A hydrogen atom moves with a velocity u and makes a head on inelastic collision with another stationary hydrogen atom. Both are in the ground state before collision. What is the minimum value of u (in x10⁴ m/s ), if one of them is to be given a minimum excitation energy? The ionization energy is 13.6 eV. Mass of the hydrogen atom is 1.0078 x 1.66 x 10^-27 kg.

Difficult

Solution

From conservation of momentum mu=2mv
$\Rightarrow v = u / 2 . . . . . . . . . . (i)$
$The energy of excitation Δ E is given by$
$Δ E = \frac{1}{2} m u^{2} - 2 \times \frac{1}{2} m (u / 2)^{2}$
$\Rightarrow Δ E = \frac{1}{4} m u^{2} . . . . . . . (i i)$
The minimum excitation energy for a H-atom is for transition n₁= 1 to n₂= 2 which corresponds to an energy of 10.2 eV
From Eq. (2)
$\frac{1}{4} \times (1.0078 \times 1.66 \times 10^{- 27}) u^{2} = 10.2 \times (1.6 \times 10^{- 19})$
$\Rightarrow u = 6.24 \times 10^{4} m / s$

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