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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{12400 \; eV \cdot \mathring A}{4500 \; \mathring A} \approx 2.76 \;

Vedclass · Accepted Answer

$1.36$

Vedclass · Answer

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{12400 \; eV \cdot \mathring A}{4500 \; \mathring A} \approx 2.76 \; eV$.The radius of the circular path of a charged particle in a magnetic field is $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.Rearranging for $K$: $K = \frac{(qBR)^2}{2m}$.Given $q = 1.6 	imes 10^{-19} \; C$,$B = 2 	imes 10^{-3} \; T$,$R = 2 	imes 10^{-3} \; m$,and $m = 9.1 	imes 10^{-31} \; kg$:$K = \frac{(1.6 	imes 10^{-19} 	imes 2 	imes 10^{-3} 	imes 2 	imes 10^{-3})^2}{2 	imes 9.1 	imes 10^{-31}} = \frac{(6.4 	imes 10^{-25})^2}{18.2 	imes 10^{-31}} = \frac{40.96 	imes 10^{-50}}{18.2 	imes 10^{-31}} \approx 2.25 	imes 10^{-19} \; J$.Converting to $eV$: $K = \frac{2.25 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 1.40 \; eV$.Using Einstein's photoelectric equation: $\phi = E - K = 2.76 \; eV - 1.40 \; eV = 1.36 \; eV$.

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