Which of the following pair of nuclei are isobars of the element?
Show Hint
- $^{236}_{92}U$ and $^{238}_{92}U$
- $^2_1H$ and $^3_1H$
- $^3_1H$ and $^3_2He$
- $^{198}_{80}Hg$ and $^{197}_{79}Au$
The Correct Option is C
Solution and Explanation
To determine which pair of nuclei are isobars, we first need to understand the term "isobars." Isobars are nuclei with the same mass number (A), but different atomic numbers (Z). This means they have different numbers of protons but the same total number of nucleons (protons + neutrons).
| Option | Nucleus 1 | Nucleus 2 | Check if Isobars |
|---|---|---|---|
| (a) | \(^{236}_{92}U\) | \(^{238}_{92}U\) | Not Isobars (same Z and different A) |
| (b) | \(^2_1H\) | \(^3_1H\) | Not Isobars (same Z and different A) |
| (c) | \(^3_1H\) | \(^3_2He\) | Isobars (same A = 3, different Z) |
| (d) | \(^{198}_{80}Hg\) | \(^{197}_{79}Au\) | Not Isobars (different A and Z) |
From the above table, we can see that option (c), \(^3_1H\) and \(^3_2He\), are isobars. They have the same mass number (3) but different atomic numbers (1 and 2, respectively). This matches the definition of isobars.
Thus, the correct answer is \(^3_1H\) and \(^3_2He\).
All other options are ruled out as they do not satisfy the condition of having the same mass number with different atomic numbers.
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