Question

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?

Answer 1

Data: λ = 4500Å = 4.5 × 10^-7 m, 

Φ = 2.0eV = 2 × 1.6 × 10^-19 J = 3.2 × 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

\(\cfrac{1}2\) mv²_max (KE_max) = \(\cfrac{hc}{\lambda}\) - \(\phi\)

Now, centripetal force, \(\cfrac{mv^2_{max}}r\) = magnetic force,

Bev_max

This is the value of the magnetic field.