A plot of the kinetic energy left(frac{1}{2} | vedclass

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(B) According to the photoelectric effect equation,the kinetic energy of the ejected electron is given by:$K.E. = h
u - W_0$where $W_0 = h
u_0$ is t

Vedclass · Accepted Answer

$Rb, K, Na$ and $Li$

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(B) According to the photoelectric effect equation,the kinetic energy of the ejected electron is given by:$K.E. = h
u - W_0$where $W_0 = h
u_0$ is the work function and $
u_0$ is the threshold frequency.The plot shows that the threshold frequency $
u_0$ increases in the order $M_1 Since the work function $W_0$ is directly proportional to the threshold frequency $
u_0$,the work function also increases in the order $M_1 For alkali metals,the ionization energy (and thus the work function) decreases down the group: $Li > Na > K > Rb$.Therefore,the order of increasing work function is $Rb Matching this with the plot,we get $M_1 = Rb, M_2 = K, M_3 = Na, M_4 = Li$.Thus,the correct option is $B$.

$A$ plot of the kinetic energy $\left(\frac{1}{2} m v^2\right)$ of ejected electrons as a function of the frequency $(\nu)$ of incident radiation for four alkali metals $(M_1, M_2, M_3, M_4)$ is shown below.
The alkali metals $M_1, M_2, M_3$ and $M_4$ are,respectively:

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$A$ plot of the kinetic energy $\left(\frac{1}{2} m v^2\right)$ of ejected electrons as a function of the frequency $(\nu)$ of incident radiation for four alkali metals $(M_1, M_2, M_3, M_4)$ is shown below.The alkali metals $M_1, M_2, M_3$ and $M_4$ are,respectively: