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THE ELECTRIC FIELD E AT A POINT ASSOCIATED WITH A LIGHT WAVE IS E=100[SIN(3.14T INTO TEN POWER 15 )+SIN(6.28 TINTO TEN POWER 15 )]V/M .IF THIS LIGHT FALLS ON A METL OF WORK FUNCTION OF 2ev MAX. K.E. OF EMITTED ELECTRONS IS?

THE ELECTRIC FIELD E AT A POINT ASSOCIATED WITH A LIGHT WAVE IS


E=100[SIN(3.14T INTO TEN POWER 15 )+SIN(6.28 TINTO TEN POWER 15 )]V/M .IF THIS LIGHT FALLS ON A METL OF WORK FUNCTION OF 2ev MAX. K.E. OF EMITTED ELECTRONS IS?

Grade:12

1 Answers

AskiitianExpert Shine
10 Points
12 years ago

Hi

The electric field stores energy. The energy density of the electric field is given by

 u = \frac{1}{2} \varepsilon |\mathbf{E}|^2

where

 \varepsilon is the permittivity of the medium in which the field exists
\mathbf{E} is the electric field vector.

The total energy stored in the electric field in a given volume V is therefore

 \int_{V} \frac{1}{2} \varepsilon |\mathbf{E}|^2 \, \mathrm{d}V

where

dV is the differential volume element

Energy associated with an electric field is 1/2ε0E2 . find the energy assosiated. 

Then , apply the formula for photoelectric effect, i.e energy given = work fn + max K.E . Convert the respective units into ev.

Photoelectric effect

In effect quantitatively using Einstein's method, the following equivalent equations are used (valid for visible and ultraviolet radiation):

Energy of photon = Energy needed to remove an electron + Kinetic energy of the emitted electron

Algebraically:

hf = \phi + E_{k_\mathrm{max}}

where

  • h is Planck's constant,
  • f is the frequency of the incident photon,
  • φ = hf0 is the work function (sometimes denoted W instead), the minimum energy required to remove a delocalised electron from the surface of any given metal,
  • E_{k_\mathrm{max}} = \frac{1}{2} m v_m^2 is the maximum kinetic energy of ejected electrons,
  • f0 is the threshold frequency for the photoelectric effect to occur,
  • m is the rest mass of the ejected electron, and
  • vm is the speed of the ejected electron.

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