✔ 最佳答案
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
