The radiation corresponding to `3 to 2` transition of hydrogen atom falls on a metal surface to produce photoelectrons.These electrons are made to enter a magnetic field of `3xx10^(-4)`T.If the radius of the largest circular path followed by these electrons is 10.0 mm the work function of the metal is close to:

Options

A. 0.8 eV
B. 1.6 eV
C. 1.8 eV
D. 1.1 eV

Answer

D

Explanation

Radius of circular path of electron entered in magnetic field is ,
`r=(mv)/(qbeta)=(sqrt(2meV))/(qbeta)[because mv=sqrt(2meV)]` and q=e
`(1)/(beta)sqrt((2m)/(e )V)`
`therefore r^(2)=(1)/(beta^(2)).(2mV)/(e )`
`therefore V=(beta^(2)r^(2)e)/(2m)=((3xx10^(-4))^(2)xx(10xx10^(-3))^(2)xx1.6x10^(-19))/(2xx9.1xx10^(-31))`
`therefore v~~0.8 V therefore hf~~0.8 eV`
Now, transition from n=3 to n=2,
`E=13.6[(1)/(4)-(1)/(9)]eV=13.6 [(5)/(36)]=1.88 eV`
`therefore` work function =E-hf
`=1.88-0.=1.08 eV~~1.1 eV`