The work functions of metal A and B are in th | vedclass

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h
u - \phi$,where $h
u$ is the ener

Vedclass · Accepted Answer

$1: 2$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h
u - \phi$,where $h
u$ is the energy of the incident photon and $\phi$ is the work function of the metal.For metal $A$: $K_{A} = hf - \phi_A$For metal $B$: $K_{B} = h(2f) - \phi_B = 2hf - \phi_B$Given the ratio of work functions $\frac{\phi_A}{\phi_B} = \frac{1}{2}$,we have $\phi_B = 2\phi_A$.Substituting this into the expression for $K_B$: $K_B = 2hf - 2\phi_A = 2(hf - \phi_A)$.Now,the ratio of kinetic energies is $\frac{K_A}{K_B} = \frac{hf - \phi_A}{2(hf - \phi_A)} = \frac{1}{2}$.

The work functions of metal $A$ and $B$ are in the ratio $1: 2$. If light of frequency $f$ and $2f$ is incident on surface $A$ and $B$ respectively,then the ratio of the maximum kinetic energies of the emitted photoelectrons is:

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