The radiation corresponding to ` 3 rarr 2` transition of hydrogen atom falls on a metal surface to produce photoelectrons . These electrons are made to enter circuit a magnetic field `3 xx 10^(-4) T` if the ratio of the largest circular path follow by these electron is `10.0 mm , the work function of the metal is close to

Options

A. `0.8eV`
B. `2.14eV`
C. `1.8eV`
D. `1.1eV`

Answer

D

Explanation

`1/2mv^(2)=eV or mv=sqrt(2meV)`
Radius of circular path of electron
`r=(mv)/(qB)=(sqrt(2meV))/(eB) =1/B sqrt((2mV)/e)`
or `V=(B^(2)er^(2))/(2m)`
`=((3xx10^(-4))^(2)xx(1.6xx10^(-19))xx(10xx10^(-3))^(2))/(2xx(9xx10^(-31)))`
`=0.8V`
KE of electron, K=eV=0.8eV
for transition between 3 to 2, we have
`E=13.6(1/(2^(2))-1/(3^(2)))=(13.6xx5)/36=1.88eV`
Work fuction, `phi_(0)=E-K=1.88-0.8`
`=1.08eV~~1.1eV`