Physics, asked by bhargav713, 5 days ago

Calculate the stopping potential for the photo- elections emitted by gold cathode if the wavelength of incident radiation is 3x10^-7 m, Given that the work function of gold is 4eV.​

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Answered by bishnupriyadas1989
0

Explanation:

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>>Dual Nature of Radiation and Matter

>>Photons and Photoelectric Effect

>>The photoelectric work function for a me

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The photoelectric work function for a metal is 4.2eV. If the stopping potential is 3V, find the threshold wavelength and maximum kinetic energy of emitted electrons.

(Velocity of light in air =3×10

8

m/s, Planck's constant =6.63×10

−34

J−s, Charge on electron =1.6×10

−19

C)

Medium

Solution

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Given : e=1.6×10

−19

C, c=3×10

8

m/s

ϕ=4.2eV, h=6.63×10

−34

J.s, V

S

=3V

ϕ=4.2×1.6×10

−19

J

=6.72×10

−19

J

(a) Threshold wavelength =λ

0

=

ϕ

hc

=

6.72×10

−19

6.63×10

−34

×3×10

8

=2.960×10

−7

m

=2960

A

˚

(b) Maximum kinetic energy,

KE

max

=eV

S

=1.6×10

−19

×3

=4.8×10

−19

J

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