Question

Identify X in each of the following nuclear reactions:
(a) ${}_{30}{}^{68}\mathrm{Zn}+{}_0{}^{1}n\to {}_{28}{}^{65}\mathrm{Ni}+\mathrm{X}$

Short answer

(a) ${}_2{}^4\mathrm{He}$

(b) ${}_1{}^0\mathrm{e}$

(c) ${}_{-1}{}^{0}e$

(d) ${}_{62}^{160}Sm$

(e) ${}_0{}^{1}n$

(f) ${}_1{}^{0}e$

Step-by-step solution

Part(a)-Step 1- Given

${}_{30}{}^{68}\mathrm{Zn}+{}_0{}^{1}n\to {}_{28}{}^{65}Ni+X$

Part (a) Step 2- Explanation

A nuclear reaction is defined as the process in which two nuclei or nucleus of atom or subatomic particle collide with an outside atom and results in the formation of two or more nuclides which are different nuclei of the parent atom.

(a)The given reaction is -

${}_{30}{}^{68}\mathrm{Zn}+{}_0{}^{1}n\to {}_{28}{}^{65}Ni+X$

The above reaction shows the alpha decay. The alpha decay of element ${}_{30}{}^{68}Zn$results in the formation of ${}_{28}{}^{65}Ni$and ${}_2{}^4\mathrm{He}$. Therefore, for the given reaction the value of X is ${}_2{}^4\mathrm{He}$. Hence the correct nuclear reaction is given as-

${}_{30}{}^{68}\mathrm{Zn}+{}_0{}^{1}n\to {}_{28}{}^{65}\mathrm{Ni}+{}_2{}^4\mathrm{He}$

Part (b) -Step 1 - Given

${}_{15}{}^{30}\mathrm{P}\to {}_{14}{}^{30}\mathrm{Si}+\mathrm{X}$

Part(b)-Step 2 -Explanation

Identify X.

number of electrons =30-30=0 and proton = 15-14= 1

So X is${}_1{}^{0}e$

Part(c)-Step 1-Given

${}_{82}{}^{214}\mathrm{Pb}\to {}_{83}{}^{214}\mathrm{Bi}+\mathrm{X}$

Identify X

Part(c)-Step2-Explanation

${}_{82}{}^{214}\mathrm{Pb}\to {}_{83}{}^{214}\mathrm{Bi}+\mathrm{X}$

It should have 0, and -1 so the

X is ${}_{-1}{}^{0}e$

Part(d)-Step 1-Given

Find the value of X

Part(d)-Step 2- Explanation

${}_{92}{}^{235}\mathrm{U}+{}_0{}^{1}n\to 4\left({}_0{}^{1}n\right)+{}_{30}{}^{72}\mathrm{Zn}+\mathrm{X}$

The X will have mass and number as

235+1 - (72+4) = 160

92 -30=62

X=${}_{62}^{160}Sm$

Part(e)-Step1-Given Information

${}_{52}{}^{130}\mathrm{Te}+{}_1{}^2\mathrm{H}\to {}_{53}{}^{131}\mathrm{I}+\mathrm{X}$

Find the value of X

Part(e)-Step 2- Explanation

${}_{52}{}^{130}\mathrm{Te}+{}_1{}^2\mathrm{H}\to {}_{53}{}^{131}\mathrm{I}+\mathrm{X}$

X will have atomic number and mass as

130+2-131 =1

52+1 -53 =0

X= ${}_0{}^{1}n$

Part (f) - Step 1 - Given Information

${}_{29}{}^{64}\mathrm{Cu}+\mathrm{X}\to {}_{28}{}^{64}\mathrm{Ni}$

Part (f)-Step 2- Explanation

${}_{29}{}^{64}\mathrm{Cu}+\mathrm{X}\to {}_{28}{}^{64}\mathrm{Ni}$

Atomic mass and number for X is

64-64=0

29-28=1

Then X=${}_1{}^{0}e$