Question

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

E = 200[sin(6 × 10¹⁵)t + sin(9 × 10¹⁵)t]Vm^-1

Given: h = 4.14 × 10^-15 eVs.

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

विकल्प

• 1.90 eV

• 3.27 eV

• 3.60 eV

• 3.42 eV

Answer 1

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

E = 200[sin(6 × 10¹⁵)t + sin(9 × 10¹⁵)t]Vm^-1

Given: h = 4.14 × 10^-15 eVs.

If this light falls on a metal surface having a work function of 2.50eV, the maximum kinetic energy of the photoelectrons will be 3.42 eV.

Explanation:

We have, according to the given equation,

`n_1 = 6/(2pi) xx 10^15` and `n_2 = 9/(2pi) xx 10^15`

Hence, the photon's energy in these waves, E = hν

`E_1 = 4.14 xx 10^-15 xx 6/(2pi) xx 10^15 eV = 3.95  eV`

`E_2 = 4.14 xx 10^-15 xx 9/(2pi) xx 10^15 eV = 5.93  eV`

Hence, the energy of the maximum energetic electron,

E = E₂ - Φ (Φ = work function)

= 5.93 – 2.50

= 3.43 eV ≈ 3.42 eV