Physics - Modern Physics Question with Solution | TestHub

Let the work function of metal A be ${\varphi }_A=\varphi$ and that of metal B be ${\varphi }_B=2\varphi$.

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K=hf-\varphi$.

For metal A, the incident frequency is $f$, so the kinetic energy is $K_A=hf-\varphi$.

For metal B, the incident frequency is $2f$, so the kinetic energy is $K_B=h(2f)-2\varphi =2(hf-\varphi )$.

The ratio of their kinetic energies is $\frac{K_A}{K_B}=\frac{hf-\varphi }{2(hf-\varphi )}=\frac{1}{2}$.

Answer: $1:2$