Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other.

Q. 13.20 Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

Answers (1)

best_answer

For a head-on collision of two deuterons, the closest  distances between their centres will be d=2\timesr

d=2\times2.0

d=4.0 fm

d=4\times10-15 m

charge on each deuteron = charge of one proton=q =1.6\times10-19 C

The maximum electrostatic potential energy of the system during the head-on collision will be E

\\=\frac{q^{2}}{4\pi \epsilon _{0}d}\\ =\frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{4\times 10^{-15}}\ J\\ =\frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{4\times 10^{-15}\times 1.6\times 10^{-19}}\ eV\\=360\ keV

The above basically means to bring two deuterons from infinity to each other would require 360 keV of work to be done or would require 360 keV of energy to be spent.

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question