When a metal surface is illuminated by light of wavelength \( \lambda \), the stopping potential is 8V. When the same surface is illuminated by light of wavelength \( 3\lambda \), the stopping potential is 2V. The threshold wavelength for this surface is:
- 5\( \lambda \)
- 3\( \lambda \)
- 9\( \lambda \)
- 4.5\( \lambda \)
The Correct Option is C
Approach Solution - 1
To solve the problem of finding the threshold wavelength for a given metal surface illuminated by light, we use the photoelectric effect equation. The photoelectric equation relates the kinetic energy of ejected electrons to the frequency of the incident light and the work function of the material:
\(E_k = h(f - f_0)\)
where:
- \(E_k\) is the kinetic energy of the ejected electron.
- \(h\) is Planck's constant.
- \(f\) is the frequency of incident light.
- \(f_0\) is the threshold frequency.
The stopping potential \(V_s\) is related to the kinetic energy by:
\(E_k = eV_s\)
where \(e\) is the elementary charge.
We also have the relationship between frequency and wavelength:
\(f = \frac{c}{\lambda}\)
where \(c\) is the speed of light.
Substitute this into the photoelectric equation:
\(eV_s = h\frac{c}{\lambda} - h\frac{c}{\lambda_0}\)
For two different stopping potentials, we can write two equations:
- For \(\lambda\) with stopping potential 8V:
- For \(3\lambda\) with stopping potential 2V:
Now, we solve these equations to find \(\lambda_0\), the threshold wavelength.
Rewrite Equation 1:
\(h \frac{c}{\lambda_0} = h \frac{c}{\lambda} - 8e\)
Rewrite Equation 2:
\(h \frac{c}{\lambda_0} = h \frac{c}{3\lambda} - 2e\)
Equating the two expressions for \(h \frac{c}{\lambda_0}\):
\(h \frac{c}{\lambda} - 8e = h \frac{c}{3\lambda} - 2e\)
Isolate terms:
\(h \frac{c}{\lambda} - h \frac{c}{3\lambda} = 8e - 2e\)
\(h c \left( \frac{1}{\lambda} - \frac{1}{3\lambda} \right) = 6e\)
Simplify:
\(h c \left( \frac{2}{3\lambda} \right) = 6e\)
\(h c = 9e \lambda\)
So, using Equation 1:
\(h \frac{c}{\lambda_0} = \frac{2}{3} \cdot 9e \lambda \Rightarrow \lambda_0 = 9\lambda\)
Thus, the threshold wavelength \(\lambda_0\) is 9\( \lambda \), and the correct answer is 9\( \lambda \).
Approach Solution -2
\[ E = \phi + K_{\text{max}} \]
\[ \phi = \frac{hc}{\lambda_0} \]
\[ K_{\text{max}} = eV_0 \]
\[ 8e = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad \text{(i)} \]
\[ 2e = \frac{hc}{3\lambda} - \frac{hc}{\lambda_0} \quad \text{(ii)} \]
On solving (i) & (ii),
\[ \lambda_0 = 9\lambda \]
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