The work functions of metals A and B are in t | vedclass

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = hf - W_0$,where $W_0$ is the work fun

Vedclass · Accepted Answer

$1 : 2$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = hf - W_0$,where $W_0$ is the work function.For metal $A$: $K_A = hf - W_A$.For metal $B$: $K_B = h(2f) - W_B = 2hf - W_B$.Given the ratio of work functions $\frac{W_A}{W_B} = \frac{1}{2}$,we can write $W_B = 2W_A$.Substituting this into the equation for $K_B$: $K_B = 2hf - 2W_A = 2(hf - W_A)$.Comparing $K_A$ and $K_B$,we get $K_B = 2K_A$.Therefore,the ratio $\frac{K_A}{K_B} = \frac{1}{2}$.

The work functions of metals $A$ and $B$ are in the ratio $1 : 2$. If light of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of photoelectrons emitted is (given that $f > f_{0A}$ and $2f > f_{0B}$).

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