Physics Problems & Solutions: #10

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Nuclear & Particle Physics - Internal Conversion

Internal conversion is the process whereby an excited nucleus transfers its energy directly to one of the most tightly bound atomic electrons, causing the electron to be ejected from the atom and leaving the atom in an excited state. The most probable process after an internal conversion electron is ejected from an atom with a high atomic number is that the

atom returns to its ground state through inelastic collisions with either atoms
atom emits one or several X-rays
nucleus emits a gamma-ray
nucleus emits an electron
nucleus emits a positron

(GR9677 #10)

Solution:

Electron transitions in atom (internal conversion) = X-ray production
→ An orbital electron is absorbed and ejected along with an X-ray

compared to:

Nuclear transitions = Gamma, γ Ray production

→ The excited nucleus jumps to a lower level and emits a photon γ

Answer: B

Note:

1. (C), (D), and (E) are products of radioactive decay which are results of unstable nuclei.

2. In internal conversion:

For low atomic number, it will produce the Auger effect and ionize the outside electron.
For high atomic number, it will only emit X-rays.

Electromagnetism - Method of Image

Two positive charged of q and 2q coulombs are located on the x-axis at x = 0.5a and 1.5a respectively, as shown above. There is an infinite grounded conducting plane at x = 0. What is the magnitude of the net force on the charge q?

A. $\frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2}$

B. $\frac{1}{4\pi \epsilon_0} \frac{3q^2}{2a^2}$

C. $\frac{1}{4\pi \epsilon_0} \frac{2q^2}{a^2}$

D. $\frac{1}{4\pi \epsilon_0} \frac{3q^2}{a^2}$

E. $\frac{1}{4\pi \epsilon_0} \frac{7q^2}{2a^2}$

(GR9277 #10)

Solution:

Using method of image:

On q:

$\begin{aligned} \vec{F}_{\text{total}}&=kq\left [ \frac{2q}{a^2}\left ( -\hat{x} \right )+ \frac{-q}{a^2}\hat{x} +\frac{-2q}{(2a)^2}\hat{x} \right ] \\&=\frac{kq^2}{a^2}\left ( -2-1-\frac{1}{2} \right )\hat{x} \\&=-\frac{7}{2}\frac{kq^2}{a^2}\hat{x} \\\left |\vec{F}_{\text{total}} \right |&=k\frac{7q^2}{2a^2}\end{aligned}$

with $k=\frac{1}{4\pi \epsilon_0}$

Answer: E

Electromagnetism - Gauss' Law

An isolated sphere of radius R contains a uniform volume distribution of positive charge. Which of the curves on the graph above correctly illustrated the dependence of the magnitude of the electric field of the sphere as a function of the distance r from its center?

A. A
B. B
C. C
D. D
E. E

(GR8677 #10)

Solution:

The graph shows different lines for condition inside the solid sphere, r

R

Gauss’ Law: $\oint \bar{E} \cdot \hat{n} dA=\frac{q_{\textup{enc}}}{\varepsilon_0}$

Inside the sphere:

$E(4\pi r^2)=\frac{\rho V}{\varepsilon_0}=\frac{\rho}{\varepsilon_0}\frac{4}{3}\pi r^3$
$E=\frac{\rho }{3\varepsilon_0}r$

E is linearly proportional to r.

Answer: C

Electromagnetism - Capacitor

A 3-microfarad capacitor is connected in series with 6-microfarad capacitor. When a 300-volt potential difference is applied across this combination, the total energy stored in the two capacitor is

A. 0.09 J
B. 0.18 J
C. 0.27 J
D. 0.41 J
E. 0.81 J

(GR0177 #10)

Solution:

C in Series:

¹/_{C_net} = ¹/_C₁ + ¹/_C₂ = ( ¹/₃ + ¹/₆ ) ( ¹/₁₀-6 ) = ¹/₂_∙ 10-6

C_net = 2 ∙ 10^-6F

U = ½ QV = ½ C_netV² = ½ (2 ∙ 10^-6) (300)²= 9 (10^-6) (10⁴) = 0.09 J

Answer: A