Light of wavelength `2475 Å` is incident on barium. Photoelectrons emitted describe a circle of radius `100 cm` by a magnetic field of flux density `(1)/(sqrt(17)) xx 10^(-5) Tesla`.
Work function of the barium is (Given `(e )/(m) = 1.7 xx 10^(11)`)

Options

A. `1.8 eV`
B. `2.1 eV`
C. `4.5 eV`
D. `3.3 eV`

Answer

C

Explanation

Radius of circular path described by a charged particle in a magnetic field is given by `r = (sqrt(2 mK))/(qB)`,
where `K` is Kinetic energy of electron
`rArr K = (q^(2) B^(2) r^(2))/( 2m) = ((e )/(m)) (eB^(2) r^(2))/(2)`
`= (1)/(2) xx 1.7 xx 10^(11) xx 1.6 xx 10^(-19) xx ((1)/(sqrt(17)) xx 10^(-5))^(2) xx (1)^(2)`
` = 8 xx 10^(-20) J = 0.5 eV`
By using `E = W_(0) + K_(max)`
`rArr W_(0) = E - K_(max) ((12375)/(2475))V - 0.5 V = 4.5 V`