Physics Problems & Solutions: #39

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Quantum Mechanics - Ionization Potential

Which of the following atoms has the lowest ionization potential?

A. ²_₄He
B. ⁷₁₄N
C. ⁸₁₆O
D. ¹⁸₄₀Ar
E. ⁵⁵₁₃₃Cs

(GR9677 #39)

Solution:

Ionization potential: the energy required to to remove the valence (outermost) electron from an atom

It is minimum for atoms (alkali metals) which have weakly bound electrons and higher for the noble gases which have closed shells.

²_₄He and Argon ⁸₄₀Ar are noble gases which have a closed-shell valence electron configuration.

⁷₁₄N → 1s² 2s² 2p⁶3s² 3p²(2 outermost electrons, almost a full orbital)

⁸₁₆O → 1s² 2s² 2p⁶⁴3s² 3p²(4 outermost electrons, almost a full orbital)

⁵⁵₁₃₃Cs has the highest atomic number and hence a lot of electrons. The valence electron is far from the nucleus (weakly bound), thus it needs the lowest ionization potential to remove it.

Answer: E

Electromagnetism - Fourier Series

If n is an integer ranging from 1 to infinity, is an angular frequency, and t is time, then the Fourier series for a square wave, as shown above, is given by which of the following?

A. $V(t)=\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n}\sin(n\omega t)$

B. $V(t)=\frac{4}{\pi}\sum_{0}^{\infty}\frac{1}{(2n+1)}\sin(\left(2n+1\right)\omega t)$

C. $V(t)=\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n}\cos(n\omega t)$

D. $V(t)=\frac{4}{\pi}\sum_{0}^{\infty}\frac{1}{(2n+1)}\cos(\left(2n+1\right)\omega t)$

E. $V(t)=-\frac{4}{\pi}+\frac{4}{\pi}\sum_{1}^{\infty}\frac{1}{n^2}\cos(n\omega t)$

(GR9277 #39)

Solution:

The sine function is an odd function.
The cosine function is an even function.
The function in this problem is an odd square wave function.
→ C, D, E are FALSE.

Both A and B are the same, thus the right solutions. However, A is more generalized and B is the simplified answer.

For Answer A, the sine term vanishes when n is even (n = 0, 2, 4, ...) → rewrite n for odd term as 2n +1 with n starts at 0 → the answer can be simplified to B.

Answer: B

Lab Methods - Graph

The gain of an amplifier is plotted versus angular frequency ω in the diagram above. If K and a are positive constants, the frequency dependence of the gain near ω = 3 × 10⁵ second⁻¹ is most accurately expressed by

A. Ke^−aω
B. Kω²
C. Kω
D. Kω⁻¹
E. Kω⁻²

(GR8677 #39)

Solution:

(A) g = Ke^−aω→ decreasing (exponential decay). FALSE.

exponential decay photo Exp decay_zps6okbzju5.png

(B) g = Kω²→ increasing (quadratic function/parabola). FALSE.

quadratic function photo Quadratic function_zps13pu3odx.png

(C) g = Kω → increasing (linear function). FALSE.
(D) g = K/ω→ decreasing (linear function). FALSE.

(E) g = K/ω²→ decreasing (quadratic function). TRUE.

Check:
At ω = 10⁶, g ≈ 10
g = K/ω²
10 = K/10¹²
K = 10¹³

At ω = 3 × 10⁵, K = 10¹³
g = 10¹³/(9 × 10¹⁰) ≈ 0.11 × 10³
10²

10³

Answer: E

Electromagnetism - High-Pass Filters

Which two of the following circuits are high-pass filters

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

(GR0177 #39)

Solution:

High-pass filter: V_out ≈ V_in for frequency, ω → ∞

Inductive reactance, X_L = ωL
Capacitive reactance, X_C = 1 / ωC

For ω → ∞
X_L → ∞
X_C → 0

Voltage divider:
$V_{\text{out}}=\left( \frac{R_2}{R_1+R_2}\right ) V_{\text{in}}$

Case I.
$V_{\text{out}}=\left( \frac{R}{R+X_L}\right) V_{\text{in}}=\left( \frac{R}{R+\infty}\right ) V_{\text{in}}=0$
→ Low-pass filter

Case II.
$V_{\text{out}}=\left( \frac{X_L}{R+X_L}\right) V_{\text{in}}=\left( \frac{\infty}{R+\infty}\right ) V_{\text{in}}=V_{\text{in}}$
Note: R ≪ ∞
→ High-pass filter

Case III.
$V_{\text{out}}=\left( \frac{R}{R+X_C}\right) V_{\text{in}}=\left( \frac{R}{R+0}\right ) V_{\text{in}}=V_{\text{in}}$
→ High-pass filter

Case IV.
$V_{\text{out}}=\left( \frac{X_C}{R+X_C}\right) V_{\text{in}}=\left( \frac{0}{R+0}\right ) V_{\text{in}}=0$
→ Low-pass filter

Answer: D