The radiation corresponding to `3to2` 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. 1.6 eV
B. 1.8 eV
C. 1.1 eV
D. 0. 8 eV

Answer

C

Explanation

`r=(mv)/(eB) rArr (r^(2)e^(2)B^(2))/(2)=(m^(2)v^(2))/(2)`
`(r^(2)e^(2)B^(2))/(2m)=(mv^(2))/(2)`
Now `hv=phi+1//2 mv^(2)`
`1.89=phi+1//2 mv^(2)`
`:.1.89-phi=(r^(2)e^(2)B^(2))/(2m) (1)/(e) eV=(r^(2)e^(2))/(2m) eV`
`=(100xx10^(-6)xx1.6xx10^(-19)xx9xx10^(-8))/(2xx9.1xx10^(-31))`
Work function `phi=1.89-(1.6xx9)/(2xx9.1)`
`=1.89-0.79~=eV`