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

The change in energy during the transition is given by:
$\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 1.9 \, \text{eV}$

The relationship between energy and wavelength is given by:
$\Delta E = \frac{hc}{\lambda}$

Rearranging for $\lambda$:
$\lambda = \frac{hc}{\Delta E}$

To find the effect of recoil, consider the conservation of momentum:
$P_i = P_f$ $0 = -mv + \frac{h}{\lambda'}$

Solving for the velocity $v$ of the recoiling atom:
$v = \frac{h}{m\lambda'}$

The total energy change $\Delta E'$ considering kinetic energy and emitted photon energy is given by:
$\Delta E' = \frac{1}{2} mv^2 + \frac{hc}{\lambda'}$

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

Using the energy conservation equation:
$\lambda'^2 \Delta E - hc\lambda' - \frac{h^2}{2m} = 0$

Solving this quadratic equation for $\lambda'$:
$\lambda' = \frac{hc \pm \sqrt{h^2c^2 + 4\Delta E h^2 / 2m}}{2\Delta E}$

Approximating for small changes:
$\frac{\lambda' - \lambda}{\lambda} \approx \frac{\Delta E}{2mc^2}$

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

The percentage change in wavelength is approximately $10^{-7}$.