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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.
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Solution
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
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