Question
A hydrogen atom moving at speed υ collides with another hydrogen atom kept at rest. Find the minimum value of υ for which one of the atoms may get ionized. The mass of a hydrogen atom $= 1.67 \times 10^{-27}kg.$
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Answer
The hydrogen atoms after collision move with speeds $v_1$ and $v_2$
$\text{mv}=\text{mv}_1+\text{mv}_2\ ....(\text{i})$
$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}^2_1+\frac{1}{2}\text{mv}^2_2+\Delta\text{E}\ ....(\text{ii})$
From (i), $\text{v}^2=(\text{v}_1+\text{v}_2)^2=\text{v}^2_1+\text{v}^2_2+2\text{v}_1\text{v}_2$
From (ii), $\text{v}^2=\text{v}^2_1+\text{v}^2_2+\frac{2\Delta\text{E}}{\text{m}}$
$=2\text{v}_1\text{v}_2=\frac{2\Delta\text{E}}{\text{m}}\ ...(\text{iii})$
$(\text{v}_1-\text{v}_2)^2=\big(\text{v}_1+\text{v}_2\big)^2-4\text{v}_1\text{v}_2$
$(\text{v}_1-\text{v}_2)=\text{v}^2-\frac{4\Delta\text{E}}{\text{m}}$
For minimum value of ‘v’
$\text{v}_1=\text{v}_2$
$\text{v}^2-\Big(\frac{4\Delta\text{E}}{\text{m}}\Big)=0$
$\text{v}^2=\frac{4\Delta\text{E}}{\text{m}}=\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}$
$\text{v}=\sqrt{\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}}=7.2\times10^4\text{m/s}$
$\text{mv}=\text{mv}_1+\text{mv}_2\ ....(\text{i})$
$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}^2_1+\frac{1}{2}\text{mv}^2_2+\Delta\text{E}\ ....(\text{ii})$
From (i), $\text{v}^2=(\text{v}_1+\text{v}_2)^2=\text{v}^2_1+\text{v}^2_2+2\text{v}_1\text{v}_2$
From (ii), $\text{v}^2=\text{v}^2_1+\text{v}^2_2+\frac{2\Delta\text{E}}{\text{m}}$
$=2\text{v}_1\text{v}_2=\frac{2\Delta\text{E}}{\text{m}}\ ...(\text{iii})$
$(\text{v}_1-\text{v}_2)^2=\big(\text{v}_1+\text{v}_2\big)^2-4\text{v}_1\text{v}_2$
$(\text{v}_1-\text{v}_2)=\text{v}^2-\frac{4\Delta\text{E}}{\text{m}}$
For minimum value of ‘v’
$\text{v}_1=\text{v}_2$
$\text{v}^2-\Big(\frac{4\Delta\text{E}}{\text{m}}\Big)=0$
$\text{v}^2=\frac{4\Delta\text{E}}{\text{m}}=\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}$
$\text{v}=\sqrt{\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}}=7.2\times10^4\text{m/s}$
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