Question

A photosensitive surface is illuminated by radiations of wavelength $\lambda_1$, $\lambda_2$ ($>\lambda_1$) and $\lambda_3$ one by one and photoemission is observed in each case. $\lambda_1, \lambda_2$ lies in UV range and $\lambda_3$ in visible range. If $V_1, V_2$ and $V_3$ are stopping potential in these cases respectively, then

• A. $V_1 = V_2 = V_3$
• B. $V_2 > V_1 > V_3$
• C. $V_1 > V_2 > V_3$
• D. $V_3 > V_1 > V_2$

Hint:

Always remember the golden rule of the electromagnetic spectrum: Higher Frequency $\implies$ Shorter Wavelength $\implies$ Higher Energy $\implies$ Higher Stopping Potential.

Answer 1

The Correct Option is C

Concept:
• The photoelectric effect describes the emission of electrons when light hits a photosensitive material.

• According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = \frac{hc}{\lambda} - \Phi$, where $\lambda$ is the wavelength of incident light and $\Phi$ is the work function.

• The stopping potential $V_0$ is directly proportional to this maximum kinetic energy via the relation $eV_0 = K_{max}$.

• Therefore, a smaller wavelength $\lambda$ precisely corresponds to a higher incident photon energy, which inevitably leads to a higher stopping potential.

Step 1: Analyze the given wavelengths in different spectral regions
The problem states that radiations $\lambda_1$ and $\lambda_2$ strictly belong to the Ultraviolet (UV) spectrum.
The radiation $\lambda_3$ belongs entirely to the visible light spectrum.
From the standard electromagnetic spectrum, we universally know that UV light possesses much shorter wavelengths than any visible light.
Therefore, we can establish the fundamental inequality: $\lambda_{UV} < \lambda_{visible}$.
This strictly implies that both $\lambda_1$ and $\lambda_2$ are distinctly smaller than $\lambda_3$.

Step 2: Order the wavelengths from smallest to largest
The problem explicitly provides a crucial condition between the two UV wavelengths: $\lambda_2 > \lambda_1$.
Combining this given fact with our spectral knowledge from Step 1, we can write a complete, uninterrupted inequality for all three wavelengths:
\[ \lambda_1 < \lambda_2 < \lambda_3 \]

Step 3: Relate wavelengths to incident photon energies
The energy of an individual photon is mathematically given by $E = \frac{hc}{\lambda}$.
Because energy is inversely proportional to wavelength, the inequality order flawlessly reverses for the photon energies.
Thus, the corresponding energies of the incident radiations follow the strict order:
\[ E_1 > E_2 > E_3 \]

Step 4: Relate photon energies to stopping potentials
Einstein's equation $eV_0 = E - \Phi$ securely connects photon energy to stopping potential.
Since the work function $\Phi$ remains perfectly constant for the same given photosensitive surface in all three cases, the stopping potential $V_0$ will simply scale directly with the incident photon energy $E$.
A higher photon energy definitively demands a correspondingly higher stopping potential to completely halt the fastest emitted photoelectrons.
Consequently, the stopping potentials must inherently follow the exact same ordering as the photon energies:
\[ V_1 > V_2 > V_3 \]

Step 5: Conclusion
Comparing our rigorously derived relationship with the given choices, we find that option (C) perfectly matches our result.