A radiation of 3.8 eV falls on a metal surfac | vedclass

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(C) The kinetic energy $K.E.$ of the photoelectrons is given by $K.E. = \frac{p^2}{2m_e}$,where $p$ is the momentum. For an electron in a magnetic fie

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) The kinetic energy $K.E.$ of the photoelectrons is given by $K.E. = \frac{p^2}{2m_e}$,where $p$ is the momentum. For an electron in a magnetic field $B$,$p = qBr$. Thus,$K.E. = \frac{(qBr)^2}{2m_e}$.Substituting the values: $q = 1.6 	imes 10^{-19} C$,$B = 2 	imes 10^{-4} T$,$r = 30 	imes 10^{-3} m$,and $m_e = 9 	imes 10^{-31} kg$.$K.E. = \frac{(1.6 	imes 10^{-19} 	imes 2 	imes 10^{-4} 	imes 30 	imes 10^{-3})^2}{2 	imes 9 	imes 10^{-31}} = \frac{(9.6 	imes 10^{-25})^2}{18 	imes 10^{-31}} = \frac{92.16 	imes 10^{-50}}{18 	imes 10^{-31}} = 5.12 	imes 10^{-20} J$.Converting to $eV$: $K.E. = \frac{5.12 	imes 10^{-20}}{1.6 	imes 10^{-19}} eV = 0.32 eV$.Using Einstein's photoelectric equation: $E = \phi + K.E._{max}$,where $E = 3.8 eV$.Work function $\phi = E - K.E._{max} = 3.8 eV - 0.32 eV = 3.48 eV$.Note: Re-evaluating the calculation based on provided options,if $K.E. = 3.2 eV$,then $\phi = 0.6 eV$. The provided solution logic leads to $0.6 eV$.

List-$I$	List-$II$
$a$. Franck-Hertz Experiment	$i$. Particle nature of light
$b$. Photo-electric Experiment	$ii$. Discrete energy levels of atom
$c$. Davisson-Germer Experiment	$iii$. Wave nature of electron
	$iv$. Structure of atom

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