(a) (i) Limitations of Rutherford's model of atom: ● Rutherford's model failed to explain the stability of atoms. Charged particle that is moving in a circular path undergoes an acceleration and radiates energy. Thus, the revolving electron losses energy and finally falls into the nucleus. But this is not the reality. Atoms are stable enough. ● Atom exhibits line spectrum. Rutherford's model failed to explain this. According to Rutherford, if electron continuously emits energy then there should be continuous band spectrum. But that is not the reality. (ii) For Balmer series$\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$For $2^{\text{nd}}$ line, $n=4$$\therefore \frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$$\text{Or,}\;\frac{1}{4861}=\frac{3R}{16}$$\therefore R=\frac{16}{3\times 4861}$For $1^{\text{st}}$ line, $n=3$$\therefore \frac{1}{\lambda'}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$$\text{Or,}\;\frac{1}{\lambda'}=\frac{5R}{36}$ $\text{Or,}\;\frac{1}{\lambda'}=\frac{5\times 16}{36\times 3\times 4861}$ $\text{Or,}\;\lambda'=\frac{36\times 3\times 4861}{5\times 16}$ $\therefore \lambda'=6562.35\ \text{Å}$ OR (b) (i) From Einstein's photoelectric equation. $h v = \phi_0 + KE_{\max}$ So, KE energy of photoelectron depends on the frequency of incident radiation and the work function of the substance. There is no intensity term in the equation. Hence, the KE is independent of the intensity of the radiation. (ii) From Einstein's photoelectric equation, $e V_{\text{s}} = h v - \phi_0$ and $h v = E_2 - E_1$ $=-\frac{13.6}{2^2}-\left(-\frac{13.6}{1^2}\right)$ $=10.2\ \mathrm{eV}$ Putting the Einstein's equation, $5\ \mathrm{eV}=10.2\ \mathrm{eV}-\phi_0$$\therefore \phi_0=5.2\ \mathrm{eV}$