JEE Physics 2024 Question with Solution

Given: A hydrogen atom makes a transition from $n = 2$ to $n = 1$. Mass of hydrogen atom is $1.6 \times 10^{-27} \, \text{kg}$.

Find: The value of $x$ if recoil speed is $x/5 \, \text{m/s}$.

The energy difference between the two levels is

$$E = 13.6\left(\frac{1}{1^2} - \frac{1}{2^2}\right) \, \text{eV}$$ $$E = 13.6\left(1 - \frac{1}{4}\right) = 10.2 \, \text{eV}$$

Convert this energy into joules:

$$E = 10.2 \times 1.6 \times 10^{-19} = 1.632 \times 10^{-18} \, \text{J}$$

For the emitted photon,

$$E = pc \Rightarrow p = \frac{E}{c}$$ $$p = \frac{1.632 \times 10^{-18}}{3 \times 10^8} = 5.44 \times 10^{-27} \, \text{kg·m/s}$$

By momentum conservation, the recoil momentum of the atom equals the photon momentum:

$$p_{\text{atom}} = p_{\text{photon}}$$

So recoil speed is

$$v = \frac{p}{m} = \frac{5.44 \times 10^{-27}}{1.6 \times 10^{-27}} = 3.4 \, \text{m/s}$$

Now compare with the given expression:

$$v = \frac{x}{5}$$ $$3.4 = \frac{x}{5}$$ $$x = 17$$

Therefore, the value of $x$ is $17$.

Given: Transition of hydrogen atom from $n = 2$ to $n = 1$.

Find: The value of $x$.

Using photon momentum $p = E/c$ and recoil speed $v = p/m$,

$$v = \frac{E}{mc}$$

Now the transition energy is

$$\Delta E = 10.2 \, \text{eV} = 1.632 \times 10^{-18} \, \text{J}$$

Hence,

$$v = \frac{1.632 \times 10^{-18}}{1.6 \times 10^{-27} \times 3 \times 10^8} = 3.4 \, \text{m/s}$$

Since recoil speed is given as $x/5 \, \text{m/s}$,

$$\frac{x}{5} = 3.4$$ $$x = 17$$

Therefore, the value of $x$ is $17$.