Question

The acceleration for electron and proton due to electrical force of their mutual attraction when they are $1\mathring{A}$ apart is

• A. $3.1\times {10}^{22}m{s}^{-2},1.3\times {10}^{19}m{s}^{-2}$
• B. $3.3\times {10}^{18}m{s}^{-2},3.2\times {10}^{16}m{s}^{-2}$
• C. $2.5\times {10}^{22}m{s}^{-2},1.4\times {10}^{19}m{s}^{-2}$
• D. $2.5\times {10}^{18}m{s}^{-2},1.3\times {10}^{16}m{s}^{-2}$

Answer 1

The correct option is C $2.5\times {10}^{22}m{s}^{-2},1.4\times {10}^{19}m{s}^{-2}$

$\text{Step 1: Force between electron and proton}$

Distance between electron and proton, $r=1\mathring{A}={10}^{-10}m$

From Coulomb's law, $F=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{r^2}$ $=\frac{9\times {10}^9\times (1.6\times {10}^{-19}C{)}^2}{({10}^{-10}m{)}^2}=2.3\times {10}^{-8}N$

$\text{Step 2: Acceleration of proton and electron}$

From newton's law, $F=ma$

Acceleration of electron $=\frac{F}{m_e}$$=\frac{2.3\times {10}^{-8}N}{9\times {10}^{-31}kg}$ $=2.5\times {10}^{22}m/s^2$

Acceleration of proton $=\frac{F}{m_p}$$=\frac{2.3\times {10}^{-8}N}{1.66\times {10}^{-27}kg}$ $=1.4\times {10}^{19}m/s^2$

Hence, Option(C) is correct.