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Question: The momentum of electron moving with 1/3<sup>rd</sup> velocity of light is (in g cm sec<sup>–1</sup>...

The momentum of electron moving with 1/3rd velocity of light is (in g cm sec–1)

A

9.69×10−89.69 \times 10^{- 8}

B

8.01×10108.01 \times 10^{10}

C

9.652×10−189.652 \times 10^{- 18}

D

None

Answer

9.652×10−189.652 \times 10^{- 18}

Explanation

Solution

Momentum of electron, ‘p’ = m′×um^{'} \times u

Where m′m^{'} is mass of electron in motion=m1−(u/c)2= \frac{m}{\sqrt{1 - (u/c)^{2}}}; Also u=c/3u = c/3

Momentu=9.108×10−281−(c3×c)2×3×10103=9.108×10−28×3×10100.94×3=9.652×10−18gcmsec⁡−1= \frac{9.108 \times 10^{- 28}}{\sqrt{1 - \left( \frac{c}{3 \times c} \right)^{2}}} \times \frac{3 \times 10^{10}}{3} = \frac{9.108 \times 10^{- 28} \times 3 \times 10^{10}}{0.94 \times 3} = 9.652 \times 10^{- 18}gcm\sec^{- 1}{}