A hydrogen atom changes its state from n = 3 to n = 2. Due to recoil, the percentage change in the wave length of emitted light is approximately 1 × 10^–n. The value of n is__________. [Given Rhc = 13.6 eV, hc = 1242 eV nm, h = 6.6 × 10^–34 J s, mass of the hydrogen atom = 1.6 × 10^–27 kg]

The correct answer is 7.00.

Step-by-Step Explanation:
Let us analyze the transition of the hydrogen atom and calculate the energy of the emitted photon without recoil.

The energy difference between the orbits $n=3$ and $n=2$ is given by:
$E=Rhc\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Given that $Rhc=13.6\text{eV}$, $n_1=2$, and $n_2=3$:
$E=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\times \frac{5}{36}\approx 1.889\text{eV}$

Let us convert the energy $E$ to Joules:
$E=1.889\times 1.6\times {10}^{-19}\text{J}\approx 3.022\times {10}^{-19}\text{J}$

When the photon is emitted, the hydrogen atom recoils in the opposite direction to conserve momentum.
The momentum of the emitted photon is:
$p=\frac{E}{c}$
By conservation of momentum, the momentum of the recoiling hydrogen atom is also $P=p=\frac{E}{c}$.
The recoil kinetic energy of the hydrogen atom ($K_R$) is given by:
$K_R=\frac{P^2}{2M}=\frac{E^2}{2M{c}^2}$
where $M=1.6\times {10}^{-27}\text{kg}$ is the mass of the hydrogen atom, and $c=3\times {10}^8\text{m/s}$ is the speed of light.

The energy of the emitted photon with recoil is $E^{\prime }=E-K_R$.
Therefore, the decrease in photon energy is:
$\Delta E=K_R=\frac{E^2}{2M{c}^2}$

Since energy of a photon is related to its wavelength by $E=\frac{hc}{\lambda }$, differentiating both sides gives:
$\frac{\Delta \lambda }{\lambda }\approx \frac{\Delta E}{E}=\frac{K_R}{E}=\frac{E}{2M{c}^2}$

Substituting the values:
$M{c}^2=\left(1.6\times {10}^{-27}\text{kg}\right)\times {\left(3\times {10}^8\text{m/s}\right)}^2=1.44\times {10}^{-10}\text{J}$
Thus, the fractional change in wavelength is:
$\frac{\Delta \lambda }{\lambda }=\frac{3.022\times {10}^{-19}\text{J}}{2\times 1.44\times {10}^{-10}\text{J}}\approx 1.05\times {10}^{-9}$

The percentage change in wavelength is:
$\frac{\Delta \lambda }{\lambda }\times 100=1.05\times {10}^{-9}\times 100\approx 1.05\times {10}^{-7}\%$

Comparing this with the given format $1\times {10}^{\mathrm{-n}}\%$:
$1\times {10}^{-7}\approx 1\times {10}^{\mathrm{-n}}$
Hence, $n=7$ (or $7.00$).