Question

A hydrogen atom moving at speed v collides with another hydrogen atom kept at rest. Find the minimum value of v for which one of the atoms may get ionized. The mass of a hyd rogen atom = $1.67\times {10}^{-27}$ kg

Answer 1

The hydrogen atoms after collision move with speeds $V_{1}and{V}_2$

mv = $m{v}_1+m{v}_2$ .............(1)

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2+\Delta E.....\left(2\right)$

From (1)

$v^2=(v_2+v_2{)}^2$

= $v_1^2+V_2^2+2v_1{v}_2$

from (2)

$v^2=v_1^2+V_1^2+2\Delta E/M........\left(3\right)$

$[\therefore 2v_1{v}_2=\frac{2\Delta E}{M}]$

$(v_1-v_2{)}^2=(v_1+v_2{)}^2-4v_1{v}_2$

$\Rightarrow (v_1-v_2)=\sqrt{(v_1+v_2{)}^2-4\Delta E/M}$

for minimum value of 'v'

$v^2=\frac{4\Delta E}{m}$

= $\frac{4\times 13.6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}$

v = $\sqrt{\frac{4\times 13.6\times 1.6\times {10}^8}{1.67\times {10}^{-27}}}$

= ${10}^4\sqrt{\frac{4\times 13.6\times 1.6}{1.67}}$

= $7.2\times {10}^{4}m/s$