An electron is accelerated through a potential difference of V=100 V. Find its de Broglie wavelength. Take h=6.63×10−34 J⋅s, me=9.1×10−31 kg, e=1.6×10−19 C.
Solution — Step by Step
eV=21mev2⟹v=me2eV
This works because we assume the electron starts from rest and the speeds are non-relativistic.
p=mev=2me⋅eV
λ=ph=2meeVh
Plugging in:
λ=2×9.1×10−31×1.6×10−19×1006.63×10−34
2×9.1×1.6×100×10−25=2912×10−25≈53.96×10−25
λ≈5.396×10−246.63×10−34≈1.23×10−10 m=1.23A˚
Final answer: λ≈1.23A˚.
Why This Works
de Broglie’s hypothesis: every particle has wave-like behaviour with λ=h/p. For an electron accelerated through V volts, there’s a beautiful shortcut formula:
λ=V12.27A˚(V in volts)
For V=100, λ=12.27/10=1.227A˚. Memorise this — it appears in NEET almost every other year.
Alternative Method
Use the shortcut directly:
λ=10012.27=1.227A˚
Same answer, no plugging-in arithmetic.
Common slip: students use λ=h/(mv) but forget that v here means the magnitude of velocity, not its component. Also, this formula is non-relativistic — for V>50 kV, you need relativistic momentum p=2mE+(E/c)2 where E=eV.
Common Mistake
Using kinetic energy directly as E=eV but then writing λ=h/E — that’s wrong. The de Broglie relation uses momentum, not energy. The correct chain is E→p (via E=p2/2m) →λ=h/p.
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