In the given figure,V represents the stopping | vedclass

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(B) From the photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \phi = \frac{hc}{\lambda} - \phi$.This implies that the st

Vedclass · Accepted Answer

$\lambda_1 > \lambda_2$

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(B) From the photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \phi = \frac{hc}{\lambda} - \phi$.This implies that the stopping potential is inversely proportional to the wavelength $\lambda$ of the incident light.Given that $|V_2| > |V_1|$,where $V_2$ and $V_1$ are the stopping potentials corresponding to wavelengths $\lambda_2$ and $\lambda_1$ respectively.Since $V_s \propto \frac{1}{\lambda}$,a higher stopping potential corresponds to a smaller wavelength.Therefore,$|V_2| > |V_1| \implies \lambda_2  \lambda_2$.

In the given figure,$V$ represents the stopping potential. If $\lambda_1$ and $\lambda_2$ are the wavelengths of the incident light and $|V_2| > |V_1|$,then:

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