A photon of energy E ejects a photoelectron from a metal surface whose work function is W0. If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is given by, (in the usual notation)

Question

A photon of energy E ejects a photoelectron from a metal surface whose work function is W0. If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is given by, (in the usual notation) (A) (√2m(E-W0)/mB) (B) √2m(E-W0)eB (C) (√2e(E-W0)/mB) (D) (√2m(E-W0)/eB)

Tardigrade Admin · Accepted Answer

From Einstein's equation E=W0+(1/2) m v2 √(2(E-W0)/m)=v or A charged particle placed in uniform magnetic field experience a force F =(m v2/r) or e v B =(m v2/r) or r =(m v/e B) or r =(m √(2(E-W0)/m)/e B) ⇒ r=(√2 m(E-W0)/e B)

$\frac{2 m ( E - W _{0} )}{m B}$

$2 m (E - W_{0}) e B$

$\frac{2 e ( E - W _{0} )}{m B}$

$\frac{2 m ( E - W _{0} )}{e B}$

mB2m(E−W0​)​​

2m(E−W0​)eB​

mB2e(E−W0​)​​

eB2m(E−W0​)​​

From Einstein's equation E=W0​+21​mv2 m2(E−W0​)​​=v or A charged particle placed in uniform magnetic field experience a force F=rmv2​ or evB=rmv2​ or r=eBmv​ or r=eBmm2(E−W0​)​​​ ⇒r=eB2m(E−W0​)​​

$\frac{2 m ( E - W _{0} )}{m B}$

$2 m (E - W_{0}) e B$

$\frac{2 e ( E - W _{0} )}{m B}$

$\frac{2 m ( E - W _{0} )}{e B}$