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Question: A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of t...

A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to the of the electron is 1.813×10-4. The mass of the particle is

(me=9.1×10−31kgm_{e} = 9.1 \times 10^{- 31}kg) :

A

1.67×10−27kg1.67 \times 10^{- 27}kg

B

1.67×10−31kg1.67 \times 10^{- 31}kg

C

1.67×10−19kg1.67 \times 10^{- 19}kg

D

1.67×10−14kg1.67 \times 10^{- 14}kg

Answer

1.67×10−27kg1.67 \times 10^{- 27}kg

Explanation

Solution

: de Broglie wavelength of a moving particle having mass m and velocity v is given by

λ=hp=hmvorm=hλv\lambda = \frac{h}{p} = \frac{h}{mv}orm = \frac{h}{\lambda v}

For an electron λe=hmeve\lambda_{e} = \frac{h}{m_{e}v_{e}}

Or me=hλevem_{e} = \frac{h}{\lambda_{e}v_{e}}

Given : vve=3andλλe=1.813×10−4\frac{v}{v_{e}} = 3and\frac{\lambda}{\lambda_{e}} = 1.813 \times 10^{- 4}

Mass of the particle, m=me(λeλ)(vev)m = m_{e}\left( \frac{\lambda_{e}}{\lambda} \right)\left( \frac{v_{e}}{v} \right)

Substituting the values we get

m=9.1×10−31×11.813×10−4×13m = 9.1 \times 10^{- 31} \times \frac{1}{1.813 \times 10^{- 4}} \times \frac{1}{3}

Or m=1.67×10−27kgm = 1.67 \times 10^{- 27}kg