Question

Element ${}_{26}^{120}Y$ is formed as a result of $3{\beta }^{+}$ and $2{\beta }^{-}$ decays. Which of the following is the parent element?

• A. ${}_{26}^{119}\Psi$
• B. ${}_{27}^{120}\iota$
• C. ${}_{25}^{120}\Omega$
• D. ${}_{14}^{60}\Delta {+}_{13}^{60}\vartheta$

Answer 1

Answer: (B)

${}_{27}^{120}\iota$

Equation involved in ${\beta }^{+}$ decay.

$\bullet p\to n+{\beta }^{+}+{\nu }_{e..}$

$\Rightarrow {}_Z^{A}X\to {Z-1}^{X^A}+{\beta }^{+}+{\nu }_{e.}$

where,

${\nu }_e$ is a position.

Equation involved in ${\beta }^{-}$ decay.

$\bullet n\to p+{\beta }^{-}+{\nu }_{e..}$

$\Rightarrow {}_Z^{A}X\to {Z+1}^{X^A}+{\beta }^{-}+{\nu }_{e.}$

Now, let the parent element be ${}_Z^{A}X$

26 = z-3+2$\to$ (1)

$\Rightarrow$ Z=27

There is no change in atomic mass in $\beta$ - decay

$\therefore$ A = 120

Hence, parent element : ${}_{27}^{120}X\equiv {}_{27}^{120}\iota$