Question

A deuteron and an alpha particle are accelerated with the same potential. Which one of the two has

1. greater value of de Broglie wavelength associated with it and
2. less kinetic energy?

Explain.

Answer 1

i. Using the de-Broglie wavelength formula, the deuteron and alpha particle are accelerated with the same potential. So, both their velocities are the same.

λ = `"h"/sqrt(2"mV"_0"q")`, `λ ∝ 1/sqrt("mq")`

So, `λ_"due" ∝ 1/sqrt(2"m"_"d""q"_"d")` and `λ_α ∝ 1/sqrt(8"m"_α"q"_α)`

`λ_α/λ_"d" = sqrt(2"m"_"d""q"_"d")/sqrt(8"m"_α"q"_α) = sqrt(2/8) = sqrt(1/4) = 1/2`

λ_d = 2λ_α

ii. For the same potential of acceleration, KE is directly proportional to the ‘q’

Charge of duetron is +e

Charge of alpha is +2e

So, K_d = `"K"_α/2`
Charge of an alpha particle is more than the deuteron.