Question

A hydrogen atom changes its state from \( n = 3 \) to \( n = 2 \). Due to recoil, the percentage change in the wavelength of emitted light is approximately \( 1 \times 10^{-n} \). The value of \( n \) is ______. \([ \text{Given: } Rhc = 13.6 \, \text{eV}, \, hc = 1242 \, \text{eV nm}, \, h = 6.6 \times 10^{-34} \, \text{J s}, \, \text{mass of the hydrogen atom} = 1.6 \times 10^{-27} \, \text{kg} ]\)

Answer 1

Correct Answer: 7

The energy difference for the transition is calculated as:
\[ \Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 1.9 \, \text{eV} \]

The energy and wavelength are related by:
 \[ \Delta E = \frac{hc}{\lambda} \]

Solving for wavelength \( \lambda \):
 \[ \lambda = \frac{hc}{\Delta E} \]

Momentum conservation is considered to account for recoil:
 \[ P_i = P_f \] \[ 0 = -mv + \frac{h}{\lambda'} \]

The velocity \( v \) of the recoiling atom is derived as:
 \[ v = \frac{h}{m\lambda'} \]

The total energy change \( \Delta E' \), including kinetic energy and photon energy, is:
 \[ \Delta E' = \frac{1}{2} mv^2 + \frac{hc}{\lambda'} \]

Substituting \( v \) and simplifying yields:
 \[ \Delta E' = \frac{1}{2} \left( \frac{h}{m\lambda'} \right)^2 + \frac{hc}{\lambda'} \]

Using the energy conservation equation, we get:
 \[ \lambda'^2 \Delta E - hc\lambda' - \frac{h^2}{2m} = 0 \]

The quadratic equation for \( \lambda' \) is solved as:
 \[ \lambda' = \frac{hc \pm \sqrt{h^2c^2 + 4\Delta E h^2 / 2m}}{2\Delta E} \]

For small changes, the approximation is:
 \[ \frac{\lambda' - \lambda}{\lambda} \approx \frac{\Delta E}{2mc^2} \]

With the given values substituted:
 \[ \frac{\lambda' - \lambda}{\lambda} = \frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times (3 \times 10^8)^2} \approx 10^{-7} \]

The approximate percentage change in wavelength is \( 10^{-7} \).