Light of wavelength 488 ;nm is produced by an | vedclass

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(B) Given: Wavelength $\lambda = 488 \;nm = 488 	imes 10^{-9} \;m$, Stopping potential $V_o = 0.38 \;V$.The energy of the incident photon is given by

Vedclass · Accepted Answer

$2.16$

Vedclass · Answer

(B) Given: Wavelength $\lambda = 488 \;nm = 488 	imes 10^{-9} \;m$, Stopping potential $V_o = 0.38 \;V$.The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $h = 6.63 	imes 10^{-34} \;J \cdot s$ and $c = 3 	imes 10^8 \;m/s$:$E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{488 	imes 10^{-9}} \approx 4.076 	imes 10^{-19} \;J$.Converting to $eV$: $E = \frac{4.076 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 2.547 \;eV$.According to Einstein's photoelectric equation: $K_{max} = E - \phi_o$, where $K_{max} = eV_o$.Thus, $\phi_o = E - eV_o$.$\phi_o = 2.547 \;eV - 0.38 \;eV = 2.167 \;eV$.Rounding to the nearest option, the work function is $2.16 \;eV$.

Similar Questions

$A$ light wave described by $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$ (in $SI$ units) falls on a metal surface of work function $2.8 \text{ eV}$. The maximum kinetic energy of the ejected photoelectron is (approximately) . . . . . . $\text{eV}$.

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