# 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, - Physics

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

Concept: The Photoelectric Effect
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 14 Dual Nature of Radiation and Matter
Exercises | Q 11 | Page 323