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 Energy Difference (\(\Delta E\)):

- The energy levels for \(n = 1\) and \(n = 2\) in a hydrogen atom are given by:

\[ E_{n=1} = -13.6 \text{ eV}, \quad E_{n=2} = -3.4 \text{ eV} \]

- The energy difference \(\Delta E\) when the atom transitions from \(n = 2\) to \(n = 1\) is:

\[ \Delta E = E_{n=1} - E_{n=2} = -13.6 \text{ eV} + 3.4 \text{ eV} = 10.2 \text{ eV} \]

Step 2: Convert Energy to Joules:- \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\), so:

\[ \Delta E = 10.2 \times 1.6 \times 10^{-19} \text{ J} = 1.632 \times 10^{-18} \text{ J} \]

Step 3: Calculate Recoil Speed (\(v\)):

- Using \(v = \frac{\Delta E}{mc}\), where \(m = 1.6 \times 10^{-27} \text{ kg}\) and \(c = 3 \times 10^8 \text{ m/s}\):

\[ v = \frac{1.632 \times 10^{-18}}{1.6 \times 10^{-27} \times 3 \times 10^8} \]

- Simplify:

\[ v = 3.4 \text{ m/s} = \frac{17}{5} \text{ m/s} \]

Step 4: Determine \(x\):

- Since the recoil speed is \(\frac{x}{5}\), we have \(x = 17\).

So, the correct answer is: \(x = 17\)