The Question is from neet-physics dual-nature-matter-radiation , and it was created by the author NAGA SAI

Question

The stopping potential for photoelectrons emitted from a metal surface illuminated by light of wavelength $\lambda$ is $V_0$. If the wavelength is reduced to $\lambda/2$, keeping the intensity constant, the new stopping potential will be:

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
 **Primary Concept**: Photoelectric effect and its relation to stopping potential.
 **Related Topics**: Planck's constant, work function, energy of a photon.
 **Prerequisites**: Understanding of the photoelectric effect equation, energy of a photon ($E = h
u = \frac{hc}{\lambda}$), and the concept of stopping potential.
## SOLUTION PATHWAY
**Given Information**:
  Initial wavelength: $\lambda$
  Initial stopping potential: $V_0$
  Final wavelength: $\lambda/2$
  Intensity remains constant.
**Method & Steps**:
 1. Relate the energy of the photon to the stopping potential using the photoelectric effect equation.
  Reasoning: The maximum kinetic energy of the emitted photoelectrons is given by $KE_{max} = h
u - \phi$, where $h$ is Planck's constant, $
u$ is the frequency of the incident light, and $\phi$ is the work function of the metal.  The stopping potential $V_0$ is related to the maximum kinetic energy by $KE_{max} = eV_0$, where $e$ is the elementary charge.
  Formula/Rule: $eV_0 = h
u - \phi = \frac{hc}{\lambda} - \phi$
 2. Apply the equation to both initial and final situations.
  Working: Initially, $eV_0 = \frac{hc}{\lambda} - \phi$.  When the wavelength is reduced to $\lambda/2$, the new stopping potential $V_1$ is given by $eV_1 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$.
  Key Point: The work function $\phi$ remains constant for the same metal.
 3. Find the relationship between $V_1$ and $V_0$.
  Result: Subtracting the initial equation from the final equation, we get $e(V_1 - V_0) = \frac{hc}{\lambda}$.  Since $eV_0 = \frac{hc}{\lambda} - \phi$, we have $\frac{hc}{\lambda} = eV_0 + \phi$. Substituting this into the previous equation, we get $e(V_1 - V_0) = eV_0 + \phi$, which simplifies to $eV_1 = 2eV_0 + \phi$.  This shows that $V_1 > 2V_0$.
  Verification: The increase in frequency (decrease in wavelength) directly increases the kinetic energy of the emitted electrons, leading to a greater stopping potential.
 ## OPTION ANALYSIS
A :x: $V_0/2$ is incorrect as the energy of the photons increases, leading to a higher stopping potential.
B :x: $V_0$ is incorrect; the energy of photons has increased.
C :x: $2V_0$ is incorrect; it doesn't account for the work function.
D ✓: Greater than $2V_0$ is correct because the increase in photon energy is more than double the initial energy due to the work function.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the work function:  Failing to account for the work function in the photoelectric equation leads to incorrect results.
  2.Incorrectly relating wavelength and frequency:  Remember that frequency is inversely proportional to wavelength ($
u = c/\lambda$).
 **Quick Tips**:
  1.Start with the photoelectric equation and carefully substitute values.
  2.Remember that the stopping potential is directly related to the maximum kinetic energy of the emitted electrons.

NAGA SAI · Answer

$V_0/2$

NAGA SAI · Answer

$V_0$

NAGA SAI · Answer

$2V_0$

NAGA SAI · Answer

The correct Option is "Greater than $2V_0$"

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