Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need clarity, kindly explain! - Atoms And Nuclei - JEE Main

An   \alpha    particle of energy 5MeV is scattered through 180^{\circ} by a fixed uranium nucleus. The distance of the closest approach is of the order of

  • Option 1)

    1\dot{A}

  • Option 2)

    10^{-10}cm

  • Option 3)

    10^{-12}cm

  • Option 4)

    10^{-15}cm

 

Answers (1)

best_answer

As we learnt in

Binding energy per nucleon -

 

- wherein

This graph shows the stability of nuclei many nuclear phenomena can be explained by this graph

 

 At distance of closest approach whole kinetic energy is converted into potential energy.

\frac{1}{2}mv^{2}=\frac{(ze)(2e)}{4\pi \epsilon_{0}r}=5\ MeV

\therefore\ \; \frac{2ze^{2}}{4\pi \epsilon_{0}(5 MeV)}=5.3\times10^{-12}cm

Correct option is 3.


Option 1)

1\dot{A}

This is an incorrect option.

Option 2)

10^{-10}cm

This is an incorrect option.

Option 3)

10^{-12}cm

This is the correct option.

Option 4)

10^{-15}cm

This is an incorrect option.

Posted by

perimeter

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

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today
anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

Latest Question