The magnetic field associated with a light wave is given, at the origin, by $$\mathrm{B}=\mathrm{B}_{0}\left[\sin \left(3.14 \times 10^{7}\right) \mathrm{ct}+\sin \left(6.28 \times 10^{7}\right) \mathrm{ct}\right]$$ If this light falls on a silver plate having a work function of 4.7 eV , what will be the maximum kinetic energy of the photo electrons?

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Modern Physics

The magnetic field associated with a light wave is given, at the origin, by

B = B_{0} [sin (3.14 \times 1 0^{7}) ct + sin (6.28 \times 1 0^{7}) ct]

If this light falls on a silver plate having a work function of 4.7 eV , what will be the maximum kinetic energy of the photo electrons?

Answer: (D)

The field is a sum of two waves, and the photoelectrons respond to whichever photon is most energetic -- here the higher angular frequency

ω = 6.28 \times 1 0^{7} c = 2 π \times 1 0^{7} c

. Its frequency is

f = \frac{ω}{2 π} = 1 0^{7} c = 1 0^{7} \times 3 \times 1 0^{8} = 3 \times 1 0^{15} Hz

, so the photon energy is

E = h f = \frac{( 6.6 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{15} )}{1.6 \times 1 0 ^{- 19}} \approx 12.4 eV

. Einstein's equation then gives the maximum kinetic energy

K_{max} = E - ϕ = 12.4 - 4.7 \approx 7.72 eV

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option