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

Question

The acceleration for electron and proton due to electrical force of their mutual attraction when they are 1 mathringA apart is (A) 3.1×1022 m s-2, 1.3×1019 m s-2 (B) 3.3×1018 m s-2, 3.2×1016 m s-2 (C) 2.5×1022 m s-2, 1.4×1019 m s-2 (D) 2.5×1018 m s-2, 1.3×1016 m s-2

Tardigrade Admin · Accepted Answer

Force of mutual attraction between an electron and a proton F=(1/4πε0) (e2/r2)=(9×109(1.6×10-19)2/(10-10)2)=2.3×10-8 N Acceleration of electron =(F/me)=(2.3×10-8/9×10-31) =2.5×1022 m s-2 Acceleration of proton =(F/mp)=(2.3×10-8/1.66×10-27) =1.4× 1019 m s-2

Q. The acceleration for electron and proton due to electrical force of their mutual attraction when they are $1 \overset{˚}{A}$ apart is

$3.1 \times 1 0^{22} m s^{- 2}$ , $1.3 \times 1 0^{19} m s^{- 2}$

$3.3 \times 1 0^{18} m s^{- 2}$ , $3.2 \times 1 0^{16} m s^{- 2}$

$2.5 \times 1 0^{22} m s^{- 2}$ , $1.4 \times 1 0^{19} m s^{- 2}$

$2.5 \times 1 0^{18} m s^{- 2}$ , $1.3 \times 1 0^{16} m s^{- 2}$

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

3.1×1022ms−2, 1.3×1019ms−2

3.3×1018ms−2, 3.2×1016ms−2

2.5×1022ms−2, 1.4×1019ms−2

2.5×1018ms−2, 1.3×1016ms−2

Force of mutual attraction between an electron and a proton F=4πε0​1​r2e2​=(10−10)29×109(1.6×10−19)2​=2.3×10−8N Acceleration of electron =me​F​=9×10−312.3×10−8​ =2.5×1022ms−2 Acceleration of proton =mp​F​=1.66×10−272.3×10−8​ =1.4×1019ms−2

Q. The acceleration for electron and proton due to electrical force of their mutual attraction when they are $1 \overset{˚}{A}$ apart is

$3.1 \times 1 0^{22} m s^{- 2}$ , $1.3 \times 1 0^{19} m s^{- 2}$

$3.3 \times 1 0^{18} m s^{- 2}$ , $3.2 \times 1 0^{16} m s^{- 2}$

$2.5 \times 1 0^{22} m s^{- 2}$ , $1.4 \times 1 0^{19} m s^{- 2}$

$2.5 \times 1 0^{18} m s^{- 2}$ , $1.3 \times 1 0^{16} m s^{- 2}$