Question

An $H$ -atom moving with speed $v$ makes a head-on collision with another $H$ -atom at rest. Both the atoms are in the ground state. The minimum value of velocity $v$ for which one of the atoms may excite is

• A. $\text{6.25}\times 10^{4}m{s}^{-1}$
• B. $\text{8}\times 10^{4}m{s}^{-1}$
• C. $\text{7.25}\times 10^{4}m{s}^{-1}$
• D. $\text{13.6}\times 10^{4}m{s}^{-1}$

Answer 1

Answer: (A)

For $v_{min}$ collision should be completely inelastic.
According to energy conservation principle,
$\therefore \frac{1}{2}m{v}_{min}^2=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\Delta E$ ...(i))
According to momentum conservation principle.
$m{v}_{min}=mv+mv$
$\therefore v=\frac{m_{min}}{2}$ ...(ii)
After solving eq. (i) an d(ii), we get
$\frac{1}{2}m{v}_{min}^2=2\Delta E$
$\therefore v_{min}=\sqrt{\left(\frac{4\Delta E}{m}\right)}$
Here, $\Delta E=$ minimum excitation energy
$=10.2eV$
$m=1.67\times 10^{-27}kg$
$\therefore v_{\text{min }}=\sqrt{\left(\frac{4\times 10.2\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\right)}$
$=\text{6.25}\times 10^{4}m{s}^{-1}$