Light of wavelength 4000Å is incident on barium. Photoelectrons emitted describe a circle of radius 50 cm by a magnetic field of flux density `5.26xx10^-6 tesla`. What is the work function of barium in eV? Given `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C, m_e=9.1xx10^(-31)kg.`

Answer

`2.5`

Explanation

Here ,`lambda = 4000 xx 10^(-10) m = 4 xx 10^(-7) m ` ,
r= 50 cm = 0.50 m , `B = 5.26 xx 10^(-6) T `
Force on electron in magnetic field is ,
F = Be v = `(mv^(2))/r or v = ("B e r")/m `
Work function of barium is given by
`phi_(0) =(hc)/lambda -1/2 mv^(2)`
`rArr (hc)/lambda -1/2 m xx ((Ber")/m)^(2) = (hc)/lambda -1/2 (B^(2)e^(2)r^(2))/m`
`rArr ((6.6 xx 10^(-34))xx(3xx10^(8)))/((4xx10^(-7))xx(1.6 xx 10^(-19)))`
`-1/2 xx((5.26 xx10^(-6))^(2)xx(1.6 xx10^(-19))^(2)xx(0.50)^(2))/((9.1xx10^(-31))xx(1.6xx10^(-19)))`
`approx 2.5 eV `