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Q.

The momentum of an electron revolving in n^th orbit is given by : (Symbols have their usual meanings)

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a

$\frac{nh}{2 πr}$

b

$\frac{nh}{2 r}$

c

$\frac{nh}{2 π}$

d

$\frac{2 πr}{nh}$

answer is A.

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Detailed Solution

detailed_solution_thumbnail

Angular momentum is integral multiple of $\frac{h}{2 π}$

$mvr = \frac{nh}{2 π}$

So momentum $mv = \frac{nh}{2 πr}$

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