The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 mathringA apart is, (me ≃ 9× 10-31 kg, e=1.6 × 10-19C) (Take (1/4πε0)=9× 109 Nm2 C-2)

Question

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 mathringA apart is, (me ≃ 9× 10-31 kg, e=1.6 × 10-19C) (Take (1/4πε0)=9× 109 Nm2 C-2) (A) 1024 m/s2 (B) 1023 m/s2 (C) 1022 m/s2 (D) 1025 m/s2

Tardigrade Admin · Accepted Answer

F=K (e2/r2) a=K (e2/mr2) a=9×109 ((1.6×10-19)2/(1.6×10-10)2(9×10-31)) a=10-29×1051=1022 m /s2

Q. The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \overset{˚}{A}$ apart is, $(m_{e} ≃ 9 \times 1 0^{- 31} k g, e = 1.6 \times 1 0^{- 19} C)$
(Take $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2})$

$1 0^{24} m / s^{2}$

$1 0^{23} m / s^{2}$

$1 0^{22} m / s^{2}$

$1 0^{25} m / s^{2}$

$F = K \frac{e ^{2}}{r ^{2}}$
$a = K \frac{e ^{2}}{m r ^{2}}$
$a = 9 \times 1 0^{9} \frac{( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 1.6 \times 1 0 ^{- 10} ) ^{2} ( 9 \times 1 0 ^{- 31} )}$
$a = 1 0^{- 29} \times 1 0^{51} = 1 0^{22} m / s^{2}$

Q. The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6A˚ apart is, (me​≃9×10−31kg,e=1.6×10−19C) (Take 4πε0​1​=9×109Nm2C−2)

1024m/s2

1023m/s2

1022m/s2

1025m/s2