In a photocell bichromatic light of wavelength `2475 Å` and `6000 Å` are incident on cathode whose work function is `4.8 eV`. If a uniform magnetic field of `3 xx 10^(-5) tesla` exists parallel to the plate, the radius of the path described by the photoelectron will be (mass of electron ` = 9 xx 10^(-31) kg`)

Options

A. 1 cm
B. 5 cm
C. 10 cm
D. 25 cm

Answer

B

Explanation

`E_1=(12375)/(lamda_1(Å))(eV)=(12375)/(2475)=5eV`
`E_2=(12375)/(6000)eV = 2.06eV`
Since, `E_2 lt phi_0 and E_1 gt phi_0`
Photoelectric emission is possible only for `lamda_1`
`K_("max") = E-phi_0 =5 -4.8 =0.2eV`
Radius `r=(sqrt(2mk_(max)))/(qB)=(sqrt(2xx9xx10^(-31)xx0.2 xx1.6 xx10^(-19)))/(1.6 xx10^(-19)xx3 xx10^(-5))`
`= 0.05m = 5 cm.`