Question

what is the kinetic energy of an electron with de broglie wavelength 0.3 NM?

Answer 1

Relation between momentum and Kinetic energy is
P = $\bold{\sqrt{2mK.E}}$, here m is mass of object, P is linear momentum of object and K.E kinetic energy of that object.

Now according to De - broglie's theory,
λ = h/P
Put above equation in de - broglie's wavelength formula
Then, λ = h/√{2mK.E}
⇒λ² = h²/2mK.E
⇒K.E = h²/2mλ²
Now, h = 6.6 × 10⁻³⁴ Js , m = 9.1 × 10⁻³¹ Kg and λ = 0.3 nm = 3 × 10⁻¹⁰ m
∴ K.E = (6.6 × 10⁻³⁴)²/2 × 9.1 × 10⁻³¹ × (3 × 10⁻¹⁰)²
= 6.6 × 6.6 × 10⁻⁶⁸ × 10³¹ × 10²⁰/9 × 9.1 × 2
= 6.6 × 6.6 × 10⁻¹⁷/18 × 9.1
= 2.659 × 10⁻¹⁸ J