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. $7.2\times {10}^{8}m/s$
• B. $2.3\times {10}^{8}m/s$
• C. $2.3\times {10}^{10}m/s$
• D. $7.2\times {10}^{10}m/s$

Answer 1

The correct option is A $7.2\times {10}^{8}m/s$

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., $DeltaE=13.6$eV
Assuming completely inelastic collision.
For an inelastic collision
$v_1=v_2=v$
By momentum conservation,
$mv=2mV$
$\Rightarrow V=\frac{v}{2}$
$\frac{1}{2}m{v}^2=\frac{1}{2}\cdot 2m{v}^2+\Delta E$
$=m{\left(\frac{v}{2}\right)}^2+\Delta E$ $[\because V=v/2]$
$\frac{1}{4}m{v}^2=\Delta E$
$v=\sqrt{\frac{4\Delta E}{m}}$
$v_{min}=\sqrt{\frac{4\times 13.6\times 16\times {10}^{-19}}{167\times {10}^{-27}}}$
$=7.2\times {10}^{8}m/s$.