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Question: Under the action of given columbic force the acceleration of an electron is \(2.5 \times 10^{22}ms^{...

Under the action of given columbic force the acceleration of an electron is 2.5×1022ms−22.5 \times 10^{22}ms^{- 2}. Then the magnitude of the acceleration of a proton under the action of same force is nearly

A

1.6×10−19ms−21.6 \times 10^{- 19}ms^{- 2}

B

9.1×1031ms−29.1 \times 10^{31}ms^{- 2}

C

1.5×1019ms−21.5 \times 10^{19}ms^{- 2}

D

1.6×1027ms21.6 \times 10^{27}ms^{2}

Answer

1.5×1019ms−21.5 \times 10^{19}ms^{- 2}

Explanation

Solution

: The acceleration of the electron due to given coulombic force F is

ae=Fmea_{e} = \frac{F}{m_{e}} …(i)

Where mem_{e}is the mass of the electron The acceleration of the proton due to same force F is ap=Fmpa_{p} = \frac{F}{m_{p}}

Where mpm_{p}is the mass of the proton Divide (ii) by (i), we get

apae=memp\frac{a_{p}}{a_{e}} = \frac{m_{e}}{m_{p}}

ap=aememp=(2.5×1022ms−2)(9.1×10−31kg)(1.67×10−27kg)a_{p} = \frac{a_{e}m_{e}}{m_{p}} = \frac{(2.5 \times 10^{22}ms^{- 2})(9.1 \times 10^{- 31}kg)}{(1.67 \times 10^{- 27}kg)}

=13.6×1018ms−2≈1.5×1019ms−2= 13.6 \times 10^{18}ms^{- 2} \approx 1.5 \times 10^{19}ms^{- 2}