Physics Problems & Solutions: #12

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Nuclear & Particle Physics - Positronium

The ground-state energy of positronium is most nearly equal to

A. − 27.2 eV
B. − 13.6 eV
C. − 6.8 eV
D. − 3.4 eV
E. 13.6 eV

(GR9677 #12)

Solution:

Energy levels of Positronium is half those of Hydrogen (See GR8677 #99)

E_n(H)=− 13.6 / n²
E_n(Ps)= ½ E_n(H)
For the ground-state n = 1 → E_Ps= − ½ × 13.6 eV= − 6.8 eV

Answer: C

Electromagnetism - Electric Potential

Two large conducting plates form a wedge of angle α as shown-in the diagram above. The plates are insulated from each other; one has a potential V₀ and the other is grounded. Assuming that the plates are large enough so that the potential difference between them is independent of the cylindrical coordinates z and ρ, the potential anywhere between the plates as s function of the angle φ is

A. V₀/α
B. V₀φ/α
C. V₀α/φ
D. V₀φ²/α
E. V₀α/φ²

(GR9277 #12)

Solution:

Boundary conditions:
V(φ = 0) = 0
V(φ = α) = V₀

(A) FALSE
V₀/α does not depend on φ

(B) TRUE
V₀φ/α = 0 for φ = 0
V₀φ/α = V₀ for φ = α

(C) FALSE. V₀α/φ = ∞ for φ = 0

(D) FALSE
For φ = α, V₀φ²/α = V₀α

(E) FALSE
For φ = α, V₀φ²/α = V₀α

Answer: B

Sound and Wave - Doppler Effect

A source of 1-kilohertz sound is moving straight toward you at a speed 0.9 times the speed of sound. The frequency you receive is

A. 0.1 kHz
B. 0.5 kHz
C. 1.1 kHz
D. 1.9 kHz
E. 10 kHz

(GR8677 #12)

Solution:

Moving toward observer
→ f increases
→ f > 1 kHz
→ A and B are FALSE

Speed 0.9 times the speed of sound
→ f increases greatly
→ E. TRUE

Answer: E

Calculation:

$f'=\frac{f_{\text{source}}}{1\pm \beta}$

$\beta = \frac{V_{\text{source}}}{V_{\text{sound}}}$

+ sign = receding
− sign = approaching

$\begin{aligned} f'&=\frac{f_{\text{source}}}{1- \frac{V_{\text{source}}}{V_{\text{sound}}}} \\&=\frac{1}{1- \frac{0.9 \text s}{\text s}}=\frac{1}{0.1}=10 \end{aligned}$

Optics - Concave Mirror

A spherical concave mirror is shown in the figure above. The focal point F and the location of the object O are indicated. At what point will the image be located?

A. I
B. II
C. III
D. IV
E. V

(GR0177 #12)

Solution:

The mirror equation:

$\\\frac{1}{d_o} + \frac{1}{d_i} =\frac{1}{f}\rightarrow \frac{1}{O} + \frac{1}{d_i} =\frac{1}{F} \\\\d_i=\frac{FO}{O-F}$

The picture shows: $O < F \rightarrow d_i$ is negative.
The image is VIRTUAL (inside the mirror) i.e. at point V

Answer: E

Note:
For concave mirror:

If an object is located BEFORE the focal point then image is VIRTUAL (inside the mirror).
If an object is located AFTER the focal point then image is REAL (outside the mirror).