Question

${ } _ { 80 } \mathrm { Hg } ^ { 208 }$ nucleus is bombarded by $\alpha$-particles with velocity $10 ^ { 7 }$m/s. If the $\alpha$-particle is approaching the Hg nucleus head-on then the distance of closest approach will be

• A. $1.115 \times 10 ^ { - 13 } \mathrm {~m}$
• B. $11.15 \times 10 ^ { - 13 } \mathrm {~m}$
• C. $111.5 \times 10 ^ { - 13 } \mathrm {~m}$
• D. Zero

Answer 1

Answer: (A)

$1.115 \times 10 ^ { - 13 } \mathrm {~m}$

Answer 2

When α particle moves towards the mercury nucleus its kinetic energy gets converted in potential energy of the system. At the distance of closest approach

$\frac { 1 } { 2 } m v ^ { 2 } = \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \frac { q _ { 1 } q _ { 2 } } { r }$

⇒ $\frac { 1 } { 2 } \times \left( 1.6 \times 10 ^ { - 27 } \right) \left( 10 ^ { 7 } \right) ^ { 2 } = 9 \times 10 ^ { 9 } \frac { ( 2 . e ) ( 80 e ) } { r }$

⇒ $r = 1.115 \times 10 ^ { - 13 }$m.