The Question is from neet-physics dual-nature-matter-radiation , and it was created by the author P.NARASIMHA

Question

In Lenard's experiment, the stopping potential for photoelectrons emitted from a metal surface illuminated by light of wavelength $\lambda_1$ is $V_1$. When the wavelength is changed to $\lambda_2$ ($\lambda_2 > \lambda_1$), the stopping potential becomes $V_2$.  If the work function of the metal is $\phi$, which of the following relations is correct?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: 1
## CONCEPT FRAMEWORK
 **Primary Concept**: Photoelectric effect and Einstein's equation for photoelectric emission.
 **Related Topics**: Work function, stopping potential, Planck's constant.
 **Prerequisites**: Understanding of energy conservation in the photoelectric effect.
## SOLUTION PATHWAY
**Given Information**:
  $\lambda_1$, $V_1$, $\lambda_2$, $V_2$, work function $\phi$
  $\lambda_2 > \lambda_1$
**Method & Steps**:
 1. Apply Einstein's photoelectric equation for both wavelengths.
  Reasoning: The energy of the incident photon is used to overcome the work function and provide kinetic energy to the emitted electron. The stopping potential is the potential required to stop the most energetic electrons.
  Formula/Rule: $E = h
u = \frac{hc}{\lambda} = \phi + KE_{max} = \phi + eV$
 2. Write the equations for both wavelengths:
  Working: For $\lambda_1$: $\frac{hc}{\lambda_1} = \phi + eV_1$ (1)
  For $\lambda_2$: $\frac{hc}{\lambda_2} = \phi + eV_2$ (2)
  Subtract equation (2) from equation (1):
  $\frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = eV_1 - eV_2$
  $hc(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = eV_1 - eV_2$
 3. Final step:
  Result: $hc(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = eV_1 - eV_2$
  Verification:  The equation shows that the difference in photon energy is equal to the difference in the kinetic energy of the emitted electrons. Since $\lambda_2 > \lambda_1$, the energy of the photon with $\lambda_2$ is less than the energy of the photon with $\lambda_1$, thus $V_2 < V_1$, which is consistent with the equation.
 ## OPTION ANALYSIS
A] ✓: $hc(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = eV_1 - eV_2$ This is the correct equation derived from Einstein's photoelectric equation.
B] :x: $hc(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}) = eV_1 - eV_2$ This equation has the wrong sign for the difference in wavelengths.
C] :x: $hc(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = eV_2 - eV_1$ This equation has the wrong sign for the difference in stopping potentials.
D] :x: $hc(\frac{1}{\lambda_1} + \frac{1}{\lambda_2}) = eV_1 + eV_2$ This equation incorrectly adds the energies instead of subtracting them.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying Einstein's equation:  Remember that the energy of the photon is used to overcome the work function and provide kinetic energy to the electron.
  2.Mixing up the wavelengths and stopping potentials: Pay close attention to which wavelength corresponds to which stopping potential.
 **Quick Tips**:
  1.Remember the formula: $\frac{hc}{\lambda} = \phi + eV$
  2.Visualize the energy levels: Draw a diagram showing the energy of the incident photon, work function, and kinetic energy of the electron to understand the energy relationships.

P.NARASIMHA · Answer

The correct Option is "$hc(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = eV_1 - eV_2$"

P.NARASIMHA · Answer

$hc(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}) = eV_1 - eV_2$

P.NARASIMHA · Answer

$hc(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = eV_2 - eV_1$

P.NARASIMHA · Answer

$hc(\frac{1}{\lambda_1} + \frac{1}{\lambda_2}) = eV_1 + eV_2$

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