Question

When photons of energy $4.25eV$ strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy of $K_A$ eV and de Broglie wavelength ${\lambda }_A$. The maximum kinetic energy of photo electrons liberated from another metal B by photons of energy $4.70eV$ is $K_B=(K_A-1.5)eV$. If the de Broglie wavelength of these photoelectrons is ${\lambda }_B=2{\lambda }_A$ then

• A. the work function of A is 2.25 eV
• B. the work function of B is 4.20 eV
• C. $K_A=2.00$ eV
• D. $K_B=2.75$ eV

Answer 1

Answer: (A), (B), (C)

The correct options are
A the work function of A is 2.25 eV
B the work function of B is 4.20 eV
C $K_A=2.00$ eV

De Broglie wavelength $\lambda =\frac{h}{p}$ and kinetic energy, $K=\frac{p^2}{2m}$

Thus, $\frac{K_A}{K_B}=\frac{P_A^2}{P_B^2}=\frac{{\lambda }_B^2}{{\lambda }_A^2}=4$ as ${\lambda }_B=2{\lambda }_A$

Given, $K_B=(K_A-1.5)=4K_B-1.5\Rightarrow K_B=0.5eV$

Thus, $K_A=4\times 0.5=2.00eV$

From photo-electric effect:

Work function of A $=4.25-K_A=4.25-2.00=2.25eV$

and work function of B $=4.70-K_B=4.70-0.5=4.20eV$

Thus, only the option D is wrong.