Question 310: Atoms & Nuclei

Question

In an alpha particle scattering experiment distance of closest approach for the $$\alpha$$ particle is $$4.5 	imes 10^{-14} \mathrm{~m}$$. If target nucleus has atomic number 80 , then maximum velocity of $$\alpha$$-particle is __________ $$	imes 10^5 \mathrm{~m} / \mathrm{s}$$ approximately.
($$\frac{1}{4 \pi \epsilon_0}=9 	imes 10^9 \mathrm{SI}$$ unit, mass of $$\alpha$$ particle $$=6.72 	imes 10^{-27} \mathrm{~kg}$$)

Accepted Answer

$$\begin{aligned}
& \frac{1}{2} m v_0^2=\frac{1}{4 \pi \epsilon_0} \frac{z(e)}{r} \
& \frac{1}{2} 	imes 6.72 	imes 10^{-27} v_0^2=9 	imes 10^9 	imes \frac{80 	imes 2 	imes 1.6 	imes 1.6 	imes 10^{-19} 	imes 10^{-19}}{4.5 	imes 10^{-14}} \
& v_0^2=\frac{2 	imes 9 	imes 80 	imes 1.6 	imes 1.6 	imes 2}{6.72 	imes 4.5} 	imes 10^{-38+14+27+9} \
& v_0^2=243.8 	imes 10^{12} \
& v_0=15.6 	imes 10^6 \
& v_0=156 	imes 10^5
\end{aligned}$$

Atoms & Nuclei

Integer / Numeric Answer Type