Atoms: NEET and JEE questions with solutions

Answer: (C) see figure

Rutherford's scattering formula gives
$$N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$$

Most $\alpha$-particles pass almost undeflected, so $N(\theta)$ is very large at small angles. It falls steeply as $\theta$ increases, and only a very few particles are scattered back ($\theta$ near $180^\circ$).

This steadily and sharply decreasing curve is option (3).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1$

The Coulomb force provides the centripetal force:
$$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2} \;\Rightarrow\; r = \frac{1}{4\pi\epsilon_0}\frac{e^2}{mv^2}$$

$$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9 \times 10^{-31} \times 25.6 \times 10^{10}} = \frac{9 \times 2.56 \times 10^{-29}}{9 \times 2.56 \times 10^{-20}} = 10^{-9}\ \text{m}$$

So $x = 1$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $r \propto n^{2/3};\ v \propto n^{1/3}$

Centripetal force: $\dfrac{mv^2}{r} = F$, so $v^2 \propto r$.

Bohr quantization: $mvr = \dfrac{nh}{2\pi}$, so $v \propto \dfrac{n}{r}$.

Combining: $\dfrac{n^2}{r^2} \propto r \;\Rightarrow\; r^3 \propto n^2 \;\Rightarrow\; r \propto n^{2/3}$.

Then $v \propto \dfrac{n}{r} \propto n^{1/3}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.67$ nm

Radius of the $n = 2$ orbit: $r_2 = n^2a_0 = 4 \times 0.052 = 0.208$ nm.

Bohr's condition means the orbit holds a whole number of wavelengths: $2\pi r_n = n\lambda$.
$$\lambda = \frac{2\pi r_2}{2} = \pi \times 0.208 \approx 0.65\ \text{nm}$$

The closest option is $0.67$ nm.

Solution by Sreeraj P, M.Sc Physics