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 \( = 1.6 \times 10^{-27} \, \text{kg} \))
Correct Answer: 17
Approach Solution - 1
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:
\[ \boxed{x = 17} \]
Approach Solution -2
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: \(x = 17\)
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