In the following diagram,if V_2 V_1,then: | vedclass

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Question

(D) The stopping potential $V_0$ is related to the frequency $
u$ of incident radiation by Einstein's photoelectric equation: $e V_0 = h
u - W_0$,wh

Vedclass · Accepted Answer

$\lambda _1 > \lambda _2$

Vedclass · Answer

(D) The stopping potential $V_0$ is related to the frequency $
u$ of incident radiation by Einstein's photoelectric equation: $e V_0 = h
u - W_0$,which can be written as $V_0 = \frac{h}{e}
u - \frac{W_0}{e}$.From the given graph,the stopping potential for the curve corresponding to $\lambda_2$ is $V_2$ and for $\lambda_1$ is $V_1$. Since $V_2 > V_1$ (in magnitude,as they are negative potentials),the stopping potential for $\lambda_2$ is greater than for $\lambda_1$.This implies that the frequency of radiation $
u_2$ is greater than $
u_1$ $(
u_2 > 
u_1)$.Since the wavelength $\lambda$ is inversely proportional to frequency $(\lambda = \frac{c}{
u})$,a higher frequency corresponds to a shorter wavelength.Therefore,$
u_2 > 
u_1$ implies $\lambda_1 > \lambda_2$.

In the following diagram,if $V_2 > V_1$,then:

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