NTA Abhyas JEE Main2020ChemistryStructure of AtomPractice
An α -paticle approaches the target nucleus of copper (Z = 29) in such a way that the value of impact parameter is zero. The distance of closest approach will be
Options
- A2 π ε 0 ( K . E . ) α 29 e 2
- B29 e 2 2 π ε 0 ( K . E . ) α
- C4 π ε 0 ( K . E . ) α 29 e 2
- D( K . E . ) α
Correct answer
B. 29 e 2 2 π ε 0 ( K . E . ) α
Step-by-step solution
Distance of closest approach K q 1 q 2 r = ( K . E . ) α K = 1 4 π ε 0 1 4 π ε 0 Z e × 2 e r = ( K . E . ) α 1 4 π ε 0 2 × 29 e 2 r = ( K . E . ) α r = 1 2 π ε 0 × 29 e 2 ( K . E . ) α