Question

An alpha particle is accelerated through a potential difference of 100 V. Calculate:

1. The speed acquired by the alpha particle, and
2. The de-Broglie wavelength is associated with it.

(Take mass of alpha particle = 6.4 × 10⁻²⁷ kg)

Answer 1

(i) `1/2 "mv"^2` = qV

or, `1/2"mv"^2` = 2e × 100

or, mv² = 400 eV

or, v = `sqrt((400  "eV")/"m")`

or, v = `sqrt((400 xx 1.6 xx 10^-19)/(6.4 xx 10^-27))`

∴ v = 10⁵ m/s

(ii) de-Broglie wavelength = λ = `"h"/sqrt(2"mqV")`

or, λ = `(6.6 xx 10^-34)/(sqrt(2 xx 6.4 xx 10^-27 xx 2 xx 1.6 xx 10^-19 xx 100))`

∴ λ = 1.03 × 10⁻¹² m