Question:
When a photosensitive metal surface is illuminated with radiation of wavelength ' $\lambda_1$ ', the stopping potential is ' $V_1$ '. If the same surface is illuminated with radiation of wavelength ' $3\lambda_1$ ', the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is
When a photosensitive metal surface is illuminated with radiation of wavelength ' $\lambda_1$ ', the stopping potential is ' $V_1$ '. If the same surface is illuminated with radiation of wavelength ' $3\lambda_1$ ', the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is
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Write Einstein's photoelectric equation for the two wavelengths and eliminate the stopping potential to get the work function.
Updated On: Nov 13, 2026
- $\frac{3}{2} \lambda_1$
- $2\lambda_1$
- $5\lambda_1$
- $6\lambda_1$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
\(\dfrac{hc}{\lambda} = \phi + eV_0\), where \(\phi = \dfrac{hc}{\lambda_0}\) and \(\lambda_0\) is the threshold wavelength.
Step 2: Two Equations
Let \(E = \dfrac{hc}{\lambda_1}\). First case: \(E = \phi + eV_1\). Second case: \(\dfrac{E}{3} = \phi + \dfrac{eV_1}{6}\).
Step 3: Eliminate V1
Multiply the second by 6: \(2E = 6\phi + eV_1\). Subtract the first: \(E = 5\phi\), so \(\phi = \dfrac{E}{5}\).
Step 4: Find the Threshold Wavelength
\(\dfrac{hc}{\lambda_0} = \dfrac{1}{5}\dfrac{hc}{\lambda_1}\), so \(\lambda_0 = 5\lambda_1\), the third option.
Final Answer:
The threshold wavelength is five times lambda one. \[ \boxed{\text{(C) }5\lambda_1} \]
\(\dfrac{hc}{\lambda} = \phi + eV_0\), where \(\phi = \dfrac{hc}{\lambda_0}\) and \(\lambda_0\) is the threshold wavelength.
Step 2: Two Equations
Let \(E = \dfrac{hc}{\lambda_1}\). First case: \(E = \phi + eV_1\). Second case: \(\dfrac{E}{3} = \phi + \dfrac{eV_1}{6}\).
Step 3: Eliminate V1
Multiply the second by 6: \(2E = 6\phi + eV_1\). Subtract the first: \(E = 5\phi\), so \(\phi = \dfrac{E}{5}\).
Step 4: Find the Threshold Wavelength
\(\dfrac{hc}{\lambda_0} = \dfrac{1}{5}\dfrac{hc}{\lambda_1}\), so \(\lambda_0 = 5\lambda_1\), the third option.
Final Answer:
The threshold wavelength is five times lambda one. \[ \boxed{\text{(C) }5\lambda_1} \]
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