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Semiconductor Electronics: Materials, Devices and Simple Circuits

Mass-Energy and Nuclear Binding Energy

A nucleus X, initially a rest, undergoes alpha-decay according

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Solving time: 4 mins

A nucleus $X$ , initially a rest, undergoes alpha-decay according to the equation, $_{92}^{A} X \to_{Z}^{228} Y + α$ (a) Find the values of $A$ and $Z$ in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius $0.11 m$ in a uniform magnetic field of 3 tesla. Find the energy (in $MeV$ ) released during the process and the binding energy of the parent nucleus $X$ . Given that: $m (Y) = 228.03 amu; m (_{0}^{1} n) = 1.009 amu$ $m (_{2}^{4} He) = 4.003 amu; m (_{1}^{1} H) = 1.008 amu$

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Book:

Physics Galaxy Optics and Modern Physics Vol 4 (Ashish Arora)

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Samuel Discussed

A nucleus

X

, initially a rest, undergoes alpha-decay according to the equation,

_{92}^{A} X \to_{Z}^{228} Y + α

(a) Find the values of

A

and

Z

in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius

0.11 m

in a uniform magnetic field of 3 tesla. Find the energy (in

MeV

) released during the process and the binding energy of the parent nucleus

X

. Given that:

m (Y) = 228.03 amu; m (_{0}^{1} n) = 1.009 amu

m (_{2}^{4} He) = 4.003 amu; m (_{1}^{1} H) = 1.008 amu

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

(a) The given equation is,

_{92}^{A} X ⟶_{2}^{228} Y +_{2}^{4} He

A = 228 + 4 = 232

92 = z + 2 ∴ z = 90

\frac{m _{α} v _{α}^{2}}{r} = q v_{α} B

v_{α} = \frac{r qB}{m _{α}}

= \frac{1.1 \times 1 0 ^{2} \times 2 \times 1.6 \times 1 0 ^{- 19} \times 3 \times 1 0 ^{3}}{4.003 \times 1.66 \times 1 0 ^{- 27}}

= 4.0 \times 1 0^{6} m / s

From conservation of linear momentum,

m_{α} v_{α} v_{y} = \frac{m _{α} v _{α}}{m _{y}} = m_{y} v_{y} = \frac{( 4.003 ) ( 4.0 \times 1 0 ^{6} )}{( 228.03 )} = 7.0 \times 1 0^{4} m / s

Therefore, energy released during the process

= \frac{1}{2} [m_{α} v_{α}^{2} + m_{y} v_{y}^{2}] = \frac{( 1.66 \times 1 0 ^{- 27} )}{( 2 \times 1.6 \times 1 0 ^{- 13} )} [(4.003) (4.0 \times 1 0^{6})^{2} + (228.03) (7.0 \times 1 0^{4})^{2}] MeV = 0.34 MeV = \frac{0.34}{931.5} amu = 0.000365 amu

Therefore, mass of

_{92}^{232} X = m_{y} + m_{α} + 0.000365 = 232.033365 u

Mass defect

Δ m = 92 (1.008) + (232 - 92) (1.009) - 232.033365 = 1.962635 amu

Binding energy

= 1.962635 \times 931.5 MeV

= 1828.2 MeV

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Question Text	A nucleus $X$ , initially a rest, undergoes alpha-decay according to the equation, $_{92}^{A} X \to_{Z}^{228} Y + α$ (a) Find the values of $A$ and $Z$ in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius $0.11 m$ in a uniform magnetic field of 3 tesla. Find the energy (in $MeV$ ) released during the process and the binding energy of the parent nucleus $X$ . Given that: $m (Y) = 228.03 amu; m (_{0}^{1} n) = 1.009 amu$ $m (_{2}^{4} He) = 4.003 amu; m (_{1}^{1} H) = 1.008 amu$
Topic	Semiconductor Electronics: Materials, Devices and Simple Circuits
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	126

Text solutionVerified

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