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Q.

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

n^3/2

d

n^1/2

answer is D.

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Detailed Solution

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Given that $2 πr = nλ; 2 πr = n \frac{h}{mV}$

$and BqV = \frac{{mV}^{2}}{r}; r = \frac{mV}{Bq} mV = Bqr; 2 πr = n \frac{h}{Bqr} \Rightarrow r^{2} \propto n \Rightarrow r \propto n^{1 / 2}$

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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