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Q.

A particle moving with a velocity of 6.626×107m/s has a de Broglie wavelength of 1Å in a circular path of radius 0.529Å. The angular momentum of particle is h=6.626×10−34J×sec :

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a

3.5×10−35 kg m2 sec−1

b

3.5×10−34 kg m2sec−1

c

1.053×10−34 kg m2 sec−1

d

3.5×10−91 kg m2 sec−1

answer is A.

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Detailed Solution

It is known that;

2πr=nλ

⇒n=2×π×0.529×10−101×10−10

L=nh2π=0.529×6.626×10−34

L=3.5×10−34

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