# A nucleus X- initially at rest, undergoes alpha- decay, according to the equation. 92XA → ZY228 + α

Q: A nucleus X- initially at rest, undergoes alpha- decay, according to the equation.
92XAZY228 + α
The α – particle in the above process is found to move in a circular track of radius 1.1 × 102 m in a uniform magnetic field of 3.0 × 103 T. The energy (in MeV) released during the process and binding energy of the parent nucleus X,
respectively.
Given : my = 228.03 amu ; mα = 4.003 amu,
m(0n1 ) = 1.009 amu ; m (0H1 ) = 1.008 amu ,
1 amu = 1.66 × 10-27 kg  ≡ 931.5 MeV / c2

Sol: The given equation is 92X AZY228 + 2He4

A = 228 + 4 = 232 ; 92 = z + 2  ⇒ z = 90

$\large m_{\alpha} v_{\alpha}^2/r = q v_{\alpha} B$

$\large v_{\alpha} = \sqrt{\frac{q B r}{m_{\alpha}}}$

$\large v_{\alpha} = \sqrt{\frac{1.6 \times 10^{-19} \times 3 \times 10^3 \times 1.1 \times 10^2 }{4.003 \times 1.66 \times 10^{-27}}}$

= 4 .0 × 106 m/s

From conservation of linear momentum, $\large m_{\alpha} v_{\alpha} = m_y v_y$

$\large v_y = \frac{m_{\alpha} v_{\alpha}}{m_y}$

$\large v_y = \frac{4.003 \times 4 \times 10^6 }{228.03}$

= 7.0 × 104 m/s

Therefore, energy released during the process.

$\large = \frac{1}{2}[m_{\alpha} v_{\alpha}^2 + m_y v_y^2]$

= 0.34 MeV = 0.34/931.5 amu = 0.000365 amu

Therefore, mass of 92X232 = my + mα + 0.000365

= 232. 033365 u

Mass defect , Δm = 92 (1.008) + (232- 92) (1.009) – 232. 03365.

Binding energy = 1.962635 × 931.5 meV.

= 1828.2 MeV.