Qalaxia Master Bot
0

I found an answer from byjus.com

Alpha Decay - Definition, Equation, Examples, Formula


The emitted alpha particle is also known as a helium nucleus. ... Thus, emitting its two protons and two neutrons in the form of an alpha particle and a forming ... In nuclear physics, Alpha decay Equation for thorium-232 can be written as – ... This decay in a nucleus causes the release of energy and matter from the nucleus.


For more information, see Alpha Decay - Definition, Equation, Examples, Formula

Pravalika
0

a)

Step 1:Formula required to calculate the energy released during the \alpha -decay of ^{238}_{92} U .

As a nucleus undergoes alpha-decay, it turns it into a separate nucleus by emitting an alpha-particle, For instance, ^{238}_{92} U transforms   ^{234}_{90}Th when it undergoes alpha decay.

^{238}_{92} U \rightarrow ^{234}_{90}Th + ^{4}_2He


The decay value for disintegration energy (Q) can be expressed by

Q = (m_X - m_Y - m_{He}) * c^2

or

Q = (m_U - m_{Th} - m_{He}) * c^2


Step 2: Replacing values of atomic masses into disintegration energy

^{238}_{92} U = 238.05079 u

^{234}_{90} Th = 234.04363 u

^{2}_{4} He = 4.00260 u

Q = (m_U - m_{Th} - m_{He}) * c^2

Q = (238.05079 - 234.04363 - 4.00260) * c^2

Q = (0.00456) * (931.5 MeV/u)

Q = 4.25 MeV


b)

Step 1: The decay process, if a proton is spontaneously emitted

^{238}_{92} U \rightarrow ^{237}_{91} Th + ^{1}_1He

Step 2: Replacing values of atomic masses into disintegration energy

^{238}_{92} U = 238.05079 u

^{1}_{1} He = 1.00783 u

^{237}_{91} Pa = 237.05121 u

Q = (m_{U} - m_{Pa} - m_{He}) * c^2

Q = (238.05079 - 237.05121 - 1.00783) * c^2

Q = - (0.00825 u) * c^2

Q = - (0.00825 u)(931.5 MeV/u)

Q = - 7.68 MeV

The Q of the process is thus negative and cannot therefore be spontaneously proceeded. To make it emit one proton, we will have to provide an energy of 7.68 MeV to a ^{238}_{92} U nucleus.