Which of the following particle have non zero e/m ratio? — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Definition of Specific Charge (\(e/m\) Ratio) The charge-to-mass ratio (also referred to as the specific charge, represented as \(e/m\)) of any subatomic particle is mathematically defined as the ratio of its net electrical charge (\(e\)) to its rest mass (\(m\)): \[\text{Specific Charge} = \frac{e}{m}\] For a particle to possess a non-zero \(e/m\) ratio, it must satisfy two conditions simultaneously: 1. It must carry a non-zero net electrical charge (\(e \neq 0\)). 2. It must have a finite, non-zero rest mass (\(m \neq 0\)). If a particle is electrically neutral (\(e = 0\)), its charge-to-mass ratio will always be exactly zero: \[\frac{e}{m} = \frac{0}{m} = 0\] --- ### Step 2 - Analysis of the Positron (\ce{e^+}) A positron is the antiparticle of the electron. It is characterized by the following properties: * **Net Electrical Charge (\(e\)):** \(+1.602 \times 10^{-19}\text{ C}\) (exactly equal in magnitude to the charge of an electron, but positive). * **Rest Mass (\(m\)):** \(9.109 \times 10^{-31}\text{ kg}\) (identical to the rest mass of an electron). Using the formula for specific charge: \[\text{Specific Charge} = \frac{e}{m}\] Substituting the values with units: \[\frac{e}{m} = \frac{1.602 \times 10^{-19}\text{ C}}{9.109 \times 10^{-31}\text{ kg}}\] Calculating the ratio: \[\frac{e}{m} \approx \boxed{+1.758 \times 10^{11}\text{ C/kg}}\] Since both the charge and the mass are non-zero, the positron possesses a **non-zero** \(e/m\) ratio. --- ### Step 3 - Analysis of the Other Subatomic Particles Let us evaluate the charge and mass of the other options: 1. **Neutron (\ce{n^0}):** A constituent nucleon of the atomic nucleus. It is electrically neutral. * Net Charge (\(e\)): \(0\text{ C}\) * Rest Mass (\(m\)): \(1.675 \times 10^{-27}\text{ kg}\) * Substituting into the specific charge formula: \[\frac{e}{m} = \frac{0\text{ C}}{1.675 \times 10^{-27}\text{ kg}} = \boxed{0\text{ C/kg}}\] 2. **Neutrino (\ce{\nu}):** An elementary particle that does not participate in electromagnetic interactions. * Net Charge (\(e\)): \(0\text{ C}\) * Rest Mass (\(m\)): \(m_{\ce{\nu}} > 0\) (extremely tiny, but non-zero rest mass). * Substituting into the specific charge formula: \[\frac{e}{m} = \frac{0\text{ C}}{m_{\ce{\nu}}} = \boxed{0\text{ C/kg}}\] 3. **Neutral Meson (e.g., neutral pion, \ce{\pi^0}):** A hadronic particle consisting of a quark-antiquark pair. * Net Charge (\(e\)): \(0\text{ C}\) * Rest Mass (\(m\)): \(m_{\text{meson}} \approx 2.4 \times 10^{-28}\text{ kg}\) * Substituting into the specific charge formula: \[\frac{e}{m} = \frac{0\text{ C}}{m_{\text{meson}}} = \boxed{0\text{ C/kg}}\] --- ### Step 4 - Systematic Analysis of the Options * **Option (A) Neutron:** Incorrect. Because it is electrically neutral, its \(e/m\) ratio is exactly zero. * **Option (B) Neutrino:** Incorrect. Since it carries zero net electrical charge, its \(e/m\) ratio is exactly zero. * **Option (C) Positron:** Correct. It carries a positive charge and has a finite mass, giving it a very large, non-zero \(e/m\) ratio. * **Option (D) Neutral meson:** Incorrect. Being uncharged, its \(e/m\) ratio is exactly zero. \[\text{Correct Option: } \boxed{\text{C}}\]