In the radioactive decay - 238Pu→234U+α, What will be the kinetic e

1.	In the radioactive decay - 238Pu→234U+α, What will be the kinetic energy of the outgoing α-particle? Given:- -Atomic masses: m(238Pu)=238.04955 u; m(234U)=234.04095 u; m(4He)=4.002603 u -Constant: c2=931MeVu
Answer» In the radioactive decay - $^{238} P u \to^{234} U + α$ , What will be the kinetic energy of the outgoing α-particle? Given:- -Atomic masses: $m (^{238} P u) = 238.04955 u; m (^{234} U) = 234.04095 u;$ $m (^{4} H e) = 4.002603 u$ -Constant: $c^{2} = 931 \frac{M e V}{u}$

In the radioactive decay - 238Pu→234U+α, What will be the kinetic energy of the outgoing α-particle? Given:- -Atomic masses: m(238Pu)=238.04955 u; m(234U)=234.04095 u; m(4He)=4.002603 u -Constant: c2=931MeVu

Answer»

In the radioactive decay - $^{238} P u \to^{234} U + α$ ,
What will be the kinetic energy of the outgoing α-particle?
Given:-

-Atomic masses:

$m (^{238} P u) = 238.04955 u; m (^{234} U) = 234.04095 u;$

$m (^{4} H e) = 4.002603 u$

-Constant: $c^{2} = 931 \frac{M e V}{u}$