Photoelectric Emission

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New answer posted

3 weeks ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

For no emission of electron

λ > λ0

 

New answer posted

4 weeks ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

According to Einstein's photoelectric equation maximum kinetic energy of photoelectrons, K E  max  = E v φ

or   2 = 5 φ φ = 3 e V

When E v = 6 e V  then, K E  max  = 6 3 = 3 e V

or e ( V c a t h o d e V a n o d e ) = 3 e V

or   V c a t h o d e V a n o d e = 3 V = V s t o p p i n g

  V s t o p p i n g = 3 V

New answer posted

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Intersity a number of photons kinetic Energy a f

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Incident energy = 2.20 e V

If ? < 2.20 e V  electron will emit.

? > 2.20 e V No electron emission

Only caesium will emit electron

New answer posted

a month ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

e V s = h v - ?

  At   V s = 0 h v = ? ? = 6.62 * 10 - 34 10 14 [ 5.5 ] ? = 6.62 * 10 - 34 10 14 [ 5.5 ] e V 1.6 * 10 - 19

= 2.27

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Relation between De-Broglie wavelength and K.E. is

λ = h 2 ( K E ) m e λ 1 K E

λ A λ B = K E B K E 1 2 = T A - 1.5 T A T A 2 e V K E B = 2 - 1.5 = 0.5 e V ? B = 4.5 - 0.5 = 4 e V

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Case I :- 2 ? ? = 1 2 m v 1 2 . . . . . . . . ( i )

Case II :- 1 0 ? ? = 1 2 m v 2 2 . . . . . . . . ( i i )

1 9 = v 1 2 v 2 2

v 1 : v 2 = 1 : 3

x = 1

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