When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the…

According to the Einstein's equation of photoelectric effect for the first case,

$\frac{hc}{\lambda }-\varphi =6 \mathrm{eV} \cdots \left(\mathrm{i}\right)$

And for the second case,

$\frac{hc}{4\lambda }-\varphi =0.6 \mathrm{eV}$

Therefore,

$\frac{3hc}{4\lambda }=5.4\mathrm{eV}$

$\Rightarrow \lambda =\frac{3hc}{4\times 5.4\mathrm{eV}}=\frac{3\times 1.24\times {10}^{-6}}{4\times 5.4}$

$\Rightarrow \lambda =1.72\times {10}^{-7} \mathrm{m}$

$\Rightarrow$ from equation $\left(\mathrm{i}\right)$

$\frac{hc}{1.72\times {10}^{-7}}\times \frac{1}{1.6\times {10}^{-19}}-\varphi =6 \mathrm{eV}$

$\frac{2\times {10}^{-25}}{2.75\times {10}^{-26}}-\varphi =6$

$\Rightarrow \varphi =\left(7.27-6\right)\cong 1.2 \mathrm{eV}$