Q: A nucleus X- initially at rest, undergoes alpha- decay, according to the equation.

_{92}X^{A} → _{Z}Y^{228} + α

The α – particle in the above process is found to move in a circular track of radius 1.1 × 10^{2} m in a uniform magnetic field of 3.0 × 10^{3} T. The energy (in MeV) released during the process and binding energy of the parent nucleus X,

respectively.

Given : m_{y} = 228.03 amu ; m_{α} = 4.003 amu,

m(_{0}n^{1} ) = 1.009 amu ; m (_{0}H^{1} ) = 1.008 amu ,

1 amu = 1.66 × 10^{-27} kg ≡ 931.5 MeV / c^{2}

Sol: The given equation is _{92}X ^{A} → _{Z}Y^{228} + _{2}He^{4}

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 × 10^{6} 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 × 10^{4} 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 _{92}X^{232} = m_{y} + m_{α} + 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.