Question

Identify the missing nuclide in the following nuclear equation:
214 Pb → 0 e + ?
82 -1
A. Pb-215.
B. Bi-214.
C. Pb-213.
D. TI-215.
E. TI-214.

Answer 1

Answer:

B. Bi-214.

Explanation:

The equation shows beta particle emission of 214/82 Pb which result into 214/83 Bi, in which the mass remain same but the the atomic number increases by one.

During this emission neutron get split into an electron and a proton which are represented as 0/e/-1.

So, the final nuclear equation becomes : 214/82 Pb => 0 e -1 + 214/83 Bi

Hence, the correct answer is "B. Bi-214."