Question

Light of wavelength ${3300}^{\circ }A$ is incident on two metals A and B, whose work functions are 4 eV and 2 eV, respectively. Then:

• A. A will emit photoelectrons but B will not
• B. B will emit photoelectrons, but A will not
• C. both A and B will not emit photoelectrons
• D. neither A nor B will emit photoelectrons

Answer 1

Answer: (B)

B will emit photoelectrons, but A will not

If ${\lambda }_0$ be the threshold wavelength, work function $\varphi =\frac{hc}{{\lambda }_0}$

Now the photoelectric equation becomes, $K_{max}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$

For emission of photoelectrons, $K_{max}>0$ so $\frac{1}{\lambda }>\frac{1}{{\lambda }_0}$ or ${\lambda }_0>\lambda$

Here , $\lambda =3300A^0$

For metal A, ${\lambda }_0\left(A\right)=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{4\times 1.6\times {10}^{-19}}=3103A^0$

For metal B, ${\lambda }_0\left(B\right)=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2\times 1.6\times {10}^{-19}}=6206A^0$

As ${\lambda }_0\left(B\right)>\lambda$ so metal B will emit photoelectrons and as ${\lambda }_0\left(A\right)<\lambda$ so metal A will not emit photoelectrons.