Question

Consider the following nuclear reactions :
${}_{92}^{238}M$ $ \to $ ${}_Y^XN$$ + $ $2{}_2^4He$ ; ${}_Y^XX$$ \to $ ${}_B^AL$ $ + $ $2{\beta ^ + }$
the number of neutrons in the element L is :
A.142
B.144
C.146
D.140

Answer 1

Hint:We know that alpha decay is the process by which an unstable atom can use to become more stable. During an alpha decay process , an atom's nucleus sheds two protons and two neutrons in a packet which scientists call an alpha particle.

Complete step by step answer:
We know that, $\alpha - $ decay formula is -
${}_Z^AX\mathop \to \limits^{ - \alpha } {}_{Z - 2}^{A - 4}Y$
And the formula for $\beta $ is –
${}_Z^AX\mathop \to \limits^{ - \beta } {}_{Z - 1}^AY$
Now for the given above question we see that , -
${}_{92}^{238}M$ $\mathop \to \limits^{ - 2\alpha } {}_{92 - 4}^{238 - 8}N$
${}_Y^XX$ $\mathop \to \limits^{ - 2{\beta ^ + }} {}_{86}^{230}L$
Therefore we can now calculate the number of neutrons in $L$ , -
$L = 230 - 86$
$L = 144$
So, as we can see the no. of neutrons present in L is 144 .

Therefore, the correct option for the above question is ,Option (b). $144$.

Additional Information :
Mostly, an alpha decay occurs in very heavy elements like uranium, thorium, and radium. They are mainly called as parent nuclei and they are basically very much unstable. It is so because the nuclei of these atoms have a lot more neutrons in their nuclei than the no. of protons present, that is, they have too large a proton to neutron ratio, which makes the likes of these elements neutron-rich.

Note:
The beta decay occurs in either one of the two ways :
a) one is when the nucleus emits an electron and an antineutrino in a process that changes a neutron to a proton
b) and another is when the nucleus emits a positron and a neutrino in a process that changes a proton to a neutron.