Radiation p with wavelength 6561 ; mathring{A | vedclass

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(A) Let the work function be $\phi$. According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is given by: $KE_{\max} = \

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(A) Let the work function be $\phi$. According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is given by: $KE_{\max} = \frac{hc}{\lambda} - \phi$. When an electron with charge $q$ and mass $m$ enters a magnetic field $B$ perpendicular to its velocity,it follows a circular path of radius $R = \frac{\sqrt{2m KE_{\max}}}{qB}$. Squaring both sides,we get $R^2 = \frac{2m KE_{\max}}{q^2 B^2}$,which implies $KE_{\max} = \frac{R^2 q^2 B^2}{2m}$. Substituting the values: $h = 6.63 	imes 10^{-34} \; J \cdot s$,$c = 3 	imes 10^8 \; m/s$,$\lambda = 6561 	imes 10^{-10} \; m$,$R = 10 	imes 10^{-3} \; m$,$q = 1.6 	imes 10^{-19} \; C$,$m = 9.1 	imes 10^{-31} \; kg$,$B = 3 	imes 10^{-4} \; T$. Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{6561 	imes 10^{-10}} \approx 3.03 	imes 10^{-19} \; J = 1.89 \; eV$. Kinetic energy $KE_{\max} = \frac{(10^{-2})^2 	imes (1.6 	imes 10^{-19})^2 	imes (3 	imes 10^{-4})^2}{2 	imes 9.1 	imes 10^{-31}} \approx 1.26 	imes 10^{-20} \; J = 0.079 \; eV$. Thus,$\phi = E - KE_{\max} = 1.89 \; eV - 0.079 \; eV \approx 1.81 \; eV$. Wait,re-evaluating the calculation: $E = 1.89 \; eV$. $KE_{\max} = 0.079 \; eV$. $\phi = 1.81 \; eV$. Given the options,$1.8 \; eV$ is the closest value.

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