Question

A hydrogen atom moving at speed υ collides with another hydrogen atom kept at rest. Find the minimum value of υ for which one of the atoms may get ionized.
The mass of a hydrogen atom = 1.67 × 10⁻²⁷ kg.

Answer 1

Given:
Mass of the hydrogen atom, M = 1.67 × 10⁻²⁷ kg
Let v be the velocity with which hydrogen atom is moving before collision.
Let v₁ and v₂ be the velocities of hydrogen atoms after the collision.
Energy used for the ionisation of one atom of hydrogen, ΔE = 13.6 eV = 13.6×(1.6×10⁻¹⁹)J
Applying the conservation of momentum, we get
mv = mv₁ + mυ₂ ...(1)

Applying the conservation of mechanical energy, we get
$\frac{1}{2}m{\upsilon }^2=\frac{1}{2}m{\upsilon }_1^2+\frac{1}{2}m{\upsilon }_2^2+∆\mathrm{E}...\left(2\right)$

Using equation (1), we get
v² = (v₁ + v₂)²
v² $={\upsilon }_1^2+{\upsilon }_2^2+2{\upsilon }_1{\upsilon }_2$ ...(3)

In equation(2), on multiplying both the sides by 2 and dividing both the sides by m.
${\mathrm{\upsilon }}^2={\mathrm{\upsilon }}_1^2+{\mathrm{\upsilon }}_1^2+2∆E/m....\left(4\right)$

On comparing (4) and (3), we get
$\left[\because 2{\upsilon }_1{\upsilon }_2=\frac{2∆E}{m}\right]$
(υ₁ − υ₂)² = (υ₁ + υ₂)² − 4υ₁υ_{2
​${\left(v_1-v_2\right)}^2=v^2-\frac{4∆E}{m}$}

For minimum value of υ,
v₁ = v₂ =0
Also, $v^2-\frac{4∆E}{m}=0$
$$\begin{gathered} \therefore v^2=\frac{4∆E}{m} \\ =\frac{4\times 13.6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}} \\ v=\sqrt{\frac{4\times 13.6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}} \\ ={10}^4\sqrt{\frac{4\times 13.6\times 1.6}{1.67}} \\ =7.2\times {10}^4\mathrm{m}/\mathrm{s} \end{gathered}$$