Home
Class 12
PHYSICS
Light of wavelength 300 Å in Photoelectr...

Light of wavelength 300 Å in Photoelectric effect gives electron of max. K.E. 0.5 eV. If wavelength change to 2000 Å then max. K.E. of emitted electrons will be

A

Less than 0.5 eV

B

0.5 eV

C

Greater than 0.5 eV

D

PEE does not occurs

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the concept of the photoelectric effect and the relationship between the energy of the incident light and the maximum kinetic energy of the emitted electrons. ### Step-by-Step Solution: 1. **Understanding the Photoelectric Effect Equation**: The energy of the incident light (E) is given by the equation: \[ E = \phi + K.E. \] where \( \phi \) is the work function of the metal and \( K.E. \) is the maximum kinetic energy of the emitted electrons. 2. **Calculating the Energy of Incident Light at 300 Å**: The energy of the incident light can be calculated using the formula: \[ E = \frac{12400}{\lambda} \] where \( E \) is in electron volts (eV) and \( \lambda \) is in angstroms (Å). For \( \lambda = 300 \, \text{Å} \): \[ E = \frac{12400}{300} = 41.33 \, \text{eV} \] 3. **Finding the Work Function (\( \phi \))**: From the problem, we know that the maximum kinetic energy (\( K.E. \)) is 0.5 eV when the wavelength is 300 Å. Therefore: \[ 41.33 \, \text{eV} = \phi + 0.5 \, \text{eV} \] Rearranging gives: \[ \phi = 41.33 \, \text{eV} - 0.5 \, \text{eV} = 40.83 \, \text{eV} \] 4. **Calculating the Energy of Incident Light at 2000 Å**: Now, we need to find the energy of the incident light when the wavelength is changed to 2000 Å: \[ E = \frac{12400}{2000} = 6.2 \, \text{eV} \] 5. **Comparing the Energy with the Work Function**: We compare the energy of the incident light at 2000 Å with the work function: - Work function \( \phi = 40.83 \, \text{eV} \) - Energy at 2000 Å \( E = 6.2 \, \text{eV} \) Since \( 6.2 \, \text{eV} < 40.83 \, \text{eV} \), the energy of the incident light is less than the work function. 6. **Conclusion**: Since the energy of the incident light at 2000 Å is less than the work function, no photoelectric effect will occur. Therefore, the maximum kinetic energy of the emitted electrons will be: \[ \text{Max K.E.} = 0 \, \text{eV} \] ### Final Answer: The maximum kinetic energy of emitted electrons when the wavelength is changed to 2000 Å will be 0 eV (no photoelectric effect occurs). ---

To solve the problem, we will use the concept of the photoelectric effect and the relationship between the energy of the incident light and the maximum kinetic energy of the emitted electrons. ### Step-by-Step Solution: 1. **Understanding the Photoelectric Effect Equation**: The energy of the incident light (E) is given by the equation: \[ E = \phi + K.E. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Wavelength of an electron is 5Å. Velocity of the electron is

Work function of a metal surface is phi=1.5eV . If a light of wavelength 5000Å falls on it then the maximum K.E. of ejected electron will be-

Radiation of wavelength lambda in indent on a photocell . The fastest emitted electron has speed v if the wavelength is changed to (3 lambda)/(4) , then speed of the fastest emitted electron will be

In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be ________eV Given E (in eV) = (1237)/(lambda ("in nm"))

In a photoelectric effect experiment the threshold wavelength of light is 380 nm.If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be : Given E (in eV) =(1237)/(lambda("in nm"))

Threshold wavelength for photoelectric effect on sodium is 5000 Å . Its work function is

Photoelectric work- function of a metal is 1 eV. Light of wavelength lambda = 3000 Å falls on it. The photoelectrons come out with maximum velocity

Photoelectric work- function of a metal is 1 eV. Light of wavelength lambda = 3000 Å falls on it. The photoelectrons come out with maximum velocity

Photo electrons are liberated by ultraviolet light of wavelength 3000 Å from a metalic surface for which the photoelectric threshold wavelength is 4000 Å . Calculate the de Broglie wavelength of electrons emitted with maximum kinetic energy.

If in a photoelectric cell , the wavelength of incident light is changed from 4000 Å to 3000 Å then change in stopping potential will be