Light of wavelength 2475 , Å is incident on b | vedclass

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(C) The radius $r$ of the circular path described by a charged particle in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$, where $K$ is the

Vedclass · Accepted Answer

$4.5$

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(C) The radius $r$ of the circular path described by a charged particle in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$, where $K$ is the kinetic energy of the electron.Rearranging for $K$, we get $K = \frac{q^2 B^2 r^2}{2m} = \left(\frac{e}{m}ight) \frac{e B^2 r^2}{2}$.Given $\frac{e}{m} = 1.7 	imes 10^{11} \, C/kg$, $e = 1.6 	imes 10^{-19} \, C$, $B = \frac{1}{\sqrt{17}} 	imes 10^{-5} \, T$, and $r = 1 \, m$:$K = \frac{1}{2} 	imes (1.7 	imes 10^{11}) 	imes (1.6 	imes 10^{-19}) 	imes \left(\frac{1}{\sqrt{17}} 	imes 10^{-5}ight)^2 	imes (1)^2$.$K = 0.5 	imes 1.7 	imes 10^{-8} 	imes \frac{1}{17} 	imes 10^{-10} 	imes 1.6 = 0.8 	imes 10^{-19} \, J$.Converting to $eV$: $K = \frac{0.8 	imes 10^{-19}}{1.6 	imes 10^{-19}} \, eV = 0.5 \, eV$.The energy of the incident photon is $E = \frac{12375}{\lambda (	ext{in } Å)} \, eV = \frac{12375}{2475} \, eV = 5.0 \, eV$.Using Einstein's photoelectric equation $E = W_0 + K_{\max}$, the work function $W_0 = E - K_{\max} = 5.0 \, eV - 0.5 \, eV = 4.5 \, eV$.

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