Question

When two monochromatic light of frequency ν and \(\frac{ν}{2}\) are incident on a photoelectric metal, their stopping potential becomes \(\frac{V_s}{2}\) and V_s respectively. The threshold frequency for this metal is:

• A. 2 v
• B. 3 v
• C. \(\frac{2}{3v}\)
• D. \(\frac{3}{2v}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is the difference between the energy of the incident photon and the work function of the metal.
Max K.E. is also equal to \(e V_0\), where \(V_0\) is the stopping potential.
Key Formula or Approach:
\[ e V_0 = h\nu - h\nu_0 \]
where \(\nu\) is incident frequency and \(\nu_0\) is threshold frequency.
Step 2: Detailed Explanation:
Let's write equations for the two given cases:
Case 1: Incident frequency = \(\nu\), Stopping potential = \(V_s/2\).
\[ e \left( \frac{V_s}{2} \right) = h\nu - h\nu_0 \quad \text{ (i)} \]
Case 2: Incident frequency = \(\nu/2\), Stopping potential = \(V_s\).
\[ e V_s = h\left(\frac{\nu}{2}\right) - h\nu_0 \quad \text{ (ii)} \]
From equation (i), we can say:
\[ e V_s = 2(h\nu - h\nu_0) = 2h\nu - 2h\nu_0 \quad \text{ (iii)} \]
Equating (ii) and (iii) since both equal \(e V_s\):
\[ \frac{h\nu}{2} - h\nu_0 = 2h\nu - 2h\nu_0 \]
Rearranging terms to solve for \(\nu_0\):
\[ 2h\nu_0 - h\nu_0 = 2h\nu - \frac{h\nu}{2} \]
\[ h\nu_0 = \frac{3h\nu}{2} \]
\[ \nu_0 = \frac{3}{2}\nu \]
Step 3: Final Answer:
The threshold frequency for the metal is \(\frac{3}{2}\nu\).