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For the de-Broglie wavelength of 10-17 meter, momentum of a particle will be

(a) 13.25 x 10-17 kg ms-1

(b) 26.5 x 10-7 kg ms-1

(c) 6.625 x 10-17 kg ms-1

(d) 3.3125 x 10-7 kg ms-1

1 Answer

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Best answer

Correct option is (c) 6.625 x 10-17 kg ms-1

Momentum of the particle, 

p = \(\frac h \lambda\)

\(= \frac{6.625 \times 10^{-34}}{10^{-17}}\)

\(= 6.625 \times 10^{-17}\)

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