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(C) According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted electron is given by: $K_{\max} = \frac{hc}{\lambda} - \p

Vedclass · Accepted Answer

$\sqrt{8 m(\frac{hc}{\lambda}-\phi)} / eB$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted electron is given by: $K_{\max} = \frac{hc}{\lambda} - \phi$. The momentum $p$ of the electron is related to kinetic energy by $p = \sqrt{2mK_{\max}} = \sqrt{2m(\frac{hc}{\lambda} - \phi)}$. When an electron enters a perpendicular magnetic field $B$,it moves in a circular path of radius $R = \frac{p}{eB}$. The electron hits the plate at point $B$ after completing a semi-circle,so the distance $AB$ is the diameter of the path: $d_{AB} = 2R = \frac{2p}{eB}$. Substituting the value of $p$: $d_{AB} = \frac{2\sqrt{2m(\frac{hc}{\lambda} - \phi)}}{eB} = \frac{\sqrt{4 \cdot 2m(\frac{hc}{\lambda} - \phi)}}{eB} = \frac{\sqrt{8m(\frac{hc}{\lambda} - \phi)}}{eB}$.

List $I$	List $II$
$A$. Planck's constant $(h)$	$I$. $[M^1 L^2 T^{-2}]$
$B$. Stopping potential $(V_s)$	$II$. $[M^1 L^1 T^{-1}]$
$C$. Work function $(\phi)$	$III$. $[M^1 L^2 T^{-1}]$
$D$. Momentum $(p)$	$IV$. $[M^1 L^2 T^{-3} A^{-1}]$

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Match List $I$ with List $II$:

List $I$ List $II$

$A$. Planck's constant $(h)$ $I$. $[M^1 L^2 T^{-2}]$

$B$. Stopping potential $(V_s)$ $II$. $[M^1 L^1 T^{-1}]$

$C$. Work function $(\phi)$ $III$. $[M^1 L^2 T^{-1}]$

$D$. Momentum $(p)$ $IV$. $[M^1 L^2 T^{-3} A^{-1}]$

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