Question

The energy of an electron having de-Broglie wavelength `λ` is ______.

(h = Plank's constant, m = mass of electron)

पर्याय

• `"h"/(2"m"lambda)`

• `"h"^2/(2"m"lambda^2)`

• `"h"^2/(2"m"^2lambda^2)`

• `"h"^2/(2"m"^2lambda)`

Answer 1

The energy of an electron having de-Broglie wavelength `λ` is `underlinebb("h"^2/(2"m"lambda^2))`.

(h = Plank's constant, m = mass of electron)

Explanation:

de-Broglie wavelength

λ = `"h"/"p" "h"/sqrt(2"m K.E")`

or, λ² = `"h"^2/(2"m"("K"."E"))` 

∴ K. E. = `"h"^2/(2"m"lambda^2)`