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1. Electric Charges and Fields
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The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \;\mathring A$ apart is,$\left(m_{e} \simeq 9 \times 10^{-31} kg , e=1.6 \times 10^{-19} C \right)$
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )
A
$10^{25} \;m / s ^{2}$
B
$10^{24} \;m / s ^{2}$
C
$10^{23} \;m / s ^{2}$
D
$10^{22} \;m / s ^{2}$
(NEET-2020)
Solution
$F=\frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}}$
$a_{e}=\frac{F}{m_{e}}$
$a_{e}=\frac{e^{2}}{4 \pi \varepsilon_{0} m_{e} r^{2}}$
$a_{e}=\frac{9 \times 10^{9} \times(1.6)^{2} \times 10^{-38}}{9 \times 10^{-31} \times(1.6)^{2} \times 10^{-20}}$
$a_{n}=10^{22} m / s ^{2}$
Standard 12
Physics
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