Question

The magnetic field strength required to excite an isolated proton to its higher spin state with an electromagnetic radiation of 300 MHz is ................ Tesla. (Round off to two decimal places)
\text{[Magnetogyric ratio of proton is } 26.75 \times 10^7 \text{ rad T}^{-1} \text{ s}^{-1}]

Hint:

To calculate the magnetic field strength required for proton excitation in NMR, use the resonance condition \(E = \gamma B\), where \(E\) is the frequency of radiation, \(\gamma\) is the magnetogyric ratio, and \(B\) is the magnetic field strength.

Answer 1

Correct Answer: 7

Step 1: Use the resonance condition.
The resonance condition for proton NMR is given by: \[ E = \gamma \cdot B \cdot I \] Where: - \(E\) is the energy (frequency) of the radiation.
- \(\gamma\) is the magnetogyric ratio of proton (26.75 × 10\(^7\) rad T\(^{-1}\) s\(^{-1}\)).
- \(B\) is the magnetic field strength.
- \(I\) is the spin quantum number for the proton (which is 1/2).
From the equation, the magnetic field strength \(B\) can be found using: \[ B = \frac{E}{\gamma \cdot I} \] Step 2: Calculation.
We know the frequency \(E = 300\) MHz \( = 300 \times 10^6\) Hz.
Substituting the values: \[ B = \frac{300 \times 10^6}{26.75 \times 10^7 \times 1} = 1.12 \text{ Tesla.} \] Step 3: Conclusion.
Thus, the required magnetic field strength is 1.12 Tesla.