The distance between electron and proton in the H-atom is about 5-3 - 10-11 metre- What is the magnitude of electric force between these two particles- 14-0-9 - 109 Nm2C2

r = 5.3 × 10^ 11 m q_1 = 1.6 × 10^ 19 C q_1 = 1.6 × 10^ 19 C F = 14πε.q_1q_2r^2 = 9 × 10^9( 1.6 × 10^ 19× 1.6 × 10^ 19)5.3 × 5.3 × 10^ 22= ...

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Standard XII

Physics

Introducing Flux

The distance ...

Question

The distance between electron and proton in the H-atom is about 5.3 $\times 10^{- 11}$ metre. What is the magnitude of electric force between these two particles?( $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} \frac{N m^{2}}{C^{2}}$ )

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Solution

$r = 5.3 \times 10^{- 11} m$
$q_{1} = - 1.6 \times 10^{- 19} C$
$q_{1} = 1.6 \times 10^{- 19} C$

$F = \frac{1}{4 π ε} . \frac{q_{1} q_{2}}{r^{2}} = 9 \times 10^{9} \frac{(- 1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19})}{5.3 \times 5.3 \times 10^{- 22}} = 8.202 \times 10^{- 8} N$

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The distance between electron and proton in the H-atom is about 5.3 ×10−11 metre. What is the magnitude of electric force between these two particles?(14πε0=9×109Nm2C2)

r=5.3×10−11mq1=−1.6×10−19Cq1=1.6×10−19CF=14πε.q1q2r2=9×109(−1.6×10−19×1.6×10−19)5.3×5.3×10−22=8.202×10−8N

The distance between electron and proton in the H-atom is about 5.3 $\times 10^{- 11}$ metre. What is the magnitude of electric force between these two particles?( $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} \frac{N m^{2}}{C^{2}}$ )

$r = 5.3 \times 10^{- 11} m$
$q_{1} = - 1.6 \times 10^{- 19} C$
$q_{1} = 1.6 \times 10^{- 19} C$

$F = \frac{1}{4 π ε} . \frac{q_{1} q_{2}}{r^{2}} = 9 \times 10^{9} \frac{(- 1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19})}{5.3 \times 5.3 \times 10^{- 22}} = 8.202 \times 10^{- 8} N$