JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 5)

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

E = 200 [sin (6 $$\times$$ 10¹⁵)t + sin (9 $$\times$$ 10¹⁵)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

1.90 eV

3.27 eV

3.60 eV

3.42 eV

Frequency of EM waves = $${6 \over {2\pi }} \times {10^{15}}$$ and $${9 \over {2\pi }} \times {10^{15}}$$

Energy of one photon of these waves

$$ = \left( {4.14 \times {{10}^{ - 15}} \times {6 \over {2\pi }} \times {{10}^{15}}} \right)$$ eV

and $$\left( {4.14 \times {{10}^{ - 15}} \times {9 \over {2\pi }} \times {{10}^{15}}} \right)$$ eV

= 3.95 eV and 5.93 eV

$$\Rightarrow$$ Energy of maximum energetic electrons

= 5.93 $$-$$ 2.50 = 3.43 eV