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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?


7 years ago

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										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.


7 years ago

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