Surface of certain metal is first illuminated with light of wavelength $\lambda_1=350 \mathrm{~nm}$

Question

Surface of certain metal is first illuminated with light of wavelength $\lambda_1=350 \mathrm{~nm}$ and then, by light of wavelength $\lambda_2=540 \mathrm{~nm}$. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2 . The work function (in eV ) of the metal is close to $\_\_\_\_$ . (Energy of photon $=\frac{1240}{\lambda(\mathrm{~nm})} \mathrm{eV}$ )

Professor PC Thomas Classes · Answer

(1.88) : $E_1=\frac{1240}{350} \mathrm{eV} ; E_2=\frac{1240}{540} \mathrm{eV}$ Also, $v_1=2 v_2$ Using Einstein photoelectric equation, \ $ \begin{aligned} & E-\phi=\frac{1}{2} m v^2 \Rightarrow \frac{E_1-\phi}{E_2-\phi}=\frac{v_1^2}{v_2^2}=4 \ & \Rightarrow \quad E_1-\phi=4 E_2-4 \phi \ & \phi=\frac{4 E_2-E_1}{3}=\frac{1240}{3}\left(\frac{4}{540}-\frac{1}{350}ight) \approx 1.88 \mathrm{eV} \end{aligned} $