A nuclide represented as $^{70}_{32}X$ has a neutron-proton ratio of

Analyze the question: We need the ratio of neutrons to protons in the nuclide ⁷⁰₃₂X.

Concept: For a nuclide written as ᴬ_ZX:

• A = mass number (protons + neutrons) = 70
• Z = atomic number (number of protons) = 32
• Number of neutrons N = A − Z

Solution:

1. Protons = Z = 32
2. Neutrons = A − Z = 70 − 32 = 38
3. Neutron-proton ratio = N / Z = 38 / 32 = 1.1875 ≈ 1.2

Checking the options:

• A (0.5): too low, since neutrons are more than protons here
• B (0.8): this is below 1, so it's wrong
• C (1.2): matches 1.1875 rounded to one decimal place ✔
• D (1.5): would need about 48 neutrons, which is not the case

Tip: Always remember "Neutrons = A − Z" (the top number minus the bottom number). Then divide by the number of protons (the bottom number) for the n/p ratio. If the options are rounded, pick the closest one: 38/32 ≈ 1.19, which rounds to 1.2.