Question

If the wavelength of the incident radiation changes from ${\lambda }_1$ to ${\lambda }_2$, then the maximum kinetic energy of the emitted photo electrons change from $K_1$ to $K_2$, then the work function of the emitter surface is

• A. $\frac{{\lambda }_1{K}_1-{\lambda }_2{K}_2}{{\lambda }_2-{\lambda }_1}$
• B. $\frac{{\lambda }_1{K}_2-{\lambda }_2{K}_1}{{\lambda }_1-{\lambda }_2}$
• C. $\frac{K_2-K_1}{{\lambda }_1{K}_1-{\lambda }_2{K}_2}$
• D. $\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_2{K}_1-{\lambda }_1{K}_2}$

Answer 1

The correct option is A $\frac{{\lambda }_1{K}_1-{\lambda }_2{K}_2}{{\lambda }_2-{\lambda }_1}$

A

Work function is the minimum energy required to remove an electron from a solid to a point in the vacuum immediately outside the solid surface. If the wavelength of the incident - radiation changes from ${\lambda }_1$ to ${\lambda }_2$ and the maximum K. E changes from $K_1$ to $K_2$ then the work function is the emitter surface is given by the equation

$W=\frac{{\lambda }_1{k}_1-{\lambda }_2{k}_2}{{\lambda }_2-{\lambda }_1}$