(a) Plot a graph showing variation of DE Broglie wavelength (λ) associated with a charged particle of mass m, versus V, where V is the accelerating potential.
(b) An electron, a proton and an alpha particle have the same kinetic energy. Which one has the shortest wavelength?
Answers
(a) Graph between accelerating potential and de-Broglie wavelength:
According to de-Broglie, the momentum is given as:
p = h/λ
Where,
h = Planck’s constant
λ = Wavelength of the matter
Now,
mv = h/λ
On squaring both sides and multiplying by 1/2m on both sides, we get,
m²v² × 1/2m = h²/λ² × 1/2m
1/2 mv² = 1/2m × h²/λ² → (equation 1)
The kinetic energy is gained by the particle to attain energy gain which is given as:
KE = 1/2 mv² = qv → (equation 2)
Where,
v = accelerating potential
On equation equation (1) and (2), we get,
qv = 1/2m × h²/λ²
λ² = 1/2m × h²/qv
λ = h/√(2mqv) → (equation 3)
λ = h/√(2mq) 1/√v
∴ λ = c 1/√v
This equation gives a hyperbolic graph.
The graph is attached as image below.
(b) An alpha particle has the shortest wavelength.
From equation 3, we get,
λ = h/√(2mqv)
Now, eliminating the constants, we get,
λ ∝ 1/√m
The wavelength is inversely proportional to the square root of mass.
The electron has the lowest mass which is followed by proton which is followed by alpha particle.
∴ λelectron > λproton > λalpha-particle
