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.
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).