Question

The magnetic field associated with a light wave is given, at the origin, by
B = B₀ [sin(3.14 × 10⁷)ct + sin(6.28 × 10⁷)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?
(A) 6.82 eV (B) 12.5 eV
(C) 8.52 eV (D) 7.72 eV

Answer 1

answer : option (D) 7.72 eV

we have to find incident energy of photon to get maximum kinetic energy of the photo electrons.

here B = B₀ [sin(3.14 × 10⁷)ct + sin(6.28 × 10⁷)ct]

for B₀ [sin(3.14 × 10⁷c)t ], f₁ = 3.14 × 10⁷C/2π = 1/2× 10⁷C [ we know, frequency = angular frequency/2π ]

for B₀ [sin(6.28 × 10⁷c)t, f₂ = 6.28 × 10⁷C/2π = 10⁷C

here we see maximum kinetic energy will be when we take f₂.

so, incident energy, E = hf₂

= 6.63 × 10^-34 × 10⁷C

= 6.63 × 10^-34 × 10⁷ × 3 × 10^8

= 19.89 × 10^-19 J

we know, 1J = 1.6 × 10^19 eV

so, E = 19.89 × 10^-19/1.6 × 10^-19 = 12.4 eV

now maximum kinetic energy = E - Φ

= 12.4 - 4.7

= 7.7 eV ≈ 7.72 eV