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Question
An alpha particle is accelerated through a potential difference of 100 V. Calculate:
- The speed acquired by the alpha particle, and
- The de-Broglie wavelength is associated with it.
(Take mass of alpha particle = 6.4 × 10−27 kg)
Solution
(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
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