JEE Main 202622 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 22 January
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Nuclear Physics
JEE Main 2026 · 22 January · Morning Shift · Physics · Nuclear Physics
7.9 MeV α -particle scatters from a target material of atomic...
-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) m. \left[\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right.
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
For Rutherford scattering, at the closest approach all kinetic energy converts to electrostatic potential energy.
For alpha particle-nucleus collision:
K.E. = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r_\min}
Where: (alpha), (gold), MeV
r_\min = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.9 \times 10^6 \times 1.6 \times 10^{-19}}
r_\min = 2.88 \times 10^{-14} m
The estimated nuclear diameter is approximately:
d = 2 \times r_\min = 5.76 \times 10^{-14} m
For alpha particle-nucleus collision:
K.E. = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r_\min}
Where: (alpha), (gold), MeV
r_\min = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.9 \times 10^6 \times 1.6 \times 10^{-19}}
r_\min = 2.88 \times 10^{-14} m
The estimated nuclear diameter is approximately:
d = 2 \times r_\min = 5.76 \times 10^{-14} m