Question:
Given \(Q\) is the electromagnetic charge and \(S\) is the strangeness quantum number, identify the particle(s) for which \((Q - S) = 0\) is satisfied.
Given \(Q\) is the electromagnetic charge and \(S\) is the strangeness quantum number, identify the particle(s) for which \((Q - S) = 0\) is satisfied.
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Work out the quark content of each particle to get its charge and strangeness directly, then just subtract; only Sigma*- and K+ end up with Q equal to S.
Updated On: Jul 28, 2026
- \(\Sigma^{*-}\)
- \(K^+\)
- \(\Omega^-\)
- \(\Delta^{++}\)
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The Correct Option is A, B
Solution and Explanation
Step 1: Understanding the Concept:
Charge \(Q\) is the particle's electric charge in units of \(e\), and strangeness \(S\) counts the number of strange quarks with a minus sign (an \(s\) quark carries \(S = -1\), an \(\bar{s}\) quark carries \(S = +1\), and non-strange quarks carry \(S = 0\)). To test \((Q-S) = 0\) for each particle, we need its quark content, which fixes both \(Q\) and \(S\) directly.
Step 2: Check \(\Sigma^{*-}\) (option A).
\(\Sigma^{*-}\) is a spin-3/2 baryon in the decuplet with quark content \(dds\). Its charge is the sum of the quark charges: \(-\frac{1}{3} - \frac{1}{3} - \frac{1}{3} = -1\), so \(Q = -1\). It contains exactly one \(s\) quark, so \(S = -1\). Then \(Q - S = -1 - (-1) = 0\). So (A) satisfies the condition and is TRUE.
Step 3: Check \(K^+\) (option B).
\(K^+\) is a meson with quark content \(u\bar{s}\). Its charge is \(+\frac{2}{3} + \frac{1}{3} = +1\), so \(Q = +1\). It contains one \(\bar{s}\) antiquark, which carries \(S = +1\). Then \(Q - S = 1 - 1 = 0\). So (B) satisfies the condition and is TRUE.
Step 4: Check \(\Omega^-\) (option C).
\(\Omega^-\) is the famous triple-strange baryon with quark content \(sss\). Its charge is \(-\frac{1}{3} \times 3 = -1\), so \(Q = -1\). It has three \(s\) quarks, so \(S = -3\). Then \(Q - S = -1 - (-3) = 2\), which is not zero. So (C) is FALSE.
Step 5: Check \(\Delta^{++}\) (option D).
\(\Delta^{++}\) has quark content \(uuu\), all up quarks with no strangeness. Its charge is \(\frac{2}{3} \times 3 = +2\), so \(Q = +2\), and \(S = 0\). Then \(Q - S = 2 - 0 = 2\), which is not zero. So (D) is FALSE.
Final Answer:
Only \(\Sigma^{*-}\) and \(K^+\) satisfy \(Q - S = 0\).\[ \boxed{\Sigma^{*-}\text{ and } K^+} \]
Charge \(Q\) is the particle's electric charge in units of \(e\), and strangeness \(S\) counts the number of strange quarks with a minus sign (an \(s\) quark carries \(S = -1\), an \(\bar{s}\) quark carries \(S = +1\), and non-strange quarks carry \(S = 0\)). To test \((Q-S) = 0\) for each particle, we need its quark content, which fixes both \(Q\) and \(S\) directly.
Step 2: Check \(\Sigma^{*-}\) (option A).
\(\Sigma^{*-}\) is a spin-3/2 baryon in the decuplet with quark content \(dds\). Its charge is the sum of the quark charges: \(-\frac{1}{3} - \frac{1}{3} - \frac{1}{3} = -1\), so \(Q = -1\). It contains exactly one \(s\) quark, so \(S = -1\). Then \(Q - S = -1 - (-1) = 0\). So (A) satisfies the condition and is TRUE.
Step 3: Check \(K^+\) (option B).
\(K^+\) is a meson with quark content \(u\bar{s}\). Its charge is \(+\frac{2}{3} + \frac{1}{3} = +1\), so \(Q = +1\). It contains one \(\bar{s}\) antiquark, which carries \(S = +1\). Then \(Q - S = 1 - 1 = 0\). So (B) satisfies the condition and is TRUE.
Step 4: Check \(\Omega^-\) (option C).
\(\Omega^-\) is the famous triple-strange baryon with quark content \(sss\). Its charge is \(-\frac{1}{3} \times 3 = -1\), so \(Q = -1\). It has three \(s\) quarks, so \(S = -3\). Then \(Q - S = -1 - (-3) = 2\), which is not zero. So (C) is FALSE.
Step 5: Check \(\Delta^{++}\) (option D).
\(\Delta^{++}\) has quark content \(uuu\), all up quarks with no strangeness. Its charge is \(\frac{2}{3} \times 3 = +2\), so \(Q = +2\), and \(S = 0\). Then \(Q - S = 2 - 0 = 2\), which is not zero. So (D) is FALSE.
Final Answer:
Only \(\Sigma^{*-}\) and \(K^+\) satisfy \(Q - S = 0\).\[ \boxed{\Sigma^{*-}\text{ and } K^+} \]
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