In a Geiger-Marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 MeV 
before it comes momentarily to rest and reverses its direction. (Z for gold nucleus = 79)
Text Solution
Verified by ExpertsThe correct answer is:
C
Led d be the distance of closest approach then by the conservation of energy. Initial kinetic energy of incoming 
particle, k
= Final electric potential energy U of the system
As 
….. (i)
Here ,
substituting these values in (i)
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
An -particle of 5 MeV energy strikes with a nucleus of uranium at stationary at an scattering angl…
The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in va…
In hydrogen atom, electron makes transition from to level. Recoil momentum of the atom will be
A sodium atom is in one of the states labeled 'Lowest excited levels'. It remains in that state for…
An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The ene…
A hydrogen atom in its ground state absorbs 10.2 eV of energy. The orbital angular momentum is incr…