In a photoelectric experiment, the wavelength | vedclass

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(A) The photoelectric equation is given by $eV_S = \frac{hc}{\lambda} - \Phi$, where $\Phi = \frac{hc}{\lambda_0}$ is the work function of the metal.T

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$3.1$

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(A) The photoelectric equation is given by $eV_S = \frac{hc}{\lambda} - \Phi$, where $\Phi = \frac{hc}{\lambda_0}$ is the work function of the metal.Thus, the stopping potential is $V_S = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$.For the first wavelength $\lambda_1 = 200 \, nm$, the stopping potential is $V_{S1} = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{\lambda_0} \right)$.For the second wavelength $\lambda_2 = 400 \, nm$, the stopping potential is $V_{S2} = \frac{1240}{e} \left( \frac{1}{400} - \frac{1}{\lambda_0} \right)$.The decrease in stopping potential is $\Delta V_S = V_{S1} - V_{S2}$.$\Delta V_S = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{\lambda_0} - \left( \frac{1}{400} - \frac{1}{\lambda_0} \right) \right)$.$\Delta V_S = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{400} \right) = \frac{1240}{e} \left( \frac{2-1}{400} \right) = \frac{1240}{400} = 3.1 \, V$.

In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)

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