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A particle is moving 5 times as fast as an electron. The ratio of the de-Brogile wavelength of the particle to that of the electron is $1.878 \times 10^{- 4}$ . The mass of the particle is close to

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$4.8 \times 10^{- 27} k g$

$9.1 \times 10^{- 31} k g$

$1.2 \times 16^{- 28} k g$

$9.7 \times 16^{- 28} k g$

answer is D.

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Detailed Solution

$A = \frac{h}{m v} m = \frac{h}{v} m α \frac{1}{x v} \frac{m_{1}}{m_{2}} = \frac{A_{2}}{A_{1}} . \frac{v_{2}}{v_{1}} = \frac{1}{1.878 \times 10^{- 4}} \times \frac{1}{5} ∴ m_{1} = 9.7 \times 10^{- 28} k g$

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