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(B) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$ ($V_s$ is the stopping potential).For

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$1.03$

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(B) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$ ($V_s$ is the stopping potential).For two different wavelengths $\lambda_1 = 4000 \mathring A$ and $\lambda_2 = 3000 \mathring A$:$eV_{s1} = \frac{hc}{\lambda_1} - \phi$$eV_{s2} = \frac{hc}{\lambda_2} - \phi$The change in stopping potential $\Delta V_s = V_{s2} - V_{s1} = \frac{hc}{e} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$.Given $hc = 12400 \, eV \cdot \mathring A$:$\Delta V_s = 12400 \left( \frac{1}{3000} - \frac{1}{4000} \right) = 12400 \left( \frac{4-3}{12000} \right) = \frac{12400}{12000} = 1.033 \, V$.Thus,the change in stopping potential is approximately $1.03 \, V$.

If in a photoelectric cell,the wavelength of incident light is changed from $4000 \mathring A$ to $3000 \mathring A$,then the change in stopping potential will be ..... $V$.

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