Question

A hydrogen atom (mass = $1.66\times {10}^{-27}$ kg, ionization potential = $13.6eV$), moving with a velocity of $6.24\times {10}^{4}m{s}^{-1}$ makes a completely inelastic head-on collision with another stationary hydrogen atom. Both atoms are in the ground state before collision. Up to what state either of one atom may be excited?

Answer 1

Given: $v=6.24\times {10}^{4}m/s$

Initial kinetic energy of the system $E_i=\frac{1}{2}m{v}^2$ ..........(1)

After the inelastic collision, the two atoms combine and move with the velocity $v_1.$

Applying the conservation of momentum: $mv+0=\left(2m\right)v_1⟹v_1=\frac{v}{2}$

Final kinetic energy of the system $E_f=\frac{1}{2}\left(2m\right)(v_1{)}^2=\frac{m{v}^2}{4}$

$\therefore$ Energy lost $\Delta E=E_i-E_f=\frac{m{v}^2}{4}$

$⟹\Delta E=\frac{1.66\times {10}^{-27}\times (6.24\times {10}^4{)}^2}{4}\approx 16.26\times {10}^{-19}J=10.2eV$

The lost energy is used up to excite the hydrogen atom.

The difference between $n=1$ and $n=2$ orbit of hydrogen atom is $10.2eV$. Hence, it is excited to the second state.