The work function of a metallic surface is 5. | vedclass

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $h = 4.14 	imes 10^{-15} \ eV \cdot s$,$c = 3 	imes 10^8 \ m/s$,and

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$1.2$

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $h = 4.14 	imes 10^{-15} \ eV \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 2000 \ \mathring{A} = 2000 	imes 10^{-10} \ m$.$E = \frac{4.14 	imes 10^{-15} 	imes 3 	imes 10^8}{2000 	imes 10^{-10}} = \frac{12.42 	imes 10^{-7}}{2 	imes 10^{-7}} = 6.21 \ eV$.The maximum kinetic energy of the emitted photo-electrons is $K_{max} = E - \phi$,where $\phi = 5.01 \ eV$ is the work function.$K_{max} = 6.21 \ eV - 5.01 \ eV = 1.2 \ eV$.The stopping potential $V_0$ is related to the maximum kinetic energy by $K_{max} = eV_0$.Therefore,$eV_0 = 1.2 \ eV$,which gives $V_0 = 1.2 \ V$.

The work function of a metallic surface is $5.01 \ eV$. Photo-electrons are emitted when light of wavelength $2000 \ \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is .............. $V$ $[h = 4.14 \times 10^{-15} \ eV \cdot s]$.

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