Given:
α particles have 2 e charge
\(\begin{array}{l} \Rightarrow E = 15 * 10^{4} \\\Rightarrow F = Eq = 15 * 10^{4} * 2 * 1.6 * 10^{-19}\\\Rightarrow 4.8 * 10^{-14} N \end{array} \)
Therefore, the force acting on it is –
\(\begin{array}{l}4.8 * 10^{-14} N\end{array} \)
Distance between electron and proton, r is 0.53 A.
As we know:
E =
\(\begin{array}{l}\frac{Kqe}{r^{2}}\end{array} \)
Here
K =
\(\begin{array}{l}9 * 10^{9} N m^{2} C^{-2}\end{array} \)
qe =
\(\begin{array}{l}1.6 * 10^{-19}\end{array} \)
r =
\(\begin{array}{l}0.53 * 10^{-10} m\end{array} \)
Now, by substituting the values, we get:
E =
\(\begin{array}{l} \frac{9 \;*\; 10^{9} \;* \;1.6 \;* \;10^{-19}}{0.53 \;* \;0.53 \;*\;10^\;{-20}}\\\Rightarrow E = \frac{144 * 10^{10}}{0.2809} \approx \frac{144 * 10^{10}}{28 * 10^{-2}} \\ \Rightarrow E = 5.14 * 10^{12} N/C\end{array} \)
Therefore, the electric field at the location of electrons is –
\(\begin{array}{l}\Rightarrow E = 5.14 * 10^{12} N/C\end{array} \)
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