Question

The radioactive decay of $_{35}{X^{88}}$ by $\beta$ – emission produces an unstable nucleus, which spontaneously emits a neutron. The final product is:
A. $_{37}{X^{88}}$
B. $_{35}{Y^{89}}$
C. $_{34}{X^{88}}$
D.$_{36}{W^{87}}$

Answer 1

$\beta$ – emission followed by a neutron emission reduces mass by one unit and increases the atomic number by one unit so the product is $_{36}{W^{87}}$

Complete step by step answer:
$\beta$ decay occurs in nuclei that are rich in neutrons, that is the nuclei contains more neutrons than stable isotopes of the same element. These "proton deficient" nuclides can sometimes be identified simply by noticing that their mass number A (the sum of neutrons and protons in the nucleus) is significantly more than twice that of the atomic number Z (number of protons in nucleus). In order to regain some stability, such a nucleus can decay by converting one of its extra neutrons into a proton, emitting an electron and an antineutrino(ν). The high energy electron that is emitted in the reaction is called a beta particle and it is represented by 0−1e−−10e− in nuclear equations. Lighter atoms having atomic number less than 60 are the most likely to undergo beta decay.
$\beta$ – emission followed by a neutron emission reduces mass by one unit and increases the atomic number by one unit so the product is $_{36}{W^{87}}$
$_{35}{X^{88}}{ \to _{36}}{W^{88}} + 1{e^0}$
$_{36}{W^{88}}{ \to _{36}}{W^{87}}{ + _0}{n^1}$
So, when the radioactive decay of $_{35}{X^{88}}$ by β – emission produces an unstable nucleus, which spontaneously emits a neutron, the final product is $_{36}{W^{87}}$

Therefore, the correct answer is option (D).

Note:
The radioactive decay by $\beta$ – emission is a process of radioactive disintegration in which a proton inside the atomic nucleus is transformed into a neutron or vice versa. We can also say that the unstable nucleus becomes stable by the beta decay. It occurs when the nucleus is unstable by the possession of extra protons or extra neutrons.