PYQ NEET- Electric Charges And Fields L-1

Question: The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \AA \AA^{\circ}$ apart is,#

$$ \left(m_e \simeq 9 \times 10^{-31} \mathrm{~kg}, e=1.6 \times 10^{-19} \mathrm{C}\right) $$

$\left(\text { take }, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)$

A) $10^{24} \mathrm{~m} / \mathrm{s}^2$

B) $10^{23} \mathrm{~m} / \mathrm{s}^2$

C) $10^{22} \mathrm{~m} / \mathrm{s}^2$

D) $10^{25} \mathrm{~m} / \mathrm{s}^2$

Answer: $10^{22} \mathrm{~m} / \mathrm{s}^2$#

Sol:#

Force of mutual attraction between the electron and proton. (when, $r=1.6 \mathrm{~A}^{\circ}=1.6 \times 10^{-10} \mathrm{~m}$ ) is given as $$ \begin{aligned} F & =9 \times 10^9 \times \frac{e^2}{r^2} \ & =9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right)^2}{\left(1.6 \times 10^{-10}\right)^2} \ & =9 \times 10^{-9} \mathrm{~N} \end{aligned} $$ $\therefore$ Acceleration of electron $$ =\frac{F}{m_e}=\frac{9 \times 10^{-9}}{9 \times 10^{-31}}=10^{22} \mathrm{~m} / \mathrm{s}^2 $$