The electric field at a point associated with a light wave is given by
$E = 200 [\sin (6 \times 10^{15}) t + \sin (9 \times 10^{15}) t] {Vm}^{- 1}$
Given: $h = 4.14 \times 10^{- 15} eVs$
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

Question

The electric field at a point associated with a light wave is given by
$E = 200 [\sin (6 \times 10^{15}) t + \sin (9 \times 10^{15}) t] {Vm}^{- 1}$

Given: $h = 4.14 \times 10^{- 15} eVs$

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

Aicharya · Accepted Answer

Given: E = 200 sin 6 &#215; 10 15 t + sin 9 &#215; 10 15 t &#160; V &#160; m - 1 Here, angular velocity, &#969; 1 = 6 &#215; 10 15 &#160; and &#969; 2 = 9 &#215; 10 15 Maximum kinetic energy of the photoelectron is KE max . = E - &#981; = h f - &#981; = h &#969; 2 &#960; - &#981; = 4 . 14 &#215; 10 - 15 &#215; 9 &#215; 10 15 2 &#215; 3 . 14 - 2 . 5 = 5 . 92 - 2 . 50 = 3 . 42 &#160; eV

The electric field at a point associated with a light wave is given by E = 200 sin 6 × 10 15 t + sin 9…

Solution

The electric field at a point associated with a light wave is given by E = 200 sin 6 &#215; 10 15 t + sin 9…

Solution

The electric field at a point associated with a light wave is given by E = 200 sin 6 × 10 15 t + sin 9…