An Alpha Particle Is Kept In An Electric Field Of 15×104Nc−1. Calculate The Force Acting On The Particle. An Electron Is Separated From A Proton Through A Distance Of 0.53A. Calculate Electric Field At The Location Of Electrons​.

Given:

α particles have 2 e charge

\( \Rightarrow E = 15 * 10^{4} \\\Rightarrow F = Eq = 15 * 10^{4} * 2 * 1.6 * 10^{-19}\\\Rightarrow 4.8 * 10^{-14} N \)

Therefore, the force acting on it is – \(4.8 * 10^{-14} N\)

Distance between electron and proton, r is 0.53 A.

As we know:

E = \(\frac{Kqe}{r^{2}}\)

Here

K = \(9 * 10^{9} N m^{2} C^{-2}\)

qe = \(1.6 * 10^{-19}\)

r = \(0.53 * 10^{-10} m\)

Now, by substituting the values, we get:

E =\( \frac{9 \;*\; 10^{9} \;* \;1.6 \;* \;10^{-19}}{0.53 \;* \;0.53 \;*\;10^\;{-20}}\\\Rightarrow E = \frac{144 * 10^{10}}{0.2809} \approx \frac{144 * 10^{10}}{28 * 10^{-2}} \\ \Rightarrow E = 5.14 * 10^{12} N/C\)

Therefore, the electric field at the location of electrons is – \(\Rightarrow E = 5.14 * 10^{12} N/C\)

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