Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?
Solution
Data: λ = 4500 Å = 4.5 × 10-7 m,
Φ = 2.0 eV = 2 × 1.6 × 10-19 J,
h = 6.63 × 10-34 J.s, c = 3 × 108 m/s,
r = 20 cm = 0.2 m, e = 1.6 × 10-19 C, m = 9.1 × 10-31 kg
`1/2 "mv"_"max"^2 ("KE"_"max") = "hc"/lambda -* phi`
`= ((6.63 xx 10^-34)(3 xx 10^8))/(4.5 xx 10^-7) - 3.2 xx 10^-19`
= 4.42 × 10-19 - 3.2 × 10-19
= 1.22 × 10-19 J
∴ `"v"_"max" = sqrt(2/"m" "KE"_"max") = sqrt((2 xx 1.22 xx 10^-19)/(9.1 xx 10^-31))`
`= sqrt(0.2681 xx 10^12) = 5.178 xx 10^5` m/s
Now, centripetal force, `("mv"_"max"^2)/"r"` = magnetic force, Bevmax
∴ B = `("mv"_"max")/"re" = ((9.1 xx 10^-31)(5.178 xx 10^5))/((0.2)(1.6 xx 10^-19))`
= 1.472 × 10-5 T
