Question

A hydrogen atom, initially at rest in the n = 4 quantum state, undergoes a transition to the ground state, emitting a photon in the process. What is the speed of the recoiling hydrogen atom? (Hint: This is similar to the explosions of Chapter 9.)

Short answer

The recoilingspeed of the hydrogen atom is 4.08 m/s.

Step-by-step solution

Find the expression for the speed of the recoiling hydrogen atom:

From the conservation of the linear momentum, the momentum of the photon equals the momentum of the recoiling hydrogen atom.

$$\begin{gathered} {\mathbf{P}}_{\mathbf{atom}}\mathbf{=}{\mathbf{P}}_{\mathbf{ph}} \\ \mathbf{mv}\mathbf{=}\frac{\mathbf{E}}{\mathbf{c}} \\ \mathbf{v}\mathbf{=}\frac{\mathbf{E}}{\mathbf{mc}} \\ \mathbf{v}\mathbf{=}\frac{\mathbf{A}}{\mathbf{mc}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \end{gathered}$$ ….. (1)

Define the speed of the recoiling hydrogen atom:

The atom is initially at rest in ${\mathrm{n}}_2=4$, and undergoes a transition to the ground state ${\mathrm{n}}_1=1$.

Substitute 1 $\mathrm{for}{\mathrm{n}}_1,4\mathrm{for}{\mathrm{n}}_2,1.67\times {10}^{-27}\mathrm{kg}\mathrm{for}\mathrm{m},3\times {10}^8\mathrm{m}/\mathrm{s}\mathrm{for}\mathrm{c},$and 13.6 eV for A in equation (1).

$$\begin{gathered} \mathrm{v}=\frac{13.6\mathrm{eV}}{\left(1.67\times {10}^{-27}\mathrm{kg}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}\left(\frac{1}{1^2}-\frac{1}{4^2}\right) \\ =\frac{13.6\times 1.602\times {10}^{-19}\mathrm{J}}{5.01\times {10}^{-19}\mathrm{kg}.\mathrm{m}/\mathrm{s}}\left(1-\frac{1}{16}\right) \\ =4.348\times \left(\frac{15}{16}\right) \\ =4.08\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the recoilingspeed of the hydrogen atom is 4.08 m/s.