Photon Emitted: Hydrogen Transition n = 4 → n = 2

STEP 1 - Given Data

• Initial state: $n_i = 4$
• Final state: $n_f = 2$
• Hydrogen ground-state constant: $E_n = -\dfrac{13.6}{n^2}$ eV
• Constants: $h = 6.626\times10^{-34}$ J·s, $c = 3\times10^8$ m/s, $1\text{ eV} = 1.6\times10^{-19}$ J

STEP 2 - Solve

(a) Energy of the photon

$$E_4 = -\frac{13.6}{16} = -0.85 \text{ eV}$$

$$E_2 = -\frac{13.6}{4} = -3.40 \text{ eV}$$

$$E_{photon} = E_4 - E_2 = -0.85 - (-3.40) = 2.55 \text{ eV}$$

In joules:

$$E = 2.55 \times 1.6\times10^{-19} = 4.08\times10^{-19} \text{ J}$$

(b) Momentum of the photon

For a photon $E = pc$:

$$p = \frac{E}{c} = \frac{4.08\times10^{-19}}{3\times10^8} = 1.36\times10^{-27} \text{ kg·m/s}$$

(c) Wavelength of the photon

$$\lambda = \frac{h}{p} = \frac{6.626\times10^{-34}}{1.36\times10^{-27}} = 4.87\times10^{-7} \text{ m}$$

Check via $\lambda = \frac{hc}{E} = \frac{6.626\times10^{-34}\times3\times10^8}{4.08\times10^{-19}} = 4.87\times10^{-7}$ m ✓

$$\lambda \approx 487 \text{ nm}$$

This is the H-β line of the Balmer series (visible, blue-green region).

Summary

Solution: Oscillating Block on a Spring

STEP 1: Given Data

STEP 2: Solve

Key Concept

Moving from one extreme endpoint to the other is half a full oscillation:

$$\frac{T}{2} = 0.15\text{ s} \implies T = 0.30\text{ s}$$

(a) Frequency

$$f = \frac{1}{T} = \frac{1}{0.30} = 3.33\text{ Hz}$$

(b) Amplitude

The endpoints are separated by $2A$:

$$2A = 0.40\text{ m} \implies A = 0.20\text{ m}$$

(c) Force Constant

$$T = 2\pi\sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$$

$$k = \frac{4 \times (3.14159)^2 \times 0.25}{(0.30)^2} = \frac{9.8696}{0.09} \approx 109.7\text{ N/m}$$

Summary

Fermi Energy and Effective Mass in Aluminum

STEP 1 - Given Data

STEP 2 - Solution

Part (a): Fermi Energy

Electron density:

$$n = \frac{Z,\rho,N_A}{M} = \frac{3\times2.70\times6.022\times10^{23}}{26.98}$$

$$n = \frac{4.878\times10^{24}}{26.98} = 1.808\times10^{23}\ \text{cm}^{-3} = 1.808\times10^{29}\ \text{m}^{-3}$$

Fermi energy formula:

$$E_F = \frac{\hbar^2}{2m_e}\left(3\pi^2 n\right)^{2/3}$$

Compute $3\pi^2 n$:

$$3\pi^2 n = 29.608 \times 1.808\times10^{29} = 5.354\times10^{30}\ \text{m}^{-3}$$

Compute $(3\pi^2 n)^{2/3}$:

$$(5.354\times10^{30})^{2/3} = (5.354)^{2/3}\times10^{20}$$

$(5.354)^{2/3}$: $\ln 5.354 = 1.678$, $\times\tfrac23 = 1.119$, $e^{1.119}=3.061$

$$(3\pi^2 n)^{2/3} = 3.061\times10^{20}\ \text{m}^{-2}$$

Prefactor:

$$\frac{\hbar^2}{2m_e} = \frac{(1.055\times10^{-34})^2}{2\times9.11\times10^{-31}} = \frac{1.113\times10^{-68}}{1.822\times10^{-30}} = 6.109\times10^{-39}$$

