Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to
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a
n2
b
n
c
n3/2
d
n1/2
answer is D.
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Detailed Solution
Given that 2πr=nλ ; 2πr=nhmVand BqV=mV2r ; r=mVBq mV=Bqr ; 2πr=nhBqr⇒r2∝n⇒r∝n1/2