The work function of a certain metal is 3.31 | vedclass

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Question

(B) The work function is given as $W_0 = 3.31 	imes 10^{-19} \,J$. The wavelength of incident radiation is $\lambda = 5000 	ext{ Å} = 5000 	imes 10

Vedclass · Accepted Answer

$0.41$

Vedclass · Answer

(B) The work function is given as $W_0 = 3.31 	imes 10^{-19} \,J$. The wavelength of incident radiation is $\lambda = 5000 	ext{ Å} = 5000 	imes 10^{-10} \,m = 5 	imes 10^{-7} \,m$. According to Einstein's photoelectric equation,the energy of the incident photon $E$ is given by $E = W_0 + KE_{max}$,where $KE_{max}$ is the maximum kinetic energy. The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{6.62 	imes 10^{-34} 	imes 3 	imes 10^8}{5 	imes 10^{-7}} \,J$. $E = \frac{19.86 	imes 10^{-26}}{5 	imes 10^{-7}} = 3.972 	imes 10^{-19} \,J$. Now,$KE_{max} = E - W_0 = 3.972 	imes 10^{-19} \,J - 3.31 	imes 10^{-19} \,J = 0.662 	imes 10^{-19} \,J$. To convert this energy into electron-volts $(eV)$,divide by the charge of an electron $e = 1.6 	imes 10^{-19} \,C$: $KE_{max} = \frac{0.662 	imes 10^{-19} \,J}{1.6 	imes 10^{-19} \,C} = 0.41375 \,eV \approx 0.41 \,eV$.

The work function of a certain metal is $3.31 \times 10^{-19} \,J$. The maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \text{ Å}$ is (Given: $h = 6.62 \times 10^{-34} \,J \cdot s$,$c = 3 \times 10^8 \,m/s$,$e = 1.6 \times 10^{-19} \,C$) (in $\text{ eV}$)

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