Question

A particle with mass m and charge e is accelerated through a potential difference (V). What is the wavelength of the particle?

Answer 1

Wavelength of the particle is
`(h)/(√(2meV))`

Solution:

As per the question:

The particle with mass, m and charge, e accelerates through V (potential difference).

The momentum of the particle,
`p_(p)` if it travels with velocity,
`v_(p)`:

`p_(p) = mv_(p)`

Now, squaring both sides and dividing by 2.

`(1)/(2)p_(p)^(2) = (1)/(2)m^(2)v_(p)^(2)`

`K.E =(1)/(2m)p_(p)^(2) = (1)/(2)mv_(p)^(2)`

`K.E = (1)/(2)mv_(p)^(2)`

Also,

`p_(p) = √(2mK.E)` (1)

Now, we know that Kinetic energy of particle accelerated through V:

K.E = eV (2)

where

e = electronic charge =
`1.6* 10^(- 19) C`

From eqn (1) and (2):

`p_(p) = √(2mK.E)` (3)

From eqn (2) and (3):

`p_(p) = √(2meV)`

From the de-Broglie relation:

`\lambda_(p) = (h)/(p_(p))` (4)

where

`\lambda_(p)` = wavelength of particle

h = Planck's constant

From eqn (3) and (4):

`\lambda_(p) = (h)/(√(2meV))`

Answer 2

The wavelength of the particle is
`\lambda=(h)/(√(2V\cdot e\cdot m))`

Step-by-step explanation:

We know that after accelerating across a potential 'V' the potential energy of the charge is converted into kinetic energy as

`(1)/(2)mv^(2)=V\cdot e\\\\\therefore v=\sqrt{(2V\cdot e)/(m)}`

now according to De-Broglie theory the wavelength associated with a particle of mass 'm' moving with a speed of 'v' is given by

`\lambda =(h)/(mv)`

where

'h' is planck's constant

'm' is the mass of particle

'v' is the velocity of the particle

Applying the values in the above equation we get

`\lambda =\frac{h}{m\cdot \sqrt{(2V\cdot e)/(m)}}`

Thus

`\lambda=(h)/(√(2V\cdot e\cdot m))`