Correct Answer - Option 1 : 3 × 10
35
CONCEPT:
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Electrostatic Force: between two charges q1 and q2 at the distance of R is given by:
\(F=\frac{1}{4\pi ϵ_0}\frac{q_1q_2}{R^2} \)
where F is the electrostatic force, q1 and q2 are the charges, and ϵ0 is the electrical permittivity of the vacuum.:
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Gravitational Force: between two masses m1 and m2 at the distance of R is given by:
\(F=G\frac{m_1m_2}{R^2} \)
where F is the gravitational force, m1 and m2 are the masses, and G is the gravitational constant.
CALCULATION:
Given that both particles are alpha particles.
So charges of both q = 2e and mass M = 4m where m is the mass of a proton and e is the charge of a proton.
\({F_e \over F_g}={\frac{1}{4\pi ϵ_0}\frac{q_1q_2}{R^2} \over G\frac{m_1m_2}{R^2} }\)
\({F_e \over F_g}={\frac{1}{4\pi ϵ_0}\frac{(2e)(2e)} {G(4m)(4m)}}\)
Put e = 1.6 × 10-19 C and m = 1.66 × 10-27 Kg
\({F_e \over F_g}={\frac{1}{4\pi ϵ_0}\frac{(2e)(2e)} {G(4m)(4m)}}\)
\({F_e \over F_g}=0.337 \times 10^{36}\)
\({F_e \over F_g}≈ 3 \times 10^{35}\)
So the correct answer is option 1.
