In the experiment of P.E.E. (Photoelectric Ef | vedclass

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

Question

(D) $1$. Saturation current $(i_s)$ is directly proportional to the intensity of incident light. Since the intensity is doubled, the saturation curren

Vedclass · Accepted Answer

$i_s = 10\,mA$ and $V_s > 20\,V$

Vedclass · Answer

(D) $1$. Saturation current $(i_s)$ is directly proportional to the intensity of incident light. Since the intensity is doubled, the saturation current becomes $2 	imes 5\,mA = 10\,mA$.$2$. Stopping potential $(V_s)$ is given by Einstein's photoelectric equation: $eV_s = h
u - \Phi$, where $\Phi$ is the work function.$3$. Initially, $eV_{s1} = h
u - \Phi = 10\,eV$.$4$. When frequency is doubled $(2
u)$, the new stopping potential $V_{s2}$ satisfies $eV_{s2} = h(2
u) - \Phi = 2h
u - \Phi$.$5$. Since $h
u = 10 + \Phi$, we have $eV_{s2} = 2(10 + \Phi) - \Phi = 20 + 2\Phi - \Phi = 20 + \Phi$.$6$. Since $\Phi > 0$, it follows that $V_{s2} > 20\,V$.$7$. Therefore, $i_s = 10\,mA$ and $V_s > 20\,V$.

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

Similar Questions

The work function of a metal is .........

Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$,respectively,are incident on a metallic surface. If $\lambda_{1} = 3 \lambda_{2}$,then:

$A$ light of wavelength $310 \,nm$ is used in a photoelectric experiment. The metal electrode of work function $2.5 \,eV$ is used in the experiment. The stopping potential for the photoelectrons will be (assume $hc = 1240 \,eV-nm$): (in $V$)

Which of the following graphs represents the variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident light for a photoelectric cell?

Two photons, each having an energy of $2.5 \ eV$, are incident on a metal surface with a work function of $4.5 \ eV$. Then . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module