An α particle and proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λpλα is:
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a
2.8
b
7.8
c
8
d
3.8
answer is A.
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Detailed Solution
Kinetic energy K=P22m=qV⇒P=2mqV⇒P ∝ mqNow, λ=hP⇒λ∝1P⇒λpλα=PαPp=mαqαmpqp=4×21×1 ⇒λpλα=22=2.8