In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number 80, then maximum velocity of $\alpha$-particle is \_\_\_\_ $\times 10^{5} \mathrm{m} / \mathrm{s}$ approximately. ($\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9}$ SI unit, mass of $\alpha$ particle $= 6.72 \times 10^{-27} \mathrm{~kg}$)

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Electrostatics

In an alpha particle scattering experiment distance of closest approach for the

α

particle is

4.5 \times 1 0^{- 14} m

. If target nucleus has atomic number 80, then maximum velocity of

α

-particle is ____

\times 1 0^{5} m / s

approximately.
(

\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}

SI unit, mass of

α

particle

= 6.72 \times 1 0^{- 27} kg

)

Answer: 156

At closest approach all KE converts to potential energy:

\frac{1}{2} m v^{2} = \frac{1}{4 π ϵ _{0}} \frac{( 2 e ) ( Z e )}{r _{min}}

, so

v = \frac{4 K Z e ^{2}}{m r _{min}}

v = \frac{4 ( 9 \times 1 0 ^{9} ) ( 80 )}{( 6.72 \times 1 0 ^{- 27} ) ( 4.5 \times 1 0 ^{- 14} )} \times (1.6 \times 1 0^{- 19}) \approx 1.56 \times 1 0^{7} = 156 \times 1 0^{5} m/s

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Integer answer