In photoelectric experiment energy of 2.48 eV irradiates a photo sensitive material. The stopping potential was measured to be 0.5 V. Work function of the photo sensitive material is :
- 0.5 eV
- 1.68 eV
- 2.48 eV
- 1.98 eV
The Correct Option is D
Solution and Explanation
The energy relation for the photoelectric effect is:
\[ eV_s = h\nu - \phi \]
where $eV_s$ is the stopping potential energy, $h\nu$ is the energy of incident photons, and $\phi$ is the work function.
Given: $h\nu = 2.48 \, \text{eV}, \, V_s = 0.5 \, \text{V}.$
\[ 0.5 = 2.48 - \phi \]
\[ \phi = 2.48 - 0.5 = 1.98 \, \text{eV}. \]
The work function of the material is $\phi = 1.98 \, \text{eV}.$
Top Questions on Photoelectric Effect
- Answer the following giving reason:
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensity.
(c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
- A light source of wavelength \( \lambda \) illuminates a metal surface, and electrons are ejected with a maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength \( \frac{\lambda}{2} \), then the maximum kinetic energy of ejected electrons will be (The work function of the metal is 1 eV).
- JEE Main - 2025
- Physics
- Photoelectric Effect
- The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope of \( \frac{h}{e} \). Explain this observation.
- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
- Explain the following observations using Einstein’s photoelectric equation:
(a) Photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than a certain minimum value.
(b) It is the frequency, and not the intensity of the incident light which affects the maximum kinetic energy of the photoelectrons.
(c) The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope \( \frac{h}{e} \).
- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
- Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): For monochromatic incident radiation, the emitted photoelectrons from a given metal have speed ranging from zero to a certain maximum value.
Reason (R): Each metal has a definite work function.
- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
Questions Asked in JEE Main exam
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:
- JEE Main - 2025
- Integration
- Given below are two statements: Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas. Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:
- JEE Main - 2025
- Organic Chemistry
- Given below are two statements about X-ray spectra of elements:
Statement (I):A plot of \(\sqrt {v}\) (v=frequency of X-rays emitted) vs atomic mass is a straight line.
Statement (II): A plot of v (v=frequency of X-rays emitted)} vs atomic number is a straight line.- JEE Main - 2025
- X-ray Spectra and Moseley's Law
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
- JEE Main - 2025
- Organic Chemistry
- The ratio of the power of a light source \( S_1 \) to that of the light source \( S_2 \) is 2. \( S_1 \) is emitting \( 2 \times 10^{15} \) photons per second at 600 nm. If the wavelength of the source \( S_2 \) is 300 nm, then the number of photons per second emitted by \( S_2 \) is \_\_\_\_\_ \( \times 10^{14} \).
- JEE Main - 2025
- Units and measurement