Question

When a hydrogen atom going from \( n = 2 \) to \( n = 1 \) emits a photon, its recoil speed is \( \frac{x}{5} \) m/s. Where \( x = \, \) _____. (Use: mass of hydrogen atom \( = 1.6 \times 10^{-27} \, \text{kg} \))

Answer 1

Correct Answer: 17

Step 1: Calculate the energy of the emitted photon.

Energy difference between two energy levels of hydrogen: \[ E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \text{eV} \] \[ E = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} = 10.2 \, \text{eV} \] Convert to joules: \[ E = 10.2 \times 1.6 \times 10^{-19} = 1.632 \times 10^{-18} \, \text{J} \]

Step 2: Relate photon energy to its momentum.

For a photon: \[ E = pc \Rightarrow p = \frac{E}{c} \] \[ p = \frac{1.632 \times 10^{-18}}{3 \times 10^8} = 5.44 \times 10^{-27} \, \text{kg·m/s} \]

Step 3: Momentum conservation for recoil.

The atom recoils with momentum equal and opposite to the photon’s: \[ p_{\text{atom}} = p_{\text{photon}} \] So, recoil speed: \[ v = \frac{p}{m} = \frac{5.44 \times 10^{-27}}{1.6 \times 10^{-27}} = 3.4 \, \text{m/s} \]

Step 4: Compare with given expression.

\[ v = \frac{x}{5} \Rightarrow 3.4 = \frac{x}{5} \] \[ x = 3.4 \times 5 = 17 \]

Final Answer:

\[ \boxed{x = 17} \]