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(C) According to Einstein's photoelectric equation,the stopping potential $V$ is given by: $eV = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work f

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Harmonic Progression

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(C) According to Einstein's photoelectric equation,the stopping potential $V$ is given by: $eV = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.Rearranging for $V$,we get: $V = \frac{hc}{e\lambda} - \frac{\phi}{e}$.Given that $V_1, V_2, V_3$ are in Arithmetic Progression ($A$.$P$.),we have $2V_2 = V_1 + V_3$.Substituting the expression for $V$:$2\left(\frac{hc}{e\lambda_2} - \frac{\phi}{e}\right) = \left(\frac{hc}{e\lambda_1} - \frac{\phi}{e}\right) + \left(\frac{hc}{e\lambda_3} - \frac{\phi}{e}\right)$.Simplifying the equation:$\frac{2hc}{e\lambda_2} - \frac{2\phi}{e} = \frac{hc}{e\lambda_1} + \frac{hc}{e\lambda_3} - \frac{2\phi}{e}$.Canceling the common terms:$\frac{2}{\lambda_2} = \frac{1}{\lambda_1} + \frac{1}{\lambda_3}$.This condition implies that $\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \frac{1}{\lambda_3}$ are in $A$.$P$.,which means $\lambda_1, \lambda_2, \lambda_3$ are in Harmonic Progression ($H$.$P$.).

Cut-off potentials for a metal in the photoelectric effect for light of wavelengths $\lambda_1$,$\lambda_2$,and $\lambda_3$ are found to be $V_1$,$V_2$,and $V_3$ volts. If $V_1$,$V_2$,and $V_3$ are in Arithmetic Progression,then $\lambda_1$,$\lambda_2$,and $\lambda_3$ will be in:

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