Question

A hydrogen atom at rest de-excites from n = 2 state to n = 1 state. Calculate the percentage error in calculation of momentum of photon if recoil of atom is not taken into account. The energy equivalent of rest mass of hydrogen atom is 940 MeV.

Answer 1

5.4 x 10^{-7}

Answer 2

The energy of the electron in the $n$-th state of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2}$ eV.

When the hydrogen atom de-excites from $n=2$ to $n=1$, the energy difference is $\Delta E = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = -3.4 + 13.6 = 10.2$ eV.

If the recoil of the atom is not taken into account, the energy of the emitted photon is assumed to be equal to this energy difference. Let $E_{calc} = 10.2$ eV. The momentum of the photon calculated without considering recoil is $p_{calc} = \frac{E_{calc}}{c} = \frac{10.2 \text{ eV}}{c}$.

Now, consider the recoil of the atom. Let the initial momentum of the atom be $P_i = 0$. Let the momentum of the emitted photon be $p_{actual}$ and the recoil momentum of the atom be $P_f$. By conservation of momentum, $P_i = P_f + p_{actual}$, so $0 = P_f + p_{actual}$, which means $P_f = -p_{actual}$. The magnitude of the recoil momentum is $|P_f| = p_{actual}$.

By conservation of energy, the initial energy of the atom (in $n=2$ state, at rest) is equal to the final energy (atom in $n=1$ state, recoiling, plus the emitted photon). $E_2 + 0 = E_1 + K_{atom} + E_{photon, actual}$, where $K_{atom}$ is the kinetic energy of the recoiling atom and $E_{photon, actual}$ is the actual energy of the emitted photon. $E_{photon, actual} = E_2 - E_1 - K_{atom} = 10.2 \text{ eV} - K_{atom}$.

The kinetic energy of the recoiling atom is $K_{atom} = \frac{P_f^2}{2M} = \frac{p_{actual}^2}{2M}$, where $M$ is the mass of the hydrogen atom. The energy of the photon is related to its momentum by $E_{photon, actual} = p_{actual} c$. Substituting these into the energy conservation equation: $p_{actual} c = 10.2 \text{ eV} - \frac{p_{actual}^2}{2M}$.

The momentum calculated without recoil is $p_{calc} = \frac{10.2 \text{ eV}}{c}$. The actual momentum is $p_{actual}$. The error in calculation is $p_{calc} - p_{actual}$. From the energy equation, $p_{actual} c = 10.2 \text{ eV} - \frac{p_{actual}^2}{2M}$. Divide by $c$: $p_{actual} = \frac{10.2 \text{ eV}}{c} - \frac{p_{actual}^2}{2Mc} = p_{calc} - \frac{p_{actual}^2}{2Mc}$. So, $p_{calc} - p_{actual} = \frac{p_{actual}^2}{2Mc}$.

The percentage error in the calculation of the momentum of the photon is defined as $\frac{|p_{calc} - p_{actual}|}{p_{actual}} \times 100\%$. Percentage error $= \frac{p_{actual}^2 / (2Mc)}{p_{actual}} \times 100\% = \frac{p_{actual}}{2Mc} \times 100\%$.

Since the recoil energy is very small compared to the photon energy, $p_{actual}$ is very close to $p_{calc}$. We can approximate $p_{actual} \approx p_{calc} = \frac{10.2 \text{ eV}}{c}$ in the expression for the percentage error. Percentage error $\approx \frac{10.2 \text{ eV}/c}{2Mc} \times 100\% = \frac{10.2 \text{ eV}}{2Mc^2} \times 100\%$.

We are given that the energy equivalent of the rest mass of the hydrogen atom is 940 MeV. This means $Mc^2 = 940 \text{ MeV} = 940 \times 10^6 \text{ eV}$. Percentage error $\approx \frac{10.2 \text{ eV}}{2 \times 940 \times 10^6 \text{ eV}} \times 100\%$. Percentage error $\approx \frac{10.2}{1880 \times 10^6} \times 100\% = \frac{10.2}{1880} \times 10^{-6} \times 100\%$. Percentage error $\approx \frac{10.2}{1880} \times 10^{-4} \%$. $\frac{10.2}{1880} \approx 0.0054255$. Percentage error $\approx 0.0054255 \times 10^{-4} \% = 5.4255 \times 10^{-7} \%$.

Rounding to two significant figures, the percentage error is $5.4 \times 10^{-7} \%$.