Two identical metal plates show the photoelec | vedclass

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W$,where $W$ is the work funct

Vedclass · Accepted Answer

${K_A} < {K_B}/2$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W$,where $W$ is the work function of the metal.Since the plates are identical,they have the same work function $W$.For plate $A$: $K_A = \frac{hc}{\lambda_A} - W$.For plate $B$: $K_B = \frac{hc}{\lambda_B} - W$.Given $\lambda_A = 2\lambda_B$,we substitute this into the equation for $K_A$:$K_A = \frac{hc}{2\lambda_B} - W \Rightarrow 2K_A = \frac{hc}{\lambda_B} - 2W \Rightarrow \frac{hc}{\lambda_B} = 2K_A + 2W$.From the equation for $K_B$,we have $\frac{hc}{\lambda_B} = K_B + W$.Equating the two expressions for $\frac{hc}{\lambda_B}$:$2K_A + 2W = K_B + W$.Rearranging the terms: $K_B = 2K_A + W$.Since $W > 0$,it follows that $K_B > 2K_A$,which implies $K_A < K_B/2$.

Two identical metal plates show the photoelectric effect when light of wavelength ${\lambda _A}$ falls on plate $A$ and ${\lambda _B}$ falls on plate $B$,where ${\lambda _A} = 2{\lambda _B}$. The maximum kinetic energy is:

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