Physics

With reference to the equation of de Brogile wavelength:

λ = h/mv

where

h = Planck's constant

m = mass of the particle

v = velocity of the particle

And also, the k.e. of a particle is given by (3/2)kT where k and T are Boltzmann's constant and absolute temperature respectively.

Therefore, we have:

(1/2)mv2 = (3/2)kT

v2 = 3kT/m

v = √(3kT/m)

λ = h/mv

= (h/m)√(m/3kT)

= h√(1/3mkT)

which is the de Brogile wavelength in related to particle's mass and velocity.

So, for (a), taking T = 298K for room temperature, mass of an H molecule = 3.32 × 10-27 kg

Hence:

λ = h√(1/3mkT)

= 1.035 × 10-10 m

For (b), mass of an O atom = 1.33 × 10-26 kg

Hence:

λ = h√(1/3mkT)

= 5.175 × 10-11 m

Then, for Q2, the k.e. of an electron accelerated over a potential difference of V is given by eV, where e is the electronic charge of an electron.

Therefore, using the relation:

(1/2)mv2 = eV where m is the mass of an electron

v = √(2eV/m)


λ = h/mv

= (h/m)√(m/2eV)

= [h√(1/2me)] / √V

With reference to the following values:

m = 9.11 × 10-31 kg

e = 1.6 × 10-19 C

h = 6.626 × 10-34 Js

We have the de Brogile wavelength of the accelerated electron given by:

λ = 1.228 × 10-9 / √V

= 1.228 / √V nm