Question:

Work function of a metallic surface is 5.01 eV. When a light of wavelength 2000Å falls on the metallic surface, it emits photo electrons. The potential difference required to stop the fastest photo electron is? (in Volt)

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Use $E = \frac{12400}{\lambda\ (\text{in \AA})}$ to get photon energy instantly in eV. Subtract the work function, and the numerical value left is your stopping voltage!
Updated On: Jun 3, 2026
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The Correct Option is A

Solution and Explanation

Step 1: Concept
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K_{\max} = E - \phi$, where $E$ is the photon energy and $\phi$ is the work function. The stopping potential $V_{0}$ is related to $K_{\max}$ by $K_{\max} = eV_{0}$.

Step 2: Meaning
The energy of the incident photon in electron-volts can be calculated using the short formula $E = \frac{12400}{\lambda\ (\text{in \AA})}$.

Step 3: Analysis
Given $\lambda = 2000\text{ \AA}$ and $\phi = 5.01\text{ eV}$: $$E = \frac{12400}{2000} = 6.2\text{ eV}$$ Now, finding the maximum kinetic energy: $$K_{\max} = 6.2\text{ eV} - 5.01\text{ eV} = 1.19\text{ eV} \approx 1.2\text{ eV}$$ Since $K_{\max} = eV_{0}$, the stopping potential is $V_{0} = 1.2\text{ V}$.

Step 4: Conclusion
The stopping potential difference required is 1.2 V.

Final Answer: (A)
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