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

`1/2 mv^2=1/2mv_1^2+ 1/2mv_2^2+DeltaE//m`  .......(2)

Using equation (1), we get

`V^2 = (V_1 + V_2)^2`

`V^2 = v_1^2 +v_2^2 + 2v_1v_2`   ...(3)

In equation(2), on multiplying both the sides by 2 and dividing both the sides by m.

`v^2 = v_1^2 + v_1^2+ 2Delta E//m    .......(4)`

On comparing (4) and (3), we get

`[therefore 2v_1v_2 = (2DeltaE)/m]`

`(U_1 - U_2)^2 = (U_1 + U_2)^2 -4U_1U_2`

`(v_1 - v_2)^2 = v^2 - (4DeltaE)/m`
For minimum value of υ,
`V_1 = V_2 = 0 `
`Also, V^2 - (4Delta)/m`

`∴ v^2 = (4DeltaE)/m`

= `(4xx13.6xx1.6xx10^-19)/(1.67xx10^-27)`

`v =sqrt(4xx13.6xx1.6xx10^-19)/(1.67xx10^-27)`

= `10^4 sqrt((4xx13.6xx1.6)/(1.67))`

= 7.2 × 10⁴ m/s