Respuesta :
Answer:
r = 3.79 10⁻¹⁵ m
Explanation:
For this exercise we can use energy conservation
Starting point. Far from the fixed core
Em₀ = K
Final point
[tex]Em_{f}[/tex] = U = k q1 q2 / r
Em₀ =Em_{f}
K = k q₁ q₂ / r
r = k q₂ q₂ / K
Data
K = 6 Mev = 6 106 eV (1.6 10⁻¹⁹ J / 1 eV) = 9.6 10⁻¹³ J
Helium
q₁ = 2 e = 2 1.6 10⁻¹⁹ C = 3.2 10⁻¹⁹ C
Gold
q2 = 79 e = 79 1.6 10⁻¹⁹ = 126.4 10⁻¹⁹ C
Let's calculate
r = 8.99 10⁸ 3.2 10⁻¹⁹ 126.4 10⁻¹⁹ / 9.6 10⁻¹³
r = 3.79 10⁻¹⁵ m
Answer:
The answer to the question is;
The alpha particle will approach the gold nucleus up to 3.79 × 10⁻¹⁴ m before turning back.
Explanation:
To solve the question, we note that
There are 79 electrons the gold tom and 2 electrons in the alpha particle which is the helium nuclear
We use the energy conservation principle to obtain
W[tex]_{net}[/tex] = [tex]\int\limits^{r_{min}}_{inf} {F_e} \, dr[/tex] = Δ KE
Where
F[tex]_e[/tex] = Coulomb force
r = Distance of the alpha particle from the gold foil
we get from F = k(Q₁·Q₂)/r²
W[tex]_{net}[/tex] = [tex]\int\limits^{r_{min}}_{inf} {F_e} \, dr[/tex] =(-1)× [tex]\int\limits^{r_{min}}_{inf} {\frac{k(2e)(79e)}{r^2} } \, dr[/tex]
= -158·k·e² × [tex]\int\limits^{r_{min}}_{inf} {\frac{dr}{r^2} } \,[/tex]⇒ -158·k·e²[tex][\frac{1}{r} ]^{r_{min}} _{inf}[/tex] = [tex]\frac{-158*k*e^2}{r_{min}}[/tex] = -KE₁
or r[tex]_{min}[/tex] = (158·k·e²)/KE₁
Where Δ KE = - KE₁ as KE₂ = 0, particle at rest
Solving gives
r[tex]_{min}[/tex] =[tex]\frac{158*(8.988*10^9\frac{Nm^2}{C^2} )(1.602*10^{-19}C)^2}{6Mev*\frac{1.602*10^{-19}J}{eV} *1000000}[/tex] = 3.79 × 10⁻¹⁴ m
