Monochromatic radiation of wavelength 640.2 ; | vedclass

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(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Using the photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$

Vedclass · Accepted Answer

$1.5$

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(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Using the photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$ is the work function.For the neon lamp: $\lambda_1 = 640.2 \;nm$,$V_{01} = 0.54 \;V$.$\phi_0 = \frac{hc}{\lambda_1} - eV_{01} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{640.2 	imes 10^{-9}} - 0.54 \;eV = 1.94 \;eV - 0.54 \;eV = 1.40 \;eV$.For the iron source: $\lambda_2 = 427.2 \;nm$.$eV_{02} = \frac{hc}{\lambda_2} - \phi_0 = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{427.2 	imes 10^{-9}} - 1.40 \;eV$.$eV_{02} = 2.90 \;eV - 1.40 \;eV = 1.50 \;eV$.Thus,the new stopping voltage is $1.5 \;V$.

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