Question

An imaginary radioactive element decays to form elements $X_1,X_2,X_3,X_4,X_{5}and{X}_6$, by ejecting two $\beta$-particles, followed by an $\alpha$-particle, and again two ($\beta$-particles followed by an $\alpha$ -particle). What are the mass number and the atomic number of $X_6$?

• A. ${}_{92}^{227}{X}_6$
• B. ${}_{92}^{226}{X}_6$
• C. ${}_{97}^{227}{X}_6$
• D. ${}_{90}^{227}{X}_6$

Answer 1

The correct option is A

${}_{92}^{227}{X}_6$

The symbolic representation of the reaction is:
${}_{92}^{235}X\stackrel{\beta -particle}{\to }{X}_1\stackrel{\beta -particle}{\to }{X}_2\stackrel{\alpha -particle}{\to }{X}_3\stackrel{\beta -particle}{\to }{X}_4\stackrel{\beta -particle}{\to }{X}_5\stackrel{\alpha -particle}{\to }{X}_6$
In complete reaction, the total number of emitted a-particles = 2.
$\therefore$ Loss of the total mass number due to two $\alpha$-particles = 4 + 4 = 8
$\therefore$ Loss of atomic number due to two $\alpha$ -particles = 2 + 2 = 4
In a complete reaction, the total number of emitted $\beta$ - particles = 4.
$\therefore$ Loss of mass number due to four $\beta$ -particles = 0
Gain of atomic number due to four $\beta -particles=1\times 4=4$
$\therefore$ Total loss of mass number = 8
Total loss or gain of atomic number
= 4 - 4 = 0
$\therefore$ Mass number of $X_6=(235-8)=227$
Atomic number of $X_6=(92-0)=92$