Question

\(\alpha\)-particles of energy \(400\ \text{keV}\) are bombarded on nucleus of \({}^{82}\text{Pb}\). In scattering of \(\alpha\)-particles, its minimum distance from nucleus will be

• A. \(0.59\ \text{nm}\)
• B. \(0.59\ \text{\AA}\)
• C. \(5.9\ \text{pm}\)
• D. \(0.59\ \text{pm}\)

Hint:

For closest approach of an \(\alpha\)-particle: \[ r(\text{fm})=\frac{1.44Z_1Z_2}{E(\text{MeV})} \] Always convert energy into MeV before using this shortcut.

Answer 1

The Correct Option is D

Concept:
When an \(\alpha\)-particle moves directly toward a positively charged nucleus, it experiences electrostatic repulsion. At the closest distance of approach, the kinetic energy of the \(\alpha\)-particle becomes completely converted into electrostatic potential energy. So, at minimum distance: \[ \text{Kinetic energy}=\text{Electrostatic potential energy} \] The formula used is: \[ E=\frac{1}{4\pi\varepsilon_0}\frac{Z_1Z_2e^2}{r} \] In nuclear physics, a useful shortcut is: \[ r(\text{fm})=\frac{1.44Z_1Z_2}{E(\text{MeV})} \]

Step 1: Write the given values.
For an \(\alpha\)-particle: \[ Z_1=2 \] For lead nucleus: \[ Z_2=82 \] Energy of \(\alpha\)-particle: \[ E=400\ \text{keV} \] Convert keV into MeV: \[ 400\ \text{keV}=0.4\ \text{MeV} \]

Step 2: Apply the closest approach formula.
\[ r=\frac{1.44Z_1Z_2}{E} \] Substitute values: \[ r=\frac{1.44\times2\times82}{0.4} \]

Step 3: Calculate the value.
First multiply the numerator: \[ 1.44\times2=2.88 \] \[ 2.88\times82=236.16 \] Now divide by \(0.4\): \[ r=\frac{236.16}{0.4}=590.4\ \text{fm} \] So: \[ r\approx590\ \text{fm} \]

Step 4: Convert femtometer into picometer.
We know: \[ 1\ \text{pm}=1000\ \text{fm} \] Therefore: \[ 590\ \text{fm}=0.590\ \text{pm} \] \[ r\approx0.59\ \text{pm} \]

Step 5: Check the options.
Option (A) \(0.59\ \text{nm}\) is too large.
Option (B) \(0.59\ \text{\AA}\) is also much larger than the calculated value.
Option (C) \(5.9\ \text{pm}\) is 10 times larger than the correct value.
Option (D) \(0.59\ \text{pm}\) matches the calculated result. Hence, the correct answer is: \[ \boxed{(D)\ 0.59\ \text{pm}} \]