If an alpha particle with energy 7.7 text{ Me | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The distance of closest approach $r_0$ is the distance where the kinetic energy of the alpha particle is completely converted into electrostatic p

Vedclass · Accepted Answer

$2.95 	imes 10^{-14}$

Vedclass · Answer

(A) The distance of closest approach $r_0$ is the distance where the kinetic energy of the alpha particle is completely converted into electrostatic potential energy.Using the principle of conservation of energy:$K_i + U_i = K_f + U_f$At the distance of closest approach, final kinetic energy $K_f = 0$.$K_i = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$Given:$K_i = 7.7 	ext{ MeV} = 7.7 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J}$$Z = 79$ (for gold)$e = 1.6 	imes 10^{-19} 	ext{ C}$$\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$Substituting the values:$7.7 	imes 10^6 	imes 1.6 	imes 10^{-19} = \frac{9 	imes 10^9 	imes (79 	imes 1.6 	imes 10^{-19}) 	imes (2 	imes 1.6 	imes 10^{-19})}{r_0}$$r_0 = \frac{9 	imes 10^9 	imes 79 	imes 2 	imes 1.6 	imes 10^{-19}}{7.7 	imes 10^6}$$r_0 \approx 2.95 	imes 10^{-14} 	ext{ m}$

If an alpha particle with energy $7.7 \text{ MeV}$ is bombarded on a thin gold foil, the closest distance from the nucleus it can reach is . . . . . . $\text{m}$. (Atomic number of gold $= 79$ and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$ in $SI$ units)

Similar Questions

Using Thomson's model of the atom,consider an atom consisting of two electrons,each of charge $-e$,embedded in a sphere of charge $+2e$ and radius $R$. In equilibrium,each electron is at a distance $d$ from the center of the atom. What is the equilibrium separation between the electrons?

Rutherford's $\alpha$-particle experiment showed that the atoms have

$A$ beam of fast-moving alpha particles was directed towards a thin gold foil. The parts $A, B$,and $C$ of the incident beam and their corresponding transmitted or reflected parts $A^{\prime}, B^{\prime}$,and $C^{\prime}$ are shown in the adjoining diagram. The number of alpha particles in:

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $P$ is $d$. What is the distance of closest approach of the alpha particle to the nucleus if the linear momentum of the alpha particle is $1.5 P$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module