CBSE – AIPMT – 2011 de Broglie wavelength associated with electron

Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 km then the de-Broglie wavelength associated with the electrons would

(a) Increase by 4 times

(b) Increase by 2 times

(c) Decrease by 2 times

(d) Decrease by 4 times

Solution

Given –

V₁ = 25 kV

V₂ = 100 kV

de Broglie relations show that the wavelength is inversely proportional to the momentum of a particle. 

i.e.

λ = (h/P)

- - - - - - - - - - (1)

where

h is a Plank’s constant 

p is a momentum of particle

p = mass X velocity

p = m X v

Kinetic energy (say K ) of any particle in motion can be given as

K = (1/2) m v² 

On multiplying both side by ‘m’ we get,

mK = (1/2) m² v² 

2 mK= m² v² 

2 mK= P²

                    P = √ (2 mK)

- - - - - - - - - - (2)

If an electron (of charge ‘e’) is accelerated by potential difference ‘V’ volt, it aquires the kinetic energy

K = eV

Putting this value in equation (2) we get,

P = √ (2 meV)

- - - - - - - - - - (3)

From equation (1) and (3)

de Broglie wavelength can be given as

λ = [h/√ (2 meV)]

As,

‘h’ is a Planks constant,

‘m’ is mass of electron i.e. constant

And 

‘e’ is charge on electron i.e. constant.

Therefore we can say that

λ α [1/√(V)]

i.e.

λ √(V) = constant

i.e.

λ ₁√(V₁) = λ ₂√(V₂)

λ ₁ = √ (V₂/ V₁ ) λ ₂

λ ₁ = √ (100 /25 ) λ ₂

λ ₁ = √ (4 ) λ ₂

λ ₁ = 2 λ ₂

If the voltage is increased from 25 kV to 100 kV, the de Broglie wavelength associated with electron is decrease by two times.