Question:
A proton and an α-particle are accelerated, using the same potential difference. How are the de Broglie wavelengths λp and λa related to each other?
Solution:
The proton and α-particle are accelerated at the same potential difference.
λ = h/√2mqv
λ is proportional to 1/√mq
λp/ λa = √maqa/mpqp = √8
Therefore, the wavelength of the proton is √8 times the wavelength of α-particle.
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