# How to Find the De-Broglie Wavelength Emitted from A Surface?

By Team Aakash Byju's | 29th January 2023

λ =

## D

An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Broglie wavelength λ', then

-1

2

mc

λ'

h

λ =

2

3mc

λ'

2h

λ =

2

2mc

λ'

h

λ =

2

5mc

λ'

h

Detailed Explanation

### The correct answer is  option C .

λ =

2

2mc

λ'

h

According to Einstein’s photoelectric equation, the maximum kinetic energy is given as

KE        =hv-O

max

Where,

h - Planck’s constant

v - frequency of the incident wave

O   - work function of the surface

o

o

Since it is given that the work function of the surface is negligible and the kinetic energy is also written as

Where,

KE =      mv

1

2

2

m and v are the mass & velocity of electrons

Also, the expression of the frequency is

v =

Where,

c is the speed of light and

c

λ

λ is the wavelength

We also know that momentum can be expressed as p = mv.

KE        =hv-O

max

o

=>

mv

1

2

2

= h

c

λ

=>

1

2m

(mv)

2

= h

c

λ

=>

1

2m

p

2

= h

c

λ

Further, the momentum can also be expressed in terms of the de Broglie wave of wavelength λ′.

p =

h

λ'

.  .

1

2m

p

2

= h

c

λ

=>

1

2m

h

c

λ =

(     )

h

λ'

2

= h

c

λ

2mλ'

=>

h

2

=

c

λ

=>

2mλ'

2

.