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
\(\Rightarrow \phi = 2\,eV\)
Which of the following statements are correct, if the threshold frequency of caesium is $ 5.16 \times 10^{14} \, \text{Hz} $?
The anode voltage of a photocell is kept fixed. The frequency of the light falling on the cathode is gradually increased. Then the correct graph which shows the variation of photo current \( I \) with the frequency \( f \) of incident light is
In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)
If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
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,