Atoms And Nuclei JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (c)

Explanation: For an electron in the $n^{\text{th}}$ Bohr orbit, the angular momentum is quantised as $ L = \frac{nh}{2\pi} $ So, $ L \propto n $ Now, the total energy of an electron in the $n^{\text{th}}$ Bohr orbit is $ E_n = -\frac{13.6}{n^2} \text{ eV} $ Since the question asks for the magnitude of total energy, $ E = |E_n| = \frac{13.6}{n^2} $ So, $ E \propto \frac{1}{n^2} $ But $L \propto n$, so $n \propto L$. Hence, $ E \propto \frac{1}{L^2} $ $ E \propto L^{-2} $ So the correct option is: $ \boxed{\text{Option C } \; E \propto L^{-2}} $

Correct Answer: (b)

Explanation: $T \propto n^3$ $ \frac{T_2}{T_3}=\left(\frac{n_2}{n_3}\right)^3 $ For the second excited state, $n_2=3$ For the third excited state, $n_3=4$ $ \begin{array}{ll} \therefore & \frac{T_2}{T_3}=\left(\frac{3}{4}\right)^3=\frac{27}{64} \\ \Rightarrow & T_2: T_3=27: 64 \end{array} $

Correct Answer: (c)

Explanation: To calculate the speed of the electron in a hydrogen atom when it is in the $ n=3 $ energy level, we use the formula: $ v_n = \frac{e^2}{2 \varepsilon_0 h n} $ where: $ e = 1.6 \times 10^{-19} \mathrm{C} $ is the charge of the electron, $ \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m} $ is the permittivity of free space, $ h = 6.64 \times 10^{-34} \mathrm{Js} $ is Planck's constant, and $ n = 3 $ is the principal quantum number for the energy level. Substituting these values into the formula, we calculate: $ v_3 = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.85 \times 10^{-12} \times 6.64 \times 10^{-34} \times 3} $ Simplifying this, we find: $ v_3 \approx 0.0073 \times 10^8 \mathrm{~m/s} $ Thus, the speed of the electron in the $ n=3 $ level is: $ v_3 = 7.3 \times 10^5 \mathrm{~m/s} $

Correct Answer: (b)

Explanation: The energy of photons is given by $E=R h c\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ where, $R$ is Rydberg constant, $h$ is Planck's constant and $c$ is the speed of light. (i) Energy of photon produced from second to first energy level, $\begin{aligned} & E_1=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ & E_1=\frac{3}{4} R h c \end{aligned}$ (ii) Energy of photon produced from highest energy level (i.e., $\infty$) to second level, $\begin{aligned} & E_2=R h c\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{1}{4} R h c \\ \therefore \quad & \frac{E_1}{E_2}=\frac{\frac{3}{4} R h c}{\frac{1}{4} R h c}=\frac{3}{1} \text { or } 3: 1 \end{aligned}$

Correct Answer: (c)

Explanation: There is no change in the proton number and the neutron number as the $\gamma$ -emission takes place as a result of excitation or de-excitation of nuclei. $\gamma$ -rays have no charge or mass.