A particle moving with a velocity of $6.626 \times 10^{7} m / s$ has a de Broglie wavelength of $1 Å$ in a circular path of radius $0.529 Å$ . The angular momentum of particle is $(h = 6 .626 \times 10^{- 34} J \times \sec)$ :

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$3.5 \times 10^{- 35} kg m^{2} \sec^{- 1}$

$3.5 \times 10^{- 34} kg m^{2} \sec^{- 1}$

$1.053 \times 10^{- 34} kg m^{2} \sec^{- 1}$

$3.5 \times 10^{- 91} kg m^{2} \sec^{- 1}$

answer is A.

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Detailed Solution

It is known that;

$2 π r = n λ$

$\Rightarrow n = \frac{2 \times π \times 0.529 \times 10^{- 10}}{1 \times 10^{- 10}}$

$L = \frac{n h}{2 π} = 0.529 \times 6.626 \times 10^{- 34}$

$L = 3.5 \times 10^{- 34}$

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