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icy_2291

The electric field associated with a monochromatic beam becomes zero 1.2x10¹⁵ times per second. find the maximum KE of the photoelectrons when this light falls on a metal surface whose work function is 2eV.

rajat

see frequency is [ 1.2 x 10^5 ] / 2

therefore max. KE = hf - 2eV

where f = frequency

icy_2291

pleaser elaborate in detail how you got 1.2x10^5 /2

icy_2291

anyone please

edison

Q) The electric field associated with a monochromatic beam becomes zero 1.2x10¹⁵ times per second. find the maximum KE of the photoelectrons when this light falls on a metal surface whose work function is 2eV.

Solution)

KE_max = eV₀ = hf -

=

-

.

Where, KE_max = Maximum kinetic energy

f = frequency of light used for photoelectric effect to take place

= Work function

V₀ = stopping potential.

thus, eV₀ = hf -

Since the electric field associated with a monochromatic beam becomes zero 1.2x10¹⁵ times per second

This means that frequency of the beam is half of 1.2x10¹⁵

Thus f = 0.6 x 10¹⁵ (as in one cycle the electric field

becomes zero twice)

Now use equation KE = eV₀ = hf -

or KE = hf -

= 6.62 x 10^-34 x 0.6 x 10¹⁵ - 2

solve this to get maximum KE.

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