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Question: The acceleration for electron and proton due to electrical force of their mutual attraction when the...

The acceleration for electron and proton due to electrical force of their mutual attraction when they are 1 A apart is.

A

3.1×1022ms−2,1.3×1019ms−23.1 \times 10^{22}ms^{- 2},1.3 \times 10^{19}ms^{- 2}

B

3.3×1018ms−2,3.2×1016ms−23.3 \times 10^{18}ms^{- 2},3.2 \times 10^{16}ms^{- 2}

C

2.5×1022ms−2,1.4×1019ms−22.5 \times 10^{22}ms^{- 2},1.4 \times 10^{19}ms^{- 2}

D

2.5×1018ms−2,1.3×1016ms−22.5 \times 10^{18}ms^{- 2},1.3 \times 10^{16}ms^{- 2}

Answer

2.5×1022ms−2,1.4×1019ms−22.5 \times 10^{22}ms^{- 2},1.4 \times 10^{19}ms^{- 2}

Explanation

Solution

: Force of mutual attraction between an electron and a proton

F=14πε0e2r2F = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{r^{2}}

=9×109(1.6×10−19)2(10−10)2=2.3×10−8N= \frac{9 \times 10^{9}(1.6 \times 10^{- 19})^{2}}{(10^{- 10})^{2}} = 2.3 \times 10^{- 8}N

Acceleration of electrons =Fme=2.3×10−89×10−31= \frac{F}{m_{e}} = \frac{2.3 \times 10^{- 8}}{9 \times 10^{- 31}}

=2.5×1022ms−2= 2.5 \times 10^{22}ms^{- 2}

Acceleration of proton =Fmp=2.3×10−81.66×10−27=1.4×1019ms−2= \frac{F}{m_{p}} = \frac{2.3 \times 10^{- 8}}{\begin{aligned} & 1.66 \times 10^{- 27} \\ & = 1.4 \times 10^{19}ms^{- 2} \end{aligned}}