Question

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1800 K. (kb is Boltzmann's constant, 1.38x10^-23 J/K). (1) Wavelength of the electron = ____ m (2) Wavelength of the proton = ______ m

Answer 1

Given:

Thermal Kinetic Energy of an electron,
`KE_(t) = (3)/(2)k_(b)T`

`k_(b) = 1.38* 10^(- 23) J/k` = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron,
`\lambda_(e)`:

`\lambda_(e) = (h)/(p_(e))`

(1)

where

h = Planck's constant =
`6.626* 10^(- 34)m^(2)kg/s`

`p_(e)` = momentum of an electron

`v_(e)` = velocity of an electron

`m_(e) = 9.1* 10_(- 31) kg` = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

`(1)/(2)m_(e)v_(e)^(2) = (3)/(2)k_(b)T`

`}v_(e) = \sqrt{2((3)/(2)k_(b)T)/(m_(e))}`

`}v_(e) = \sqrt{(3* 1.38* 10^(- 23)* 1800)/(9.1* 10_(- 31))}`

`v_(e) = 2.86* 10^(5) m/s` (2)

Using eqn (2) in (1):

`\lambda_(e) = (6.626* 10^(- 34))/(9.1* 10_(- 31)* 2.86* 10^(5)) = 2.55 nm`

Now, to calculate the de-Broglie wavelength of proton,
`\lambda_(e)`:

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

(3)

where

`m_(p) = 1.6726* 10_(- 27) kg` = mass of proton

`v_(p)` = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

`(1)/(2)m_(p)v_(p)^(2) = (3)/(2)k_(b)T`

`}v_(p) = \sqrt{2((3)/(2)k_(b)T)/(m_(p))}`

`}v_(p) = \sqrt{(3* 1.38* 10^(- 23)* 1800)/(1.6726* 10_(- 27))}`

`v_(p) = 6.674* 10^(3) m/s` (4)

Using eqn (4) in (3):

`\lambda_(p) = (6.626* 10^(- 34))/(1.6726* 10_(- 27)* 6.674* 10^(3)) = 5.94* 10^(- 11) m = 0.0594 nm`