Question:
In photo electric experiment, if the wavelength of incident light on the metal changes from 200 nm to 300 nm. The decrease in stopping potential is about ($\frac{hc}{e} = 1240$ eV - nm)
In photo electric experiment, if the wavelength of incident light on the metal changes from 200 nm to 300 nm. The decrease in stopping potential is about ($\frac{hc}{e} = 1240$ eV - nm)
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Using the shortcut $\Delta V_s \approx 1240 \Delta \left(\frac{1}{\lambda}\right)$ is highly effective for photoelectric numericals.
Always keep units in eV and nm to avoid complex conversions with Joules.
Always keep units in eV and nm to avoid complex conversions with Joules.
Updated On: Jul 22, 2026
- 2.1 V
- 4.2 V
- 3.1 V
- 6.2 V
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question is about the photoelectric effect and how stopping potential varies with the wavelength of incident light.
We need to find the decrease in stopping potential when the wavelength increases from 200 nm to 300 nm.
Step 2: Key Formula or Approach:
According to Einstein's photoelectric equation, the stopping potential $V_s$ and incident wavelength $\lambda$ are related by:
\[ eV_s = \frac{hc}{\lambda} - \phi \] where $e$ is the charge of an electron, $h$ is Planck's constant, $c$ is the speed of light, and $\phi$ is the work function of the metal.
For two different wavelengths, the difference in stopping potential is given by:
\[ e(V_{s1} - V_{s2}) = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]
Step 3: Detailed Explanation:
• Let the initial wavelength $\lambda_1 = 200$ nm and the final wavelength $\lambda_2 = 300$ nm.
• The given value for the constant ratio is $\frac{hc}{e} = 1240$ eV-nm.
• We can express the change in stopping potential $\Delta V_s = V_{s1} - V_{s2}$ directly in volts as:
\[ \Delta V_s = \frac{hc}{e} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]
• Substituting the given values into the equation:
\[ \Delta V_s = 1240 \left( \frac{1}{200} - \frac{1}{300} \right) \] \[ \Delta V_s = 1240 \left( \frac{300 - 200}{200 \times 300} \right) \] \[ \Delta V_s = 1240 \left( \frac{100}{60000} \right) \] \[ \Delta V_s = \frac{1240}{600} \approx 2.067\text{ V} \]
• This value is approximately equal to 2.1 V.
Step 4: Final Answer:
The decrease in the stopping potential is about 2.1 V.
The question is about the photoelectric effect and how stopping potential varies with the wavelength of incident light.
We need to find the decrease in stopping potential when the wavelength increases from 200 nm to 300 nm.
Step 2: Key Formula or Approach:
According to Einstein's photoelectric equation, the stopping potential $V_s$ and incident wavelength $\lambda$ are related by:
\[ eV_s = \frac{hc}{\lambda} - \phi \] where $e$ is the charge of an electron, $h$ is Planck's constant, $c$ is the speed of light, and $\phi$ is the work function of the metal.
For two different wavelengths, the difference in stopping potential is given by:
\[ e(V_{s1} - V_{s2}) = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]
Step 3: Detailed Explanation:
• Let the initial wavelength $\lambda_1 = 200$ nm and the final wavelength $\lambda_2 = 300$ nm.
• The given value for the constant ratio is $\frac{hc}{e} = 1240$ eV-nm.
• We can express the change in stopping potential $\Delta V_s = V_{s1} - V_{s2}$ directly in volts as:
\[ \Delta V_s = \frac{hc}{e} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]
• Substituting the given values into the equation:
\[ \Delta V_s = 1240 \left( \frac{1}{200} - \frac{1}{300} \right) \] \[ \Delta V_s = 1240 \left( \frac{300 - 200}{200 \times 300} \right) \] \[ \Delta V_s = 1240 \left( \frac{100}{60000} \right) \] \[ \Delta V_s = \frac{1240}{600} \approx 2.067\text{ V} \]
• This value is approximately equal to 2.1 V.
Step 4: Final Answer:
The decrease in the stopping potential is about 2.1 V.
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