MHT CET 2021 Physics Solved Paper

(c) According to Einstein's photoelectric equation, maximum energy of emitted electrons is given by$K_{\max} = \frac{1}{2} m v_{\max}^{2} = h v - h v_{0}$$\Rightarrow \frac{1}{2} m v_{\max}^{2} = h(v - v_{0}) \cdots (i)$When, frequency of incident radiation, becomes $2v$, then velocity of emitted electron is $v_{1}$, hence from Eq. (i), we get$\frac{1}{2} m v_{1}^{2} = h(2v - v_{0})$$\frac{1}{2} m v_{1}^{2} = h v_{0} \cdots (ii)$Again, when frequency becomes $5v_{0}$, the velocity of emitted electron is $v_{2}$.$\therefore$ From Eq. (i), we have$\frac{1}{2} m v_{2}^{2} = h(5v - v)$$\Rightarrow \frac{1}{2} m v_{2}^{2} = 4 h v$Dividing Eq. (ii) by Eq. (iii), we get$\frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}} = \frac{h v}{4 h v}$$\Rightarrow \left(\frac{v_{1}}{v_{2}}\right)^{2} = \left(\frac{1}{2}\right)^{2}$$\Rightarrow v_{1} : v_{2} = 1 : 2$