Atoms, on the exam screen.

For the hydrogen atom the magnitude of the total energy equals half the magnitude of the potential energy, so $\frac{KQ^2}{2r} = 3.4$ eV. Solving, $r = \frac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times3.4\times1.6\times10^{-19}} \approx 2.1\times10^{-10}$ m, option (3). This is the $n=2$ Bohr orbit radius, i.e. $4a_0 \approx 2.1$ angstrom.

Rutherford's scattering formula gives $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$, so the count is enormous for small deflections and drops by several orders of magnitude as $\theta$ increases - which is exactly why the plots are drawn on a logarithmic vertical axis. The experimental finding was that the overwhelming majority of $\alpha$-particles pass almost undeviated, with only a tiny fraction turned through large angles by the nucleus. NCERT Class 12, Atoms (alpha-particle scattering).

Why the other options are wrong:

- (1) reverses the physics, implying most particles suffer large deflections and the atom is uniformly dense
- (2) captures the falling trend but far too weakly - it misses that the drop spans many orders of magnitude
- (4) invents a preferred scattering angle, as if the nucleus deflected particles by a characteristic amount

In a Bohr orbit the Coulomb attraction supplies the centripetal force: $\frac{ke^2}{r^2} = \frac{mv^2}{r}$, which rearranges to $r = \frac{ke^2}{mv^2}$. Substituting $v^2 = 25.6 \times 10^{10} = 2.56 \times 10^{11}$, the numerator is $9 \times 10^9 \times (1.6 \times 10^{-19})^2 = 2.304 \times 10^{-28}$ and the denominator is $9 \times 10^{-31} \times 2.56 \times 10^{11} = 2.304 \times 10^{-19}$. Their ratio is $1 \times 10^{-9}\ \text{m}$, so $x = 1$. NCERT Class 12, Atoms (Bohr model).

The Bohr model is valid for hydrogen-like atoms, which have one electron. Hydrogen, singly ionised helium, and deuteron all fit this criterion. Singly ionised neon, however, has more than one electron, making the Bohr model inapplicable. Chapter: Atoms.