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Question
A deuteron and an alpha particle are accelerated with the same potential. Which one of the two has
- greater value of de Broglie wavelength associated with it and
- less kinetic energy?
Explain.
Solution
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`
λd = 2λα
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, Kd = `"K"_α/2`
Charge of an alpha particle is more than the deuteron.
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