A stationary hydrogen atom de excites from first excited state to ground state. Find recoil speed of hydrogen atom up to nearest integral value. (mass of hydrogen atom = 1.8 × 10^–27 kg)

The correct answer is 3.

To find the recoil speed of the hydrogen atom, we can use the principles of energy conservation and momentum conservation.

Step 1: Calculate the energy of the emitted photon during de-excitation
The hydrogen atom de-excites from the first excited state ($n=2$) to the ground state ($n=1$).

The energy levels of a hydrogen atom are given by the formula:
$E_n=-\frac{13.6}{n^2}\text{eV}$

Energy of the first excited state ($n=2$):
$E_2=-\frac{13.6}{2^2}=-3.4\text{eV}$

Energy of the ground state ($n=1$):
$E_1=-13.6\text{eV}$

The energy of the emitted photon ($\mathrm{\Delta E}$) is the difference between these two energy levels:
$\mathrm{\Delta E}=E_2-E_1=-3.4-(-13.6)=10.2\text{eV}$

Converting this energy into Joules:
$\mathrm{\Delta E}=10.2\times 1.6\times {10}^{-19}\text{J}=1.632\times {10}^{-18}\text{J}$

Step 2: Calculate the momentum of the emitted photon
The relation between the momentum ($p$) and energy of a photon is:
$p=\frac{\mathrm{\Delta E}}{c}$
where $c$ is the speed of light ($c=3\times {10}^8\text{m/s}$).

$p=\frac{1.632\times {10}^{-18}}{3\times {10}^8}=5.44\times {10}^{-27}\text{kg m/s}$

Step 3: Apply conservation of linear momentum to find the recoil speed
Since the hydrogen atom was initially stationary, the magnitude of the recoil momentum of the atom must equal the momentum of the emitted photon:
$Mv=p$
where:
- $M$ is the mass of the hydrogen atom ($1.8\times {10}^{-27}\text{kg}$)
- $v$ is the recoil speed of the hydrogen atom

Solving for $v$:
$v=\frac{p}{M}$

$v=\frac{5.44\times {10}^{-27}}{1.8\times {10}^{-27}}\approx 3.022\text{m/s}$

Rounding to the nearest integral value, the recoil speed of the hydrogen atom is 3 m/s.