In photoelectric effect experiment, the stopping potential for a given metal is \(\mathrm{V}_0\) (in volt) when radiation of wavelength \(\lambda_0\) is used. If radiation of wavelength \(5 \lambda_0\) is used with the same metal, then the stopping potential is (in volt) ( \(\mathrm{h}=\) Planck's constant, \(\mathrm{c}=\) velocity of light, \(\mathrm{e}=\) electronic charge)
In photoelectric effect experiment, the stopping potential for a given metal is \(\mathrm{V}_0\) (in volt) when radiation of wavelength \(\lambda_0\) is used. If radiation of wavelength \(5 \lambda_0\) is used with the same metal, then the stopping potential is (in volt) ( \(\mathrm{h}=\) Planck's constant, \(\mathrm{c}=\) velocity of light, \(\mathrm{e}=\) electronic charge)
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\(\mathrm{V}_{0}-\frac{4\mathrm{hc}}{5\mathrm{e}\lambda _{0}}\)
\(\mathrm{V}_{0}+\frac{4\mathrm{hc}}{5\mathrm{e}\lambda _{0}}\)
\(\mathrm{V}_{0}+\frac{2\mathrm{hc}}{3\mathrm{e}\lambda _{0}}\)
\(\mathrm{V}_{0}-\frac{2\mathrm{hc}}{3\mathrm{e}\lambda _{0}}\)
The Correct Option is A
Solution and Explanation
Einstein's equation: \(eV_0 = \dfrac{hc}{\lambda} - \phi\).
Step 2: First Case
\(eV_0 = \dfrac{hc}{\lambda_0} - \phi\), so \(\phi = \dfrac{hc}{\lambda_0} - eV_0\).
Step 3: Second Case
\(eV' = \dfrac{hc}{5\lambda_0} - \phi = \dfrac{hc}{5\lambda_0} - \dfrac{hc}{\lambda_0} + eV_0 = eV_0 - \dfrac{4hc}{5\lambda_0}\).
Step 4: Result
\(V' = V_0 - \dfrac{4hc}{5e\lambda_0}\), option A.
Final Answer:
The new stopping potential is V0 minus 4hc over 5 e lambda0. \[ \boxed{\text{(A) }\mathrm{V}_{0}-\dfrac{4\mathrm{hc}}{5\mathrm{e}\lambda_{0}}} \]
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