Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An particle of energy is scattered through <img alt="180^

An \alpha - particle of energy 5 \; \mathrm{MeV} is scattered through 180^{\circ} by a gold nucleus. The distance of the closest approach is of the order of: \mathrm{(For \; gold, Z=79 )}

Option: 1

\mathrm{10^{-10}\; cm}


Option: 2

\mathrm{10^{-12}\; cm}


Option: 3

\mathrm{10^{-14}\; cm}


Option: 4

\mathrm{10^{-16}\; cm}


Answers (1)

At the distance of closest approach d,

Kinetic energy of an alpha particle = Potential energy of alpha particle and the gold nucleus

\mathrm{\mathrm{K}=\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{d}} ; \mathrm{d}=\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{Ze}^2}{\mathrm{~K}}}

Here, \mathrm{\mathrm{K}=5 \mathrm{MeV}=5 \times 10^6 \mathrm{eV}=5 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}=5 \times 1.6 \times 10^{-13} \mathrm{~J}}

Z = 79

\therefore \quad \mathrm{d}=\frac{9 \times 10^9 \times 2 \times 79 \times\left(1.6 \times 10^{-19}\right)^2}{5 \times 1.6 \times 10^{-13}}=4.55 \times 10^{-14} \sim 10^{-12} \mathrm{~cm}

Posted by

Ramraj Saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates
anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'
anam-khan

BTech Admissions 2025 Live: JEE Mains, SRMJEE, VITEEE dates out; AEEE, MET registrations underway

Latest Question