# ^(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

${}^{241}Am$ 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)
${}^{235}Bk$
${}^{245}Bk$
${}^{239}Np$
not given
${}^{237}Np$
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fitte8b
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:
${}_{Z}^{A}X{=}_{Z-2}^{A-4}Y{+}_{2}^{4}He$
here, X is the parent nuclei and Y is the daughter nuclei.
Taking above reaction into account the alpha decay of ${}_{93}^{241}Am$ will be:${}_{93}^{241}Am{=}_{91}^{237}Np{+}_{2}^{4}He$
The daughter nuclei is ${}_{91}^{237}Np$
Hence, option 5 is correct.