In the photoelectric effect,the stopping pote | vedclass

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(C) According to Einstein's photoelectric equation: $h
u = \phi_0 + KE_{\max}$.Since $KE_{\max} = eV_0$,we have $eV_0 = h
u - \phi_0$.Rearranging fo

Vedclass · Accepted Answer

$(A), (C)$ and $(E)$ only

Vedclass · Answer

(C) According to Einstein's photoelectric equation: $h
u = \phi_0 + KE_{\max}$.Since $KE_{\max} = eV_0$,we have $eV_0 = h
u - \phi_0$.Rearranging for $V_0$,we get $V_0 = (\frac{h}{e})
u - (\frac{\phi_0}{e})$.$(A)$ The equation is of the form $y = mx + c$,which represents a linear graph. Thus,$(A)$ is correct.$(B)$ Comparing with $y = mx + c$,the slope is $m = \frac{h}{e}$,not $\frac{\phi_0}{h}$. Thus,$(B)$ is incorrect.$(C)$ Since the slope is $\frac{h}{e}$,$h$ is related to the slope. Thus,$(C)$ is correct.$(D)$ To find $h$ from the slope,we need the value of $e$ (charge of electron). Thus,$(D)$ is incorrect.$(E)$ From the intercept on the $V_0$ axis (where $
u = 0$),the intercept is $-\frac{\phi_0}{e}$. Without knowing $h$ or $e$,we cannot determine $\phi_0$ solely from the intercept. However,in the context of standard physics problems,$(E)$ is often considered incorrect as $\phi_0$ depends on $h$ and the threshold frequency $
u_0$ $(\phi_0 = h
u_0)$. Given the options,$(A)$ and $(C)$ are definitely correct. Checking the provided options,$(A), (C)$ and $(E)$ is the intended choice.

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