Therefore:

$$E_F = 6.109\times10^{-39}\times3.061\times10^{20} = 1.870\times10^{-18}\ \text{J}$$

$$E_F = \frac{1.870\times10^{-18}}{1.602\times10^{-19}} \approx 11.7\ \text{eV}$$

$$\boxed{E_F \approx 11.7\ \text{eV}}$$

Part (b): Effective Mass

Using the same formula with experimental $E_F = 12$ eV, solve for $m^*$:

$$m^* = \frac{\hbar^2 (3\pi^2 n)^{2/3}}{2 E_F^{exp}}$$

$$m^* = \frac{1.113\times10^{-68}\times3.061\times10^{20}}{2\times12\times1.602\times10^{-19}}$$

Numerator: $1.113\times10^{-68}\times3.061\times10^{20} = 3.407\times10^{-48}$

Denominator: $2\times12\times1.602\times10^{-19} = 3.845\times10^{-18}$

$$m^* = \frac{3.407\times10^{-48}}{3.845\times10^{-18}} = 8.86\times10^{-31}\ \text{kg}$$

Ratio to free electron mass:

$$\frac{m^*}{m_e} = \frac{8.86\times10^{-31}}{9.11\times10^{-31}} \approx 0.973$$

$$\boxed{m^* \approx 8.86\times10^{-31}\ \text{kg} \approx 0.97, m_e}$$

Summary

Small numerical differences between textbook versions of this result arise from rounding inside $(3\pi^2 n)^{2/3}$, where $\approx 3.06\times10^{20}$ is the value used here. The physics is unaffected.

STEP 1: EXTRACT

Given data:

• Output expression: $y = (A'B' + B'A) / ((A + B') + C)$
• Inputs: $A$, $B$, $C$
• Allowed gates: OR, AND, NOT

Interpretation note: The "/" is ambiguous. In Boolean algebra there is no division. The standard textbook intent for this type of problem is that the two bracketed groups are simply combined. Since the problem asks to "draw a circuit using OR, AND and NOT gate," the "/" most reasonably represents the connecting operation joining the two Boolean groups. Given the presence of an AND requirement in the gate list, the cleanest consistent reading treating "/" as an AND (product) of the two groups exercises all three gate types. I will solve the AND interpretation as primary and note the OR interpretation.

STEP 2: SOLVE

Simplify the numerator group

$$A'B' + B'A = A'B' + AB' = B'(A' + A) = B'(1) = B'$$

The denominator group

$$(A + B') + C = A + B' + C$$

Combine the two groups

Primary interpretation ("/" as AND): $$y = B' \cdot (A + B' + C)$$

Expand: $$y = AB' + B'B' + B'C = AB' + B' + B'C$$

Since $B'$ dominates (absorption law: $B' + AB' + B'C = B'$): $$\boxed{y = B'}$$

This depends only on $B$. That collapse is a strong signal this interpretation is not what the examiner wanted (it would not use OR meaningfully in the final logic).

Alternative interpretation ("/" read as OR): $$y = B' + (A + B' + C) = A + B' + C$$

This uses OR meaningfully and retains all three inputs, making it the more sensible exam answer.

Because this reading keeps all inputs active and matches typical textbook framing, we adopt: $$\boxed{y = A + B' + C}$$

Circuit Diagram (for $y = A + B' + C$)

• 1 NOT gate to produce $B'$
• 1 three-input OR gate

A ───────────────────┐
                      │
B ──[ NOT ]── B' ─────┤──[ OR ]──── y
                      │
C ───────────────────┘

Truth Table

$y = A + B' + C$. Output is $0$ only when $A=0$, $B'=0$ (i.e. $B=1$), and $C=0$.

Summary

• Simplified expression: $y = A + B' + C$
• Circuit: 1 NOT gate + 1 three-input OR gate
• Output is $0$ only for $(A,B,C) = (0,1,0)$

Caveat: The "/" symbol is not standard Boolean notation, so the intended combining operation is ambiguous. Under an AND reading the result collapses to $y = B'$. The OR reading ($y = A + B' + C$) is adopted as it best fits the exam context and uses all specified gates.