What is the relationship between radiopharmaceuticals, tracers, and radionuclides?

1 answer:

4 0

These three words are the just the same. The radiopharmaceuticals are defined as a radioactive compound used for the treatment of diseases. While this also commonly called in other termed known as Radionuclides or Tracer. It is a radioactive material.

You might be interested in

The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie

telo118 [61]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The uncertainty in inverse frequency is $\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s$

Explanation:

From the question we are told that

The value of the proportionality constant is $k = 5 \frac{Hz }{T}$

The strength of the magnetic field is $B = 20 \ T$

The change in this strength of magnetic field is $\Delta B = 3 \ T$

The magnetic field is given as

$B = \frac{k}{\frac{1}{w} }$

Where $w$ is frequency

The uncertainty or error of the field is given as

$\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ]$

The uncertainty in inverse frequency is given as

$\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}$

$\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }$

substituting values

$\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }$

$\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s$

6 0

3 years ago

PLEASE HELP WITH THIS!

jok3333 [9.3K]

Answer: A

Explanation:

Parallel circuit. The source current is three times the current of a single bulb

7 0

3 years ago

A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th

Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

$E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$