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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi_0$.Since $K_{

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$1.5$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi_0$.Since $K_{max} = eV_0$,where $V_0$ is the stopping potential,we have $eV_0 = \frac{hc}{\lambda} - \phi_0$.First,calculate the energy of the incident photon in $eV$:$E = \frac{hc}{\lambda} = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{200 	imes 10^{-9} 	imes 1.6 	imes 10^{-19}} \ eV$.$E = \frac{19.878 	imes 10^{-26}}{320 	imes 10^{-28}} \ eV = \frac{19.878}{3.2} \ eV \approx 6.21 \ eV$.Now,calculate the stopping potential:$eV_0 = 6.21 \ eV - 4.71 \ eV = 1.50 \ eV$.Therefore,the stopping potential $V_0 = 1.50 \ V$,which corresponds to $1.5 \ eV$ in terms of energy units.

Ultraviolet light of wavelength $200 \ nm$ is incident on a freshly polished surface of iron. The work function of the surface is $4.71 \ eV$. What will be the stopping potential in $eV$? $(h = 6.626 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J, c = 3 \times 10^8 \ m/s)$

Similar Questions

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