The electric field at the point associated with a light wave is given by $\mathrm{E}=200\left[\sin\left(6 \times 10^{15}\right) \mathrm{t}+\sin\left(9 \times 10^{15}\right) \mathrm{t}\right] \mathrm{Vm}^{-1}$ Given: $\mathrm{h}=4.14 \times 10^{-15} \mathrm{eVs}$ If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be :

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Modern Physics

The electric field at the point associated with a light wave is given by

E = 200 [sin (6 \times 1 0^{15}) t + sin (9 \times 1 0^{15}) t] Vm^{- 1}

Given:

h = 4.14 \times 1 0^{- 15} eVs

If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be :

Answer: (D)

The electric field contains two frequency components. For maximum kinetic energy, we use the higher angular frequency

ω = 9 \times 1 0^{15}

rad/s.

Frequency:

f = \frac{ω}{2 π} = \frac{9 \times 1 0 ^{15}}{2 π}

Hz

Using Einstein's photoelectric equation:

K_{ma x} = h f - ϕ = \frac{9 \times 1 0 ^{15} \times 4.14 \times 1 0 ^{- 15}}{2 π} - 2.50

K_{ma x} = \frac{37.26}{2 π} - 2.50 \approx 5.93 - 2.50 = 3.43

eV

The nearest option is (D) 3.42 eV.

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option