In the photoelectric effect,the stopping pote | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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.

Similar Questions

The graph of maximum kinetic energy $(K.E._{max})$ against the frequency $(\nu)$ of incident light is as shown in the figure. The slope of the graph and the intercept on the $X$-axis respectively are:

The maximum wavelength of radiation that can produce the photoelectric effect in a certain metal is $200 \ nm$. The maximum kinetic energy acquired by an electron due to radiation of wavelength $100 \ nm$ will be .............. $eV$.

The figure represents a graph of the maximum kinetic energy $(K)$ of photoelectrons (in $eV$) versus the frequency $(\nu)$ of incident radiation for a metal used as a cathode in a photoelectric experiment. The work function of the metal is ............. $eV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module