Question

A beam of some kind of particle of velocity $2\times {10}^7$m/s is scattered by a gold (Z=79) foil. The specific charge of this particle (charge/mass) if the distance of closest approach is $7.9\times {10}^{-14}$ m will be:

• A. $1.39\times {10}^{14}$ C/kg
• B. $1.39\times {10}^{10}$ C/kg
• C. $1.39\times {10}^8$ C/kg
• D. $1.39\times {10}^{12}$ C/kg

Answer 1

The correct option is C $1.39\times {10}^8$ C/kg

Let the particle be a proton. At the distance of the closest approach, the whole kinetic energy (K) of the proton will be converted to electrostatic potential energy.

$K=\frac{1}{4\pi {ϵ}_o}\times \frac{Ze\times e^{\prime }}{r_o}$

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_o}\times \frac{Ze\times e^{\prime }}{r_o}$

$e^{\prime }/m=\frac{v^2\times r_o}{2Ze\times \frac{1}{4\pi {ϵ}_o}}$

$\frac{1}{4\pi {ϵ}_o}=9\times {10}^{9}N{m}^2{C}^{-2}$

$\Rightarrow e^{\prime }/m=\frac{{(2\times {10}^7)}^2\times 7.9\times {10}^{-14}}{2\times 79\times 1.6\times {10}^{-19}\times 9\times {10}^9}$

$\Rightarrow e^{\prime }/m=1.39\times {10}^{8}C/kg$