JEE Challenger
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PhyHardAdv 2013 • Paper 2

Kinetic Energy Calculation of Alpha Decay Product from Po210

Comprehension Passage

The mass of a nucleus ZAX_Z^A X is less than the combined mass of its (A−Z)(A-Z) constituent neutrons and ZZ protons. The energy equivalent of this mass difference represents the binding energy of the nucleus. A heavy nucleus with mass MM can disintegrate into two lighter nuclei of masses m1m_1 and m2m_2 provided that (m1+m2)<M(m_1+m_2) < M. Likewise, two light nuclei having masses m3m_3 and m4m_4 can undergo fusion to form a heavy nucleus of mass M′M' provided that (m3+m4)>M′(m_3+m_4) > M'. Atomic masses of certain neutral atoms are given in the following table:

11H1.007825 u12H2.014102 u13H3.016050 u24He4.002603 u36Li6.015123 u37Li7.016004 u3070Zn69.925325 u3482Se81.916709 u64152Gd151.919803 u82206Pb205.974455 u83209Bi208.980388 u84210Po209.982876 u\begin{array}{|c|c|c|c|c|c|c|c|} \hline _1^1 H & 1.007825\text{ u} & _1^2 H & 2.014102\text{ u} & _1^3 H & 3.016050\text{ u} & _2^4 He & 4.002603\text{ u} \\ \hline _3^6 Li & 6.015123\text{ u} & _3^7 Li & 7.016004\text{ u} & _{30}^{70} Zn & 69.925325\text{ u} & _{34}^{82} Se & 81.916709\text{ u} \\ \hline _{64}^{152} Gd & 151.919803\text{ u} & _{82}^{206} Pb & 205.974455\text{ u} & _{83}^{209} Bi & 208.980388\text{ u} & _{84}^{210} Po & 209.982876\text{ u} \\ \hline \end{array}

Use 1 u=932 MeV/c21\text{ u} = 932\text{ MeV}/c^2.

When an initial stationary nucleus of _{84}^{210} Po undergoes alpha decay, determine the kinetic energy (expressed in keV\text{keV}) possessed by the released alpha particle.

Options

A

5319

Correct
B

5422

C

5707

D

5818

Topics & Concepts

NucleiRadioactivityNuclear Reactions

Step-by-Step Solution

To determine the kinetic energy of the alpha particle (α\alpha-particle) produced during the alpha decay of a stationary 84210Po_{84}^{210}\text{Po} nucleus, we begin by writing the nuclear reaction equation:

84210Po⟶82206Pb+24He_{84}^{210}\text{Po} \longrightarrow _{82}^{206}\text{Pb} + _{2}^{4}\text{He}

From the given table of atomic masses:

  • Mass of parent nucleus, m(84210Po)=209.982876 um\left(_{84}^{210}\text{Po}\right) = 209.982876\text{ u}
  • Mass of daughter nucleus, m(82206Pb)=205.974455 um\left(_{82}^{206}\text{Pb}\right) = 205.974455\text{ u}
  • Mass of alpha particle, m(24He)=4.002603 um\left(_{2}^{4}\text{He}\right) = 4.002603\text{ u}

Step 1: Calculate the Mass Defect (Δm\Delta m)

The mass defect of the reaction is given by: Δm=m(84210Po)−[m(82206Pb)+m(24He)]\Delta m = m\left(_{84}^{210}\text{Po}\right) - \left[ m\left(_{82}^{206}\text{Pb}\right) + m\left(_{2}^{4}\text{He}\right) \right]

Substitute the given values into the equation: Δm=209.982876 u−(205.974455 u+4.002603 u)\Delta m = 209.982876\text{ u} - (205.974455\text{ u} + 4.002603\text{ u}) Δm=209.982876 u−209.977058 u=0.005818 u\Delta m = 209.982876\text{ u} - 209.977058\text{ u} = 0.005818\text{ u}

Step 2: Calculate the Total Energy Released (QQ-value)

Given 1 u=932 MeV/c21\text{ u} = 932\text{ MeV}/c^2: Q=Δm×932 MeVQ = \Delta m \times 932\text{ MeV} Q=0.005818×932 MeV=5.422376 MeV=5422.376 keVQ = 0.005818 \times 932\text{ MeV} = 5.422376\text{ MeV} = 5422.376\text{ keV}

Step 3: Calculate the Kinetic Energy of the Alpha Particle (KαK_\alpha)

Since the initial nucleus is stationary, conservation of linear momentum requires that the daughter nucleus and the alpha particle recoil with equal and opposite momentum (pα=pPbp_\alpha = p_{\text{Pb}}).

The kinetic energy is shared between the alpha particle and the recoil daughter nucleus according to their mass ratio: Kα=(A−4A)QK_\alpha = \left( \frac{A - 4}{A} \right) Q

Where A=210A = 210 is the mass number of the parent nucleus (84210Po_{84}^{210}\text{Po}): Kα=(206210)×5422.376 keVK_\alpha = \left( \frac{206}{210} \right) \times 5422.376\text{ keV} Kα≈5319.09 keV≈5319 keVK_\alpha \approx 5319.09\text{ keV} \approx 5319\text{ keV}

Thus, the kinetic energy possessed by the released alpha particle is approximately 5319 keV5319\text{ keV}.

Correct Answer: A

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