To find the recoil speed of a hydrogen atom when it emits a photon during an electron transition from \( n = 5 \) to \( n = 1 \), we can follow these steps:
### Step 1: Calculate the energy of the emitted photon
The energy of the photon emitted during the transition can be calculated using the formula:
\[
E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]
where \( n_1 = 1 \) and \( n_2 = 5 \).
Substituting the values:
\[
E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = 13.6 \, \text{eV} \left( 1 - \frac{1}{25} \right) = 13.6 \, \text{eV} \left( \frac{24}{25} \right)
\]
Calculating this gives:
\[
E = 13.6 \times \frac{24}{25} = 13.6 \times 0.96 = 13.056 \, \text{eV}
\]
### Step 2: Convert energy from eV to Joules
To convert the energy from electron volts to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[
E = 13.056 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.08896 \times 10^{-18} \, \text{J}
\]
### Step 3: Calculate the momentum of the emitted photon
The momentum \( p \) of the photon can be calculated using the relation:
\[
p = \frac{E}{c}
\]
where \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)):
\[
p = \frac{2.08896 \times 10^{-18} \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 6.962 \times 10^{-27} \, \text{kg m/s}
\]
### Step 4: Apply conservation of momentum
According to the conservation of momentum, the momentum of the emitted photon will be equal to the momentum of the hydrogen atom (which recoils):
\[
mv = p
\]
where \( m \) is the mass of the hydrogen atom (mass of proton \( \approx 1.6 \times 10^{-27} \, \text{kg} \)) and \( v \) is the recoil speed.
Substituting the values:
\[
1.6 \times 10^{-27} \, \text{kg} \cdot v = 6.962 \times 10^{-27} \, \text{kg m/s}
\]
### Step 5: Solve for the recoil speed \( v \)
Rearranging the equation gives:
\[
v = \frac{6.962 \times 10^{-27}}{1.6 \times 10^{-27}} = 4.35125 \, \text{m/s}
\]
### Conclusion
The recoil speed of the hydrogen atom is approximately \( 4 \, \text{m/s} \).
