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Question

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

Vedclass · Accepted Answer

$\frac{1}{\lambda_0} = \frac{1}{\lambda} - \frac{B_0^2 e^2 r^2}{2mhc}$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by: $K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ When a charged particle moves in a magnetic field $B_0$ in a circular path of radius $r$,the magnetic force provides the centripetal force: $evB_0 = \frac{mv^2}{r} \implies v = \frac{eB_0r}{m}$ The kinetic energy can be expressed as: $K_{max} = \frac{1}{2}mv^2 = \frac{1}{2}m \left( \frac{eB_0r}{m} \right)^2 = \frac{e^2 B_0^2 r^2}{2m}$ Equating the two expressions for $K_{max}$: $\frac{e^2 B_0^2 r^2}{2m} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ Rearranging to solve for $\frac{1}{\lambda_0}$: $\frac{hc}{\lambda_0} = \frac{hc}{\lambda} - \frac{e^2 B_0^2 r^2}{2m}$ Dividing both sides by $hc$: $\frac{1}{\lambda_0} = \frac{1}{\lambda} - \frac{e^2 B_0^2 r^2}{2mhc}$

When a metal is irradiated by light having wavelength $\lambda$ $(\lambda < \lambda_0)$,all the photoelectrons emitted are bent in a circle of radius $r$ by a magnetic field of flux density $B_0$. Find $\frac{1}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.

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