Give step-by-step solution with explanation and final answer:In many MR experiments it is necessary to flip a spin which is initially aligned along the z axis
into the X’-y’ plane by using an appropr

Question

Give step-by-step solution with explanation and final answer:In many MR experiments it is necessary to flip a spin which is initially aligned along the z axis
into the X’-y’ plane by using an appropriate rotating magnetic field B1 during a certain amount of
time (pulse) (in the rotating frame, Bl is a static field orthogonal to BO). This is referred to as a 90
degrees pulse. If the desired pulse time is 1.0 ms, what B1 magnitude is required for
a. A proton spin?
b. A carbon spin?
c. At which frequency are rotating magnetizations originating from these two atoms?
d. AtaBo of 5 T, at which frequency must B, rotate (in the lab frame) in resonance
conditions for a proton spin?
e. What is then the wavelength and energy of a photon of this field? Compare it to what is
used of X-ray imaging.

VIPSolutions · Accepted Answer

Let's solve each part step-by-step:

---

### **Given Data**
- 90° pulse time ($ t_{90} $) = 1.0 ms = $1.0 \times 10^{-3}$ s
- $ B_0 = 5\,T $ (for part d)
- For 90° pulse: The magnetization must rotate by π/2 radians.
- Gyromagnetic ratio ($ \gamma $):
  - Proton ($ ^1H $): $ \gamma_p = 2.675 \times 10^8\, \text{rad/s/T} $
  - Carbon-13 ($ ^{13}C $): $ \gamma_c = 6.728 \times 10^7\, \text{rad/s/T} $

---

## **a. B1 Magnitude for a Proton Spin**

For a 90° pulse:
$$
\theta = \gamma B_1 t_{p}
$$
For 90° ($ \theta = \frac{\pi}{2} $) and $ t_{p} = 1.0 $ ms:
$$
\frac{\pi}{2} = \gamma_p B_1 t_p
$$
$$
B_1 = \frac{\pi}{2\gamma_p t_p}
$$

Plug in the values:
$$
B_1 = \frac{\pi}{2 \cdot (2.675 \times 10^8) \cdot (1.0 \times 10^{-3})}
$$
$$
B_1 = \frac{3.1416}{2 \cdot 2.675 \times 10^5}
$$
$$
B_1 = \frac{3.1416}{5.35 \times 10^5}
$$
$$
B_1 \approx 5.88 \times 10^{-6}\, \text{T} = 5.88\, \mu\text{T}
$$

---

## **b. B1 Magnitude for a Carbon Spin**

Use the same formula but with $ \gamma_c $:
$$
B_1 = \frac{\pi}{2 \gamma_c t_p}
$$
$$
B_1 = \frac{3.1416}{2 \cdot 6.728 \times 10^7 \cdot 1.0 \times 10^{-3}}
$$
$$
B_1 = \frac{3.1416}{1.3456 \times 10^5}
$$
$$
B_1 \approx 2.34 \times 10^{-5}\, \text{T} = 23.4\, \mu\text{T}
$$

---

## **c. Larmor Frequency for Protons and Carbons**

Larmor frequency:
$$
f = \frac{\gamma}{2\pi} B_0
$$

For protons ($ \gamma_p $):
$$
f_p = \frac{2.675 \times 10^8}{2\pi} \cdot 5
$$
$$
f_p \approx 4.257 \times 10^7 \cdot 5 = 2.1285 \times 10^8\, \text{Hz} = 212.85\, \text{MHz}
$$

For carbon-13 ($ \gamma_c $):
$$
f_c = \frac{6.728 \times 10^7}{2\pi} \cdot 5
$$
$$
f_c \approx 1.072 \times 10^7 \cdot 5 = 5.36 \times 10^7\, \text{Hz} = 53.6\, \text{MHz}
$$

---

## **d. Frequency Needed for Resonance for Protons at $ B_0 = 5\,T $**

Already calculated above:
$$
f_p = 212.85\, \text{MHz}
$$

---

## **e. Wavelength and Energy of a Photon of This Field**

Wavelength ($ \lambda $):
$$
\lambda = \frac{c}{f}
$$
where $ c = 3.00 \times 10^8\, \text{m/s} $.

$$
\lambda = \frac{3.00 \times 10^8}{2.1285 \times 10^8}
$$
$$
\lambda \approx 1.41\, \text{m}
$$

Energy ($ E $):
$$
E = h f
$$
where $ h = 6.626 \times 10^{-34}\, \text{J s} $.

$$
E = 6.626 \times 10^{-34} \cdot 2.1285 \times 10^8
$$
$$
E \approx 1.41 \times 10^{-25}\, \text{J}
$$

### **Compare to X-rays**
- X-ray energies: $ \sim 10^{-16} $ to $ 10^{-15} $ J (tens of keV)
- X-ray wavelengths: $ \sim 10^{-10} $ m (angstrom scale)

**Conclusion:** MRI photons are much lower in energy and much longer in wavelength than X-rays.

---

## **Summary Table**

| Part | Calculation | Value |
|------|-------------|-------|
| a | $ B_1 $ for proton | $ 5.88\, \mu\text{T} $ |
| b | $ B_1 $ for carbon | $ 23.4\, \mu\text{T} $ |
| c | Larmor freq (proton) | $ 212.85\, \text{MHz} $ |
|   | Larmor freq (carbon) | $ 53.6\, \text{MHz} $ |
| d | Resonance freq (proton, 5 T) | $ 212.85\, \text{MHz} $ |
| e | Wavelength (proton, 5 T) | $ 1.41\, \text{m} $ |
|   | Energy (photon, proton, 5 T) | $ 1.41 \times 10^{-25}\, \text{J} $ |

Let me know if you need further explanation for any step!

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers