \qquad \; V=\large\frac{u}{2}$

The loss in kinetic energy $\; \bigtriangleup \varepsilon\;$ due to collision is

$\bigtriangleup \varepsilon=\large\frac{1}{2} Mu^2-\large\frac{1}{2}\;(2M) V^2$

As $V=\large\frac{u}{2}\;,$ we have $\;\bigtriangleup \varepsilon=\large\frac{1}{2}\;Mu^2-\large\frac{1}{2}\;(2M)(\large\frac{u}{2})^2$

$=\large\frac{1}{2} Mu^2-\large\frac{1}{4}Mu^2=\large\frac{1}{4}Mu^2$

This loss in energy is due to excitation of one of hydrogen atoms . The ground state (n=1) energy of a hydrogen atom is $\;\varepsilon_{1}=-13.6eV$

The energy of first excited level (n=2) is : $\varepsilon_{2}=-3.4eV$

Thus minimum energy required to excite a hydrogen atom from ground state to first excited state is :

$\varepsilon_{2}-\varepsilon_{1}=[-3.4-(-13.6)] eV =10.2 eV$

$=10.2 \times 1.6 \times 10^{-19} J =16.32 J $

As per problem with the loss in kinetic energy in collision is due to energy used up in exciting one of atoms

Thus $\;\bigtriangleup \varepsilon =\varepsilon_{2}-\varepsilon_{1}$

or $\;\large\frac{1}{4}\;Mu^2=16.32\times10^{-19}\;$ or $\;u^2=\large\frac{4 \times 16.32\times10^{-19}}{M}$

The mass of hydrogen atom is 1.0078 amu or $\;1.0078\times1.66\times10^{-27}\;Kg$

$u^2=\large\frac{4 \times 16.32 \times 10^{-19}}{1.0078\times1.66\times10^{-27}}=39.02\times10^{8}$

or $\;u=6.24 \times 10^{4} ms^{-1}\;.$

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