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(B) Using Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - \phi$.Given $hc = 20 	imes 10^{-26} \, J \cdot m$,$\lambda = 500 	imes

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(B) Using Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - \phi$.Given $hc = 20 	imes 10^{-26} \, J \cdot m$,$\lambda = 500 	imes 10^{-9} \, m$,and $\phi = 1.25 \, eV = 1.25 	imes 1.6 	imes 10^{-19} \, J = 2 	imes 10^{-19} \, J$.$K_{\max} = \frac{20 	imes 10^{-26}}{500 	imes 10^{-9}} - 2 	imes 10^{-19} = 4 	imes 10^{-19} - 2 	imes 10^{-19} = 2 	imes 10^{-19} \, J$.The radius of the circular path in a magnetic field is $r = \frac{\sqrt{2 m_e K_{\max}}}{eB}$.Rearranging for $B$: $B = \frac{\sqrt{2 m_e K_{\max}}}{er}$.Substituting values: $B = \frac{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 2 	imes 10^{-19}}}{1.6 	imes 10^{-19} 	imes 0.3} = \frac{\sqrt{36 	imes 10^{-50}}}{0.48 	imes 10^{-19}} = \frac{6 	imes 10^{-25}}{0.48 	imes 10^{-19}} = 12.5 	imes 10^{-6} \, T = 125 	imes 10^{-7} \, T$.Thus,the value of $B$ is $125 	imes 10^{-7} \, T$.

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