KCET2015PhysicsElectromagnetic Waves
An a-particle of energy ( 5 MeV ) is scattered through ( 180^ ) by gold nucleus. The distance of closest approach is of the order of
Options
- A( 10⁻¹⁰ ~cm )
- B( 10⁻¹² ~cm )
- C( 10⁻¹⁴ ~cm )
- D( 10⁻¹⁶ ~cm )
Correct answer
B. ( 10⁻¹² ~cm )
Step-by-step solution
Distance of closest approach is given as [ array l d= 1 4 ₀ (Z₁ e ) (Z₂ e ) kinetic energy d= (9 10⁹ ) (7.9 1.6 10⁻¹⁹ ) (2 1.6 10⁻¹⁹ ) 5 10⁶ 1.6 10⁻¹¹ =4.55 10⁻¹⁵ ~m 0.45 10⁻¹² ~cm 10⁻¹² ~cm array ] Therefore, distance of closest approach is of the order of ( 10⁻¹² ~cm )