An electron, a proton and an alpha particle have kinetic energies of 16E,4E and E respectively. What is the qualitative order of their de Broglie wavelengths ?
a)Ie>Ip=Ia
b)Ip=Ia>Ie
c)Ip>Ie>Ia
d)Ia<Ie>>Ip
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2 Answers
996λ is proportional to 1/√(Km)
For the electron m = 1/2000 amu approx and K =16E.
1/√(Km) = 1/√(16E x (1/2000)) = 11.2/√E approx
For the proton m = 1 amu approx and K =16E.
1/√(Km) = 1/√(16E x 1) = 0.25/√E
For the alpha particle m = 4 amu approx and K =4E
1/√(Km) = 1/√(4E x 4) = 0.25/√E
Raunaq Saha Thanks a lot :DUpvote·0· Reply ·2013-06-19 05:42:04
996λ = h / √2mK.E
i.e Ie>Ip=Ia
(A)should be the answer
- Raunaq Saha but why is Ip=Ia ?