Question

A stationary hydrogen atom of mass $m$ in the ground state achieves minimum excitation energy after head-on inelastic collision with a moving hydrogen atom. Find the velocity of moving hydrogen atom before collision.

• A. $[\frac{10.2\left(eV\right)}{m}{]}^{\frac{1}{2}}$
• B. $[\frac{40.8\left(eV\right)}{m}{]}^{\frac{1}{2}}$
• C. $[\frac{20.4\left(eV\right)}{m}{]}^{\frac{1}{2}}$
• D. $[\frac{40.8\left(eV\right)}{1.0078m}{]}^{\frac{1}{2}}$

Answer 1

The correct option is B $[\frac{40.8\left(eV\right)}{m}{]}^{\frac{1}{2}}$

Let $u$ be velocity of moving hydrogen atom befor collision
$v$ be velocity of combined mass after perfectly inelastic collision

Applying conservation of linear momentum,

$mu=2mv;v=\frac{u}{2}$

Loss of Energy

$\Delta E=\frac{1}{2}m{u}^2-\frac{1}{2}\times 2m\times v^2=\frac{1}{4}m{u}^2$

Minimum excitation energy is to excite a hydrogen atom from ground state to first excited state is

$=-3.4-(-13.6)eV=10.2eV$

$\therefore \frac{1}{4}m{u}^2=10.2eV$

$\Rightarrow u=[\frac{40.8\left(eV\right)}{m}{]}^{\frac{1}{2}}$