Consider the following nuclear reactions :
$$_{92}^{238}M \to \,_y^xN + 2_2^4He;\,\,\,_y^xN \to _B^AL + 2{\beta ^ + }$$
The number of neutrons in the element $$L$$ is
A.
140
B.
144
C.
142
D.
146
Answer :
144
Solution :
$$\eqalign{
& _{92}^{238}M \to _{88}^{230}N + 2_2^4He \cr
& _{88}^{230}N \to _{86}^{230}L + 2{\beta ^ + } \cr} $$
The number of neutrons in element $$L = 230 - 86 = 144$$
Releted MCQ Question on Physical Chemistry >> Chemical Kinetics
Releted Question 1
If uranium (mass number 238 and atomic number 92) emits an $$\alpha $$ -particle, the product has mass no. and atomic no.