Question

When radiations of wavelength 1 = 3000 Å and wavelength 2 = 6000 Å falls on a metal surface then ratio of velocity of emitted electron is v1/v2 = 3/1 obtain. Find out V1, V2, and work function of surface.
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Answer 1

Explanation:

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12th

Physics

Dual Nature of Radiation and Matter

Photons and Photoelectric Effect

The work function of a meta...

PHYSICS

The work function of a metal is 2 eV. If a radiation of wavelength 3000

A

˚

is incident on it, the maximum kinetic energy of the emitted photoelectrons is :

(Planck's constant h=6.6×10

−34

Js; Velocity of light c=3×10

8

m/s;1eV=1.6×10

−19

J)

Answer 2

Answer:

Energy of incident radiation is

$E=λ | hc=3×10−76.6×10−34×3×108 | =6.6×10 \: −19 J.</p><p></p><p>$

Also, work function is ϕ=2×1.6×10 −19

J=3.2×10 −19 J.

Hence, maximum kinetic energy of emitted photo electrons will be

$6.6×10−19J−3.2×10−19J=3.4×10−19J.$