The stopping potential \( V_0 \) is related to the maximum kinetic energy of the emitted photoelectrons through the equation:
\( KE_{\text{max}} = h\nu - \phi_0 = eV_0, \)
where:
- \( h \) is Planck’s constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi_0 \) is the work function of the emitter material,
- \( e \) is the elementary charge.
Key points to note:
1. The stopping potential \( V_0 \) depends on the frequency of the incident light (\( \nu \)) but is independent of the intensity of the light. Increasing the intensity of the incident light increases the number of emitted photoelectrons but does not affect their maximum kinetic energy or the stopping potential.
2. The stopping potential is also influenced by the nature of the emitter material since different materials have different work functions (\( \phi_0 \)).
Therefore, statement (3) is incorrect, as \( V_0 \) does not increase with an increase in the intensity of the incident light.
Which of the following statements are correct, if the threshold frequency of caesium is $ 5.16 \times 10^{14} \, \text{Hz} $?
Choose the correct set of reagents for the following conversion:
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):