When 0.50; mathring{A} X-rays strike a materi | vedclass

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(B) The velocity $v$ of the photoelectrons is given by the magnetic force providing the centripetal force: $evB = m\frac{v^2}{R} \implies v = \frac{eB

Vedclass · Accepted Answer

$6.2$

Vedclass · Answer

(B) The velocity $v$ of the photoelectrons is given by the magnetic force providing the centripetal force: $evB = m\frac{v^2}{R} \implies v = \frac{eBR}{m}$.The kinetic energy $K$ of the photoelectrons is $K = \frac{1}{2}mv^2 = \frac{e^2B^2R^2}{2m}$.Substituting the values: $e = 1.6 	imes 10^{-19}\; C$,$B = 2 	imes 10^{-2}\; T$,$R = 23 	imes 10^{-3}\; m$,$m = 9.1 	imes 10^{-31}\; kg$:$K = \frac{(1.6 	imes 10^{-19})^2 	imes (2 	imes 10^{-2})^2 	imes (23 	imes 10^{-3})^2}{2 	imes 9.1 	imes 10^{-31}} \approx 2.97 	imes 10^{-15}\; J$.Converting to $keV$: $K = \frac{2.97 	imes 10^{-15}}{1.6 	imes 10^{-19}} \approx 18.56\; keV$.The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{12.4\; keV \cdot \mathring{A}}{0.50\; \mathring{A}} = 24.8\; keV$.The binding energy $BE$ is given by $BE = E - K = 24.8\; keV - 18.6\; keV = 6.2\; keV$.

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