Write balanced nuclear equations for these reactions and identify \(X:(a){ }_{34}^{80} S e(d, p) X,(b) X(d, 2 p)_{3}^{9} L i,\) (c) \({ }_{5}^{10} \mathrm{~B}(\mathrm{n}, \alpha) \mathrm{X}\).

Short Answer

Expert verified
The unknown nuclei in the given reactions are (a) \( _{34}^{81}Se \), (b) \( _{4}^{9}Be \), and (c) \( _{3}^{7}Li \).

Step by step solution

01

Notation review

Nuclear reactions are usually written in the form (a, b) where a is the incident particle and b is the particle ejected from the reaction. A nucleus X is typically written as \(_{Z}^{A}X\), where Z represents the atomic number (number of protons) and A is the mass number (number of protons + neutrons).
02

Equation (a)

For the reaction \( _{34}^{80} Se (d,p) X \), a deuteron particle (d) is absorbed and a proton (p) is emitted. The deuteron particle has mass number 2 and charge 1. Thus, the mass number and atomic number of \(X\) are computed as follows: \[ A_X = A_{Se} + A_{d} - A_{p} = 80 + 2 - 1 = 81 \] \[ Z_X = Z_{Se} + Z_{d} - Z_{p} = 34 + 1 - 1 = 34 \] Therefore, X is \( _{34}^{81}Se \).
03

Equation (b)

For the reaction \( X(d, 2p) _{3}^{9} Li \), two protons (2p) are emitted. The deuteron particle (d) has mass number 2 and charge 1. Thus, the mass number and the atomic number of \(X\) are computed as follows: \[ A_X = A_{Li} + A_{2p} - A_{d} = 9 + 2 - 2 = 9 \] \[ Z_X = Z_{Li} + Z_{2p} - Z_{d} = 3 + 2 - 1 = 4 \] Therefore, X is \(_{4}^{9}Be \).
04

Equation (c)

For the reaction \( _{5}^{10} B (n,\alpha) X \), a neutron (n) is absorbed and a helium nucleus (alpha particle, \(\alpha\)) is emitted. The neutron has mass number 1 and charge 0, and the alpha particle has mass number 4 and charge 2. Thus, the mass number and atomic number of \(X\) are computed as follows: \[ A_X = A_{B} + A_{n} - A_{\alpha} = 10 + 1 - 4 = 7 \] \[ Z_X = Z_{B} + Z_{n} - Z_{\alpha} = 5 + 0 - 2 = 3 \] Therefore, X is \( _{3}^{7} Li \).

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