Correct Answer: D
Explanation:
The given equation represents a beta-minus (β-) decay, where a neutron is converted into a proton, emitting an electron (e-) and an antineutrino (νe).
The equation is as follows:
\(_{83}^{210}\text{Bi} \rightarrow X + e^{-1} + \bar{v}\)
In β- decay, a neutron in the nucleus of the atom is transformed into a proton. The atomic number (Z) increases by 1, while the mass number (A) remains the same because a neutron is converted to a proton without a change in the total nucleon count.
In the equation for bismuth-210 (\(_{83}^{210}\text{Bi}\)), the atomic number is 83 and the mass number is 210. After the β- decay, the resulting element will have an atomic number of 84 (one proton more than 83), but the mass number will remain the same (210). This corresponds to the element polonium (Po), which has atomic number 84 and mass number 210.
The number of neutrons in the resulting nucleus can be found by subtracting the atomic number from the mass number:
Neutrons = Mass number - Atomic number = 210 - 84 = 126
Thus, the number of neutrons in the nucleus X is 126, which corresponds to the correct answer D.
In beta decay, a neutron is converted into a proton, emitting an electron (e⁻) and an antineutrino (\(\bar{\nu}\)). The number of protons in the nucleus increases by one while the number of neutrons decreases by one. In the given equation: \[ ^{210}_{83} \text{Bi} \rightarrow X + e^{-} + \bar{\nu} \] The atomic number of bismuth (Bi) is 83, and its mass number is 210. After beta decay, the new element \(X\) will have an atomic number of 84 (since one proton is added), and the mass number will remain 210 (since a neutron is converted into a proton). Therefore, the number of neutrons in the nucleus of \(X\) will be: \[ \text{Number of neutrons in } X = 210 - 84 = 126 \] Thus, the number of neutrons in the nucleus \(X\) is 126.
\(\textbf{Correct Answer:}\) \((D) \ 126\)
Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \)
When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
Given: