de Broglie Wavelength Equation

02/05/2021 Vinod 0 Comments

de Broglie Wavelength Equation :-

In 1924 Louis de Broglie proposed that just as light has both wave like and particle like properties, electrons also have wave like particles.

de Broglie proposed that the wavelength λ is associated with a particle of momentum p is given as

λ = h/p

Or

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The equation is known as the de Broglie wavelength equation and wavelength λ of the matter wave is called de Broglie wavelength.

  • For a photon, it moves as speed of light c Thus momentum of photon may be

p = mc = hν/c

since ν = c/λ

Therefore

p = hc/cλ = h/λ

λ = h/p

  • Consider an electron (mass m, charge e) accelerated from rest through a potential V, the kinetic energy K of the electron equates the work done (eV) on it by the electric field –

K = eV

Now K = 1/2 mv² = p²/2m

p² = 2mK

p = √2meV

since λ = h/p

hence

λ = h/√2meV = h/√2mK

Substituting the numerical values of h, m and e we get

λ = 1·227/√V nm

Where V is the magnitude of accelerating potential in volts.

Numerical :-

  1. What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4 X 10⁶ m/s and (b) a ball of mass 150 g travelling at 30 m/s ?

Solution.

(a) m = 9·1 x 10⁻³¹ Kg

v = 5·4 x 10⁶ m/s

λ = ?

p = mv

From de Broglie wavelength equation

λ = h/p = h/mv

= (6·63 x 10⁻³⁴ Js)/(9·1 x 10⁻³¹ Kg x 5·4 x 10⁶ ms⁻¹)

= (6·63 x 10⁻³⁴ Js)/(4·92 x 10⁻²⁴ Kgs⁻¹)

= 0·135 nm

(b) m = 150 g = 0·150 Kg

v = 30 m/s

λ = h/mv

= (6·63 x 10⁻³⁴ Js)/(9·1 x 10⁻³¹ Kg x 30 m/s)

λ = 6·63 x 10⁻³⁴ Js/4·50 Kgs⁻¹

= 1·45 x 10⁻³⁴ m

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