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

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: $x = 17$