Question

To conserve momentum, an atom emitting a photon must recoil, meaning that not all of the energy made available in the downward jump goes to the photon. (a) Find a hydrogen atom's recoil energy when it emits a photon in a n = 2 to n = 1 transition. (Note: The calculation is easiest to carry out if it is assumed that the photon carries essentially all the transition energy, which thus determines its momentum. The result justifies the assumption.) (b) What fraction of the transition energy is the recoil energy?

Short answer

(a) Recoil energy of the hydrogen atom when it emits a photon in a n = 2 to n = 1 transition $5.5\times {10}^{-8}\mathrm{eV}$.

(b) Recoil energy is the $5\times {10}^{-9}$of the photon’s energy.

Step-by-step solution

A concept:

Energy levels or energy state is any discrete value from a set of total energy values for a force-limited subatomic particle in a limited space, or for a system of such particles, such as an atom or a nucleus.

(a) Energy difference from n = 2 to n = 1 :

According to the law of conservation of momentum, the total momentum of two isolated bodies remains same unless an external force is applied.

As you know that, energy difference,

${\mathrm{E}}_{\mathrm{n}}=13.6\mathrm{eV}\left(\frac{1}{{{\mathrm{n}}_{\mathrm{f}}}^2}-\frac{1}{{{\mathrm{n}}_{\mathrm{i}}}^2}\right)$

Where, ${\mathrm{n}}_{\mathrm{f}}$is the final orbit and ${\mathrm{n}}_{\mathrm{i}}$is the initial orbit.

Now, energy difference for transition from n = 2 to n = 1 is

role="math" localid="1659178231672" $$\begin{gathered} ∆\mathrm{E}={\mathrm{E}}_1-{\mathrm{E}}_2 \\ =\left(13.6\mathrm{eV}\right)\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right) \\ =\left(13.6\mathrm{eV}\right)\left(1-0.25\right) \\ =10.2\mathrm{eV} \\ ∆\mathrm{E}=10.2\mathrm{eV}\times 1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV} \\ =1.63\times {10}^{-18}\mathrm{J} \end{gathered}$$

Finding momentum of the photon:

As you know that, Energy of emitted photon,

$E=\frac{hc}{\lambda }$

Where, h is Plank’s Constant, c is the speed of light, and $\lambda$is the wavelength.

The relation between wavelength and momentum is as below.

$\frac{\mathrm{hc}}{\mathrm{\lambda }}=\mathrm{pc}$

Where, p is the momentum.

Therefore,

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}} \\ =\frac{1.63\times {10}^{-18}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =5.44\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Recoil Energy of the photon

As you know that,

$\mathrm{Momentum}\left(\mathrm{p}\right)=\mathrm{mass}\left(\mathrm{m}\right)\times \mathrm{Velocity}\left(\mathrm{v}\right)$

Also, the kinetic energy is,

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

Hence, by comparing both the above equation, you have

$\mathrm{KE}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}$

Since, momentum must be conserved, the atom will have an equal and opposite momentum to balance that out.

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{KE}=\frac{{(5.44\times {10}^{-27}\mathrm{kg}.\mathrm{m}/\mathrm{s})}^2}{2\times 1.67\times {10}^{-27}\mathrm{kg}} \\ =8.9\times {10}^{-27}\mathrm{J} \\ =5.5\times {10}^{-8}\mathrm{eV} \end{gathered}$$

Hence, the recoil energy is $\mathrm{KE}=5.5\times {10}^{-8}\mathrm{eV}$.

(b) Recoil energy and photon’s energy:

From Step 1 you, have,

Photon’s Energy, E = 10.2 eV

Recoil energy, $\mathrm{KE}=5.5\times {10}^{-8}\mathrm{eV}$

Take a ratio of the recoil energy and photon’s energy as below.

$$\begin{gathered} \frac{\mathrm{Recoil}\mathrm{Energy}}{\mathrm{Photon}\text{'}\mathrm{s}\mathrm{Energy}}=\frac{5.5\times {10}^{-8}\mathrm{eV}}{10.2\mathrm{eV}} \\ =5\times {10}^{-9} \end{gathered}$$

Hence, the recoil energy is the $5\times {10}^{-9}$of the photon’s energy.