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Physics · Atomic Physics

JEE Main 2026 — 22 January, Morning Shift — Question 26

7.9MeVα7.9 \mathrm{MeV} \alpha-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. [14πϵo=9×109Nm2/C2\left[\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right. and electron charge =1.6×10−19C]\left.=1.6 \times 10^{-19} \mathrm{C}\right]

  1. Option A:

    5.76×10−145.76 \times 10^{-14}

    Correct
  2. Option B:

    1.44×10−131.44 \times 10^{-13}

  3. Option C:

    2.88×10−142.88 \times 10^{-14}

  4. Option D:

    1.69×10−121.69 \times 10^{-12}

Answer: A

Step-by-step solution

By mechanical energy conservation (Me)i=(Me)f(\mathrm{Me})_{\mathrm{i}}=(\mathrm{Me})_{\mathrm{f}} PEi+KEi=PEf+KEf\mathrm{PE}_{\mathrm{i}}+\mathrm{KE}_{\mathrm{i}}=\mathrm{PE}_{\mathrm{f}}+\mathrm{KE}_{\mathrm{f}} 0+7.9×106×1.6×10−19=k(2e)(Ze)r+00+7.9 \times 10^{6} \times 1.6 \times 10^{-19}=\frac{\mathrm{k}(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}}+0 r=9×109×2×(1.6×10−19)2×797.9×106×1.6×10−19=2.88×10−14 m\mathrm{r}=\frac{9 \times 10^{9} \times 2 \times\left(1.6 \times 10^{-19}\right)^{2} \times 79}{7.9 \times 10^{6} \times 1.6 \times 10^{-19}}=2.88 \times 10^{-14} \mathrm{~m} For diameter ⇒ D = 2r=5.76×10−14 m2 \mathrm{r}=5.76 \times 10^{-14} \mathrm{~m}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Rutherford's and Bohr's Model of Atom
7.9 MeV α -particle scatters from a target material of atomic number… | JEE Main 2026 PYQ with Solution · DhiX AI