Question

Calculate the de Broglie wavelength associated with an electron with energy 1 Mega electron volt?

Answer 1

Given conditions ⇒

Energy of the Electron = 10 MeV
 = 10 × 10⁶ eV.
 = 10⁷ × 1.6 × 10⁻¹⁹ J.
 = 16 × 10⁻¹³ J.

Now, We know the Mass of the Electron = 9.1 × 10⁻³¹ kg. 

∵ E = 1/2 × mv²
where, E is the Energy, m is the mass of the electrons, and v is the velocity of the Electrons.

∴ 16 × 10⁻¹³ = 1/2 × 9.1 × 10⁻³¹ × v²
∴ v² = (32 × 10³¹ × 10⁻¹³)÷9.1
⇒ v² = 3.51 × 10¹⁸
⇒ v = √(3.51 × 10¹⁸)
⇒ v = 1.86 × 10⁹ m/s.

Momentum(p) = m × v
 = 9.1 × 10⁻³¹ × 1.86 × 10⁹
 = 16.926 × 10⁻²²

Now, Using the Formula,
 λ = h/p
where h is the plank's constant = 6.63 × 10⁻³⁴ J-s.

⇒ λ = (6.63 × 10⁻³⁴) ÷ (16.926 × 10⁻²²)
⇒ λ = 0.39 × 10⁻¹² m.
∴ λ = 0.39 pm. [∵ 1 pm = 10⁻¹².]


Hence, the de-broglie wavelength of electron is 0.39 pm.