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 _____. [Given Rhc = 13.6 eV, hc = 1242 eVnm, h = 6.6×10⁻³⁴ Js, mass of hydrogen atom = 1.6×10⁻²⁷ kg]

A hydrogen atom transitions from $$n = 3$$ to $$n = 2$$. We need to find $$n$$ in the expression $$1 \times 10^{-n}$$ for the percentage change in wavelength due to recoil.

Find the energy of the emitted photon.

$$ E = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{5}{36} = 1.889 \text{ eV} $$

Find the photon momentum.

$$ p_{photon} = \frac{E}{c} = \frac{1.889 \times 1.6 \times 10^{-19}}{3 \times 10^8} = \frac{3.022 \times 10^{-19}}{3 \times 10^8} = 1.007 \times 10^{-27} \text{ kg m/s} $$

Find the recoil kinetic energy of the atom.

By conservation of momentum, the atom recoils with momentum equal to the photon's momentum:

$$ KE_{recoil} = \frac{p^2}{2M} = \frac{(1.007 \times 10^{-27})^2}{2 \times 1.6 \times 10^{-27}} = \frac{1.014 \times 10^{-54}}{3.2 \times 10^{-27}} \approx 3.17 \times 10^{-28} \text{ J} $$

Converting to eV: $$KE_{recoil} = \frac{3.17 \times 10^{-28}}{1.6 \times 10^{-19}} \approx 1.98 \times 10^{-9}$$ eV

Find the percentage change in wavelength.

Due to recoil, the actual photon energy is reduced by $$KE_{recoil}$$. Since $$E = hc/\lambda$$, a small change in energy gives:

$$ \frac{\Delta \lambda}{\lambda} = \frac{\Delta E}{E} = \frac{1.98 \times 10^{-9}}{1.889} \approx 1.05 \times 10^{-9} $$

Percentage change: $$\approx 1.05 \times 10^{-7}\%$$, which is $$\approx 10^{-7}\%$$.

Therefore $$n = \boxed{7}$$.