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
\(\Rightarrow \phi = 2\,eV\)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ 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,