Photoelectric Effect Dual Nature of Radiation and Matter Class 12th Physics

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

E = 200 $[s i n (6 * 1 0^{15}) t + s i n (9 * 1 0^{15}) t] V m^{- 1}$

**Given : h = 4.14 * 10^-¹⁵ eVs

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

Option 1 - <p>1.90 eV</p>

Option 2 - <p>3.27 eV</p>

Option 3 - <p>3.60 eV</p>

Option 4 - <p>3.42 eV</p>

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Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 4

Detailed Solution:

The light wave contains two lights of different frequencies, so

$E_{1} = h υ_{1} = \frac{4.14 * 1 0^{- 15} * 6 * 1 0^{15}}{2 π} = 3.96 e V, a n d$

$E_{2} = h υ_{2} = \frac{4.14 * 1 0^{- 15} * 9 * 1 0^{15}}{2 π} = 5.92 e V$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

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Physics Dual Nature of Radiation and Matter 2021

Physics Dual Nature of Radiation and Matter 2021

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