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Question: A Photon of energy E ejects a Photoelectron from a metal surface whose work function is \(\phi\) if...

A Photon of energy E ejects a Photoelectron from a metal surface whose work function is ϕ\phi if this electron enters into a uniform magnetic field B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is (in the usual notation):

A

2 m(E−ϕ0)eB\sqrt { \frac { 2 \mathrm {~m} \left( \mathrm { E } - \phi _ { 0 } \right) } { \mathrm { eB } } }

B

2 m(E−ϕ0)eB\sqrt { 2 \mathrm {~m} \left( \mathrm { E } - \phi _ { 0 } \right) \mathrm { eB } }

C

2e(E−ϕ0)mB\sqrt { \frac { 2 \mathrm { e } \left( \mathrm { E } - \phi _ { 0 } \right) } { \mathrm { mB } } }

D

Answer

Explanation

Solution

: As the electron describes a circular path of radius r in the magnetic field , therefore

mv2r=evB\frac { m v ^ { 2 } } { r } = e v B

From Einstein’s photoelectric equation

K=E−ϕ0K = E - \phi _ { 0 }

∴r=2m(E−ϕ0)eB\therefore r = \frac { \sqrt { 2 m \left( E - \phi _ { 0 } \right) } } { e B }