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 ionised. The mass of a hydrogen atom $=1.67\times 10^{-27}kg.$ Assume completely inelastic collision.

• A. $7.2\times 10^{4}m{s}^{-1}$
• B. $2.3\times 10^{4}m{s}^{-1}$
• C. $2.3\times 10^{5}m{s}^{-1}$
• D. $7.2\times 10^{5}m{s}^{-1}$

Answer 1

Answer: (A)

Inelastic collision will take place, if a part of incident kinetic energy is utilised in exciting the atom. Here, one atom is to be ionised, i.e. Assuming completely inelastic collision.
$\Delta E=13.6eV$

For an inelastic collision
$v_1=v_2=v$
By momentum conservation,
$mv=2mV$
$\Rightarrow V=\frac{v}{2}$
$\frac{1}{2}m{v}^2=\frac{1}{2}\cdot 2m{V}^2+\Delta E$
$=m{\left(\frac{v}{2}\right)}^2+\Delta E[\because V=v/2]$
$\frac{1}{4}m{v}^2=\Delta E$
$v=\sqrt{\frac{4\Delta E}{m}}$
$v_{\min }=\sqrt{\frac{4\times 13.6\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}$
$=7.2\times 10^{4}m/s$