
Based on given graph between stopping potential and frequency of irradiation, work function of metal is equal to
1 eV
3 eV
2 eV
4 eV
The correct answer is option (C) : 2 eV
\(eV_5=h_v-\phi\)
On extrapolating the graph , the graph cuts the yaxis at -2
\(\Rightarrow \) at v=0,vs=-2v
\(\Rightarrow \phi = 2\,eV\)
Given below are two statements: one is labelled as Assertion (A) and the other one is labelled as Reason (R).
Assertion (A): Emission of electrons in the photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance.
Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with the frequency of incident radiation.
In light of the above statements, choose the most appropriate answer from the options given below:

Which of the following statements are correct, if the threshold frequency of caesium is $ 5.16 \times 10^{14} \, \text{Hz} $?
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
According to Einstein’s explanation of the photoelectric effect :
The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron
i.e. hν = W + E
Where,