A metal having a work function of 4 , eV is e | vedclass

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(A) Energy of incident photon: $E = \frac{12400}{\lambda(\mathring{A})} \, eV = \frac{12400}{1240} \, eV = 10 \, eV$. Work function: $W = 4 \, eV$. Ma

Vedclass · Accepted Answer

$2.18 	imes 10^6 \, m/s$

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(A) Energy of incident photon: $E = \frac{12400}{\lambda(\mathring{A})} \, eV = \frac{12400}{1240} \, eV = 10 \, eV$. Work function: $W = 4 \, eV$. Maximum kinetic energy of emitted electron: $KE_{max} = E - W = 10 \, eV - 4 \, eV = 6 \, eV$. On applying an accelerating potential of $V_{acc} = 7.6 \, V$,the additional kinetic energy gained is $q 	imes V_{acc} = 7.6 \, eV$. Final kinetic energy: $KE_f = KE_{max} + 7.6 \, eV = 6 \, eV + 7.6 \, eV = 13.6 \, eV$. Converting $KE_f$ to Joules: $13.6 	imes 1.6 	imes 10^{-19} \, J = 2.176 	imes 10^{-18} \, J$. Using $KE = \frac{1}{2}mv^2$,where $m = 9.1 	imes 10^{-31} \, kg$: $v = \sqrt{\frac{2 	imes KE_f}{m}} = \sqrt{\frac{2 	imes 2.176 	imes 10^{-18}}{9.1 	imes 10^{-31}}} \approx \sqrt{0.478 	imes 10^{13}} \approx 2.18 	imes 10^6 \, m/s$.

$A$ metal having a work function of $4 \, eV$ is exposed to a photon of $\lambda = 1240 \, \mathring{A}$. If an accelerating potential of $7.6 \, V$ is applied,then the electron with maximum kinetic energy will have a speed equal to -

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