What should be the momentum (in gram centimetre per second) of a particle if its de-Broglie wavelength is 1 angstrom and the value of his $6.6252 \times 10^{- 27}$ erg second ?

6.6252 \times 10^{- 19} g c m / s

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6.6252 \times 10^{- 21} g c m / s

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6.6252 \times 10^{- 24} g c m / s

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6.6252 \times 10^{- 27} g c m / s

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Solution

The correct option is A $6.6252 \times 10^{- 19} g c m / s$
Given that,
$λ = 1 A^{o} = 1 \times 10^{- 8} c m, h = 6.6252 \times 10^{- 27} e r g s e c o n d$
or $p = \frac{6.6252 \times 10^{- 27}}{10^{- 8}} = 6.6252 \times 10^{- 19} g r a m . c m / s e c$

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6.6252 \times 10^{- 34} k g m^{2} / s ?

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The de Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is

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