Question

An MRI signal acquired from a tissue has a transverse relaxation time (\(T_2\)) of 100 milliseconds (ms). If the signal intensity at time \(t = 0\) ms is 500 a.u., the signal intensity at \(t = 50\) ms is a.u. (Round off to one decimal place).

Hint:

Use S(t) = S0 e^(-t/T2). Here t is exactly half of T2, so the decay factor is e^(-0.5), not e^(-1).

Answer 1

Correct Answer: 303.3

Step 1: Recall the T2 decay equation.
In MRI, transverse magnetization (and the signal it produces) does not vanish instantly after excitation, it fades away exponentially over time because spins in the transverse plane slowly lose their phase coherence with each other. This is described by the transverse relaxation time \(T_2\), through the equation
\[ S(t) = S_0 \, e^{-t/T_2} \]
where \(S_0\) is the signal right after excitation (at \(t=0\)) and \(S(t)\) is the signal remaining after a time \(t\) has passed.

Step 2: Plug in the given numbers.
Here \(S_0 = 500\) a.u., \(T_2 = 100\) ms, and we want the signal at \(t = 50\) ms.
\[ \frac{t}{T_2} = \frac{50}{100} = 0.5 \]
\[ S(50) = 500 \, e^{-0.5} \]

Step 3: Evaluate the exponential.
\[ e^{-0.5} \approx 0.6065 \]
\[ S(50) = 500 \times 0.6065 = 303.25 \]
Rounded to one decimal place, this is 303.3 a.u.

Step 4: Sanity check the size of the answer.
At \(t = T_2 = 100\) ms, the signal would have dropped to \(500 \times e^{-1} \approx 184\) a.u., about 37 percent of the start value. At the halfway point \(t = 50\) ms, we expect a bit more than half the signal left, and 303.3 out of 500 (about 61 percent) fits that expectation.

Final Answer:
The signal intensity at \(t = 50\) ms is about 303.3 a.u. \[ \boxed{S(50) \approx 303.3 \text{ a.u.}} \]