Respuesta :
Answer:
Following are the responses to the given question:
Explanation:
In point a:
Nucleus denoting, he refers to those same nuclei of the helium ( that is alpha particle)
[tex]{z}^{N^{A}}\to z-2^{N^{A-4}} + 2^{He^{4}}[/tex]
In point b:
Let the kinetic energy of [tex]\alpha[/tex] particle = [tex]K_1[/tex]
[tex]\to K_1 = \frac{mass \times (velocity)^2}{2}[/tex]
The velocity of [tex]\alpha[/tex] particle [tex](v_1)= (2 \times \frac{k_1}{m_1})^\frac{1}{2}[/tex]
Let daughter core velocity be =[tex]v_2[/tex]
Preserving linear acceleration since there was no external factor we can write
[tex]\to m_2 \times v_2 = m_1 \times v_1\\\\\to v_2 = \frac{m_1\times v_1}{m_2}\\\\\to v_2 = \frac{m_1\times (2 \times \frac{k_1}{m_1})^{\frac{1}{2}}}{m_2}[/tex]
Daughter nuclei energy can be written as:
[tex]\to \frac{mass \times (velocity)^2}{2}\\\\\to \frac{m_2 \times (m_1)^2 \times 2 \times k_1}{(2\times m_1 \times (m_2)^2)}\\\\ \to (\frac{m_1}{m_2}) \times k_1[/tex]
In point c:
Initial weight = M
Total product weight [tex]= m_1 + m_2[/tex]
You can use the total release energy (Q-factor) as =[tex](M - (m_1+m_2)) c^2[/tex]
