1. Concept

When a nucleus emits a gamma photon, it recoils to conserve momentum. This recoil takes away a small amount of energy, so the photon energy $E_\gamma$ is slightly less than the transition energy $\Delta E$.

Similarly, for another nucleus to absorb a photon, the incident photon must provide the transition energy $\Delta E$ plus the kinetic energy required for the absorbing nucleus to recoil. Because of this “double recoil” loss (once at emission, once at absorption), a stationary nucleus cannot absorb a photon emitted by another stationary nucleus.

To compensate, the absorbing nucleus must move towards the source so that the Doppler effect increases the photon’s apparent energy.

2. Energy Conservation & Recoil

Let $\Delta E$ be the transition energy. Let $R$ be the recoil energy of the nucleus.

At Emission:
$E_\gamma = \Delta E – R$

At Absorption:
The photon needs energy $E’_{req} = \Delta E + R$ (in the absorber’s frame) to excite the nucleus and provide recoil.

The energy gap to be bridged is roughly $2R$. The absorbing nucleus moves with velocity $v$ towards the source, creating a Doppler shift.

Using the Doppler shift formula for $v \ll c$:

$$E’_{absorbed} = E_\gamma \left( 1 + \frac{v}{c} \right)$$ $$(\Delta E – R)\left( 1 + \frac{v}{c} \right) = \Delta E + R$$

Expanding and ignoring the small term $R \frac{v}{c}$:

$$\Delta E + \Delta E \frac{v}{c} – R \approx \Delta E + R$$ $$\Delta E \frac{v}{c} \approx 2R$$

3. Calculating Velocity

Recoil energy $R$ is given by momentum conservation ($p_{nucleus} = p_{\gamma} = E/c$):

$$R = \frac{p^2}{2M} = \frac{E_\gamma^2}{2Mc^2}$$

Substitute $2R$ into the Doppler equation:

$$\Delta E \frac{v}{c} = 2 \left( \frac{E_\gamma^2}{2Mc^2} \right)$$

Approximating $\Delta E \approx E_\gamma$:

$$E_\gamma \frac{v}{c} = \frac{E_\gamma^2}{Mc^2}$$ $$v = \frac{E_\gamma}{Mc}$$

Numerical Calculation:

• $E_\gamma = 22.6 \text{ keV} = 22.6 \times 10^3 \text{ eV}$
• $Mc^2 = 113 \text{ GeV} = 113 \times 10^9 \text{ eV}$
• $c = 3 \times 10^8 \text{ m/s}$

$$v = c \cdot \frac{E_\gamma}{Mc^2} = (3 \times 10^8) \cdot \frac{22.6 \times 10^3}{113 \times 10^9}$$ $$v = (3 \times 10^8) \cdot (0.2 \times 10^{-6}) = 0.6 \times 10^2 = 60 \text{ m/s}$$