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NEET PHYSICSMedium

In the nuclear emission stated above: ${}_{82}^{290}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q$ , the mass number and atomic number of the product Q respectively, are:

A

286, 80

B

288, 82

C

286, 81

D

280, 81

Step-by-Step Solution

Initial Nucleus ( $X$ ): Mass Number ( $A$ ) = 290, Atomic Number ( $Z$ ) = 82.
Step 1: Alpha Decay ( $\alpha$ ): Emission of a Helium nucleus ( ${}_{2}^{4}He$ ).

$A' = 290 - 4 = 286$
$Z' = 82 - 2 = 80$
Nucleus $Y$ is ${}_{80}^{286}Y$ .

Step 2: Positron Emission ( $e^+$ ): Emission of a positron ( ${}_{+1}^{0}e$ ). $Z$ decreases by 1, $A$ remains unchanged.

$A'' = 286$
$Z'' = 80 - 1 = 79$
Nucleus $Z$ is ${}_{79}^{286}Z$ .

Step 3: Beta Minus Decay ( $\beta^-$ ): Emission of an electron ( ${}_{-1}^{0}e$ ). $Z$ increases by 1, $A$ remains unchanged.

$A''' = 286$
$Z''' = 79 + 1 = 80$
Nucleus $P$ is ${}_{80}^{286}P$ .

Step 4: Electron Emission ( $e^-$ ): Another $\beta^-$ decay. $Z$ increases by 1.

$A_{final} = 286$
$Z_{final} = 80 + 1 = 81$
Nucleus $Q$ is ${}_{81}^{286}Q$ .

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