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Physics · Atomic Physics

JEE Main 2026 — 21 January, Morning Shift — Question 37

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____\_\_\_\_ m . (Atomic number of gold =79=79 and 14πϵ0=9×109\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} in SI units)

  1. Option A:

    2.95×10−142.95 \times 10^{-14}

    Correct
  2. Option B:

    2.95×10−162.95 \times 10^{-16}

  3. Option C:

    3.85×10−163.85 \times 10^{-16}

  4. Option D:

    3.85×10−143.85 \times 10^{-14}

Answer: A

Step-by-step solution

Energy conservation Ki+Ui=Kf+Uf\mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}} 7.7×106×1.6×10−19+07.7 \times 10^{6} \times 1.6 \times 10^{-19}+0 =0+9×109(1.6×10−19)(79×1.6×10−19)r=0+\frac{9 \times 10^{9}\left(1.6 \times 10^{-19}\right)\left(79 \times 1.6 \times 10^{-19}\right)}{\mathrm{r}} r=2.95×10−14\mathrm{r}=2.95 \times 10^{-14}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Rutherford's and Bohr's Model of Atom

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