^(241)Am decays by alpha particle emission. The daughter isotope in this case is (the atomic numbers of Np, Pu, Am, Cm, and Bk are 91, 92, 93, 94, and 95 respectively) a) ^(235) Bk b) ^(245) Bk c) ^(239) Np d) not given e) ^(237) Np

Antwan Perez

Antwan Perez

Answered question

2022-10-22

241 A m decays by alpha particle emission. The daughter isotope in this case is (the atomic numbers of Np, Pu, Am, Cm, and Bk are 91, 92, 93, 94, and 95 respectively)
235 B k
245 B k
239 N p
not given
237 N p

Answer & Explanation

fitte8b

fitte8b

Beginner2022-10-23Added 12 answers

A radioactive decay follows the conservation of charge and the conservation of mass.
In an alpha decay a parents nuclei converts into daughter nuclei with the emission of the helium nuclei ( alpha particle ). The general reaction of an alpha decay is given below:
Z A X = Z 2 A 4 Y + 2 4 H e
here, X is the parent nuclei and Y is the daughter nuclei.
Taking above reaction into account the alpha decay of 93 241 A m will be: 93 241 A m = 91 237 N p + 2 4 H e
The daughter nuclei is 91 237 N p
Hence, option 5 is correct.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Atomic Physics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?