In an alpha particle scattering experiment distance of closest approach for the \alpha particle is 4.5 \times 10^-14 ~m . If target nucleus has atomic number 80 , then maximum velocity of \alpha -particle is \underline\hspace2cm\times 10^5 ~m / s approximately. ( \frac14 \pi \epsilon_0=9 \times 10^9 SI unit, mass of \alpha particle =6.72 \times 10^-27 ~kg )

Atoms and Nuclei PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q90 with Solution

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Correct answer: 156

At the closest approach, the initial kinetic energy of the $\alpha$ -particle gets completely converted into electrostatic potential energy.

For an $\alpha$ -particle:

So,

\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r}

Here,

r = 4.5\times 10^{-14}\ \text{m}

m = 6.72\times 10^{-27}\ \text{kg}

\frac{1}{4\pi\varepsilon_0}=9\times 10^9

e=1.6\times 10^{-19}\ \text{C}

\frac{1}{2}mv^2 = 9\times 10^9\cdot \frac{2\cdot 80\cdot (1.6\times 10^{-19})^2}{4.5\times 10^{-14}}

First,

2\cdot 80 = 160

and

(1.6\times 10^{-19})^2 = 2.56\times 10^{-38}

Thus,

160\times 2.56\times 10^{-38} = 409.6\times 10^{-38} = 4.096\times 10^{-36}

Now,

9\times 10^9 \times 4.096\times 10^{-36} = 36.864\times 10^{-27} = 3.6864\times 10^{-26}

Divide by $4.5\times 10^{-14}$ :

\frac{3.6864\times 10^{-26}}{4.5\times 10^{-14}} = 0.8192\times 10^{-12} = 8.192\times 10^{-13}\ \text{J}

So,

\frac{1}{2}mv^2 = 8.192\times 10^{-13}

v^2 = \frac{2\times 8.192\times 10^{-13}}{6.72\times 10^{-27}}

v^2 = \frac{1.6384\times 10^{-12}}{6.72\times 10^{-27}} = 2.438\times 10^{14}

Therefore,

v = \sqrt{2.438\times 10^{14}} = 1.561\times 10^7\ \text{m/s}

v \approx 156\times 10^5\ \text{m/s}

So the required integer is:

\boxed{156}