A metal surface is illuminated by light of two different wavelengths $$248\,nm$$  and $$310\,nm.$$  The maximum speeds of the photoelectrons corresponding to these wavelengths are $${u_1}$$ and $${u_2},$$  respectively. If the ratio $${u_1}:{u_2} = 2:1$$    and $$hc = 1240\,eV\,nm$$    the work function of the metal is nearly

Solution :
$$\eqalign{ & \frac{{hC}}{{{\lambda _1}}} - W = \frac{1}{2}mu_1^2 \cr & {\text{and }}\frac{{hC}}{{{\lambda _2}}} - W = \frac{1}{2}mu_2^2 \cr} $$
Dividing the above two equations, we get
$$\eqalign{ & \frac{{\frac{{hC}}{{{\lambda _1}}} - W}}{{\frac{{hC}}{{{\lambda _2}}} - W}} = \frac{{u_1^2}}{{u_2^2}} \cr & \therefore \frac{{\frac{{1240}}{{248}} - W}}{{\frac{{1240}}{{310}} - W}} = \frac{4}{1} \cr & \therefore \frac{{1240}}{{248}} - W = \frac{{4 \times 1240}}{{310}} - 4W \cr & \therefore W = 3.7eV \cr} $$