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(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For the first source: $e(0.

Vedclass · Accepted Answer

$1.25$

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(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For the first source: $e(0.48) = \frac{1240}{670.5} - \phi$  --- $(1)$For the second source: $eV_0' = \frac{1240}{474.6} - \phi$  --- $(2)$Subtracting equation $(1)$ from $(2)$:$e(V_0' - 0.48) = 1240 \left( \frac{1}{474.6} - \frac{1}{670.5} ight)$$V_0' - 0.48 = 1240 \left( \frac{670.5 - 474.6}{474.6 	imes 670.5} ight)$$V_0' - 0.48 = 1240 \left( \frac{195.9}{318223.26} ight)$$V_0' - 0.48 \approx 1240 	imes 0.0006156 \approx 0.763$$V_0' = 0.48 + 0.763 = 1.243\, V$Rounding to the nearest option,$V_0' \approx 1.25\, V$.

$A$ monochromatic neon lamp with a wavelength of $670.5\, nm$ illuminates a photo-sensitive material which has a stopping voltage of $0.48\, V$. What will be the stopping voltage if the source light is changed with another source of wavelength $474.6\, nm$? (In $V$)

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