Taking the values of atomic masses from the tables, find the maximum kinetic energy of beta particles emitted by Be 10 nuclei and the corresponding kinetic energy of the recoiling daughter nuclei formed directly in the ground state. Be 10 = 10 . 016711 amu B 10 = 10 . 016114 amu

After emitting a β - particle, the Be 4 10 turns into B 5 10 as shown in the below reaction. Be 4 10 → B 5 10 + e - + v ¯ e The electrons emitted will have maximum K E , only when the energy of the emitted anti-neutrino v ¯ e is zero. The atomic masses from the periodic table are, Be 10 = 10 . 016711 amu B 10 = 10 . 016114 amu The energy liberated in any nuclear process is always due to the loss of an equivalent amount of mass and in this case, it comes as the kinetic energy of the emitted electrons. The mass defect, ∆ m = m reactants ~ m products ∆ m = 597 × 10 - 6 amu So, the K E of the electrons is E = ∆ m c 2 , using Einstein's energy-mass relation. 597 × 10 - 6 amu × c 2 = 0 . 56 MeV ( c 2 = 931 . 5 Mev amu - 1 ) The momentum of electrons with this K E is, using the relation, p = 2 m K E = 0 . 941 MeV c Since the initial momentum is zero, the daughter nucleus must recoil with the same momentum. Therefore, the recoil energy of the daughter is given by, K E = p 2 2 m = ( 0 . 941 ) 2 2 × M d c 2 = ( 0 . 941 ) 2 2 × 10 × 938 MeV = 47 . 2 eV

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IMPORTANT

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Find the energy $Q$ liberated in $β^{-}$ and $β^{+}$ decays and in $K$ capture, if the masses of the parent atom $M_{p},$ the daughter atom $M_{d}$ and an electron $m$ are known.( $c$ is speed of light)

Important Questions on ATOMIC AND NUCLEAR PHYSICS

HARD

JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>RADIOACTIVITY>Q 6.239.

Taking the values of atomic masses from the tables, find the maximum kinetic energy of beta particles emitted by

Q = (M_{X} - M_{Y}) c^{2} + m_{e} c^{2}

nuclei and the corresponding kinetic energy of the recoiling daughter nuclei formed directly in the ground state.

${Be}^{10} = 10.016711 amu$
$B^{10} = 10.016114 amu$

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>RADIOACTIVITY>Q 6.240.

Evaluate the amount of heat produced during a day by $β^{-}$ active ${Na}^{24}$ preparation of mass $m = 1.0 mg$ . The beta-particles are assumed to possess an average kinetic energy equal to $1 / 3$ of the highest possible energy of the given decay. The half-life of ${Na}^{24}$ is $T = 15$ hours.

Mass of ${Na}^{24}$ is $24.00903 amu$ and of ${Mg}^{24}$ is $24.01496 amu$ , energy equivalent of neutrinos or electron is $0.511 MeV$

$1 amu = 1.67 \times 10^{- 24} g$ and $1 amu$ is equivalent to $931.5 MeV$ of energy, speed of light $3 \times 10^{8} {ms}^{- 1}$ .

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>RADIOACTIVITY>Q 6.241.

Taking the values of the atomic masses from the tables, calculate the kinetic energies of a positron and a neutrino emitted by $C^{11}$ nucleus for the case, when the daughter nucleus does not recoil.

Mass defect due to positron is $0.00213 amu$ , energy equivalent of neutrinos or electron is $0.511 MeV$

$1 amu = 1.67 \times 10^{- 24} g$ and $1 amu$ is equivalent to $931.5 MeV$ of energy, speed of light $3 \times 10^{8} {ms}^{- 1}$ .

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>RADIOACTIVITY>Q 6.242.

Find the kinetic energy of the recoil nucleus in the positron decay of $N^{13}$ nucleus for the case when the energy of the positron is maximum.

energy equivalent of neutrinos or electron is $0.511 MeV$

$1 amu = 1.67 \times 10^{- 24} g$ and $1 amu$ is equivalent to $931 MeV$ of energy, speed of light $3 \times 10^{8} {ms}^{- 1}$ .

HARD

JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>RADIOACTIVITY>Q 6.243.

From the tables of atomic masses, determine the velocity of a nucleus appearing as a result of

K

capture in a

{Be}^{7}

atom, provided the daughter nucleus turns out to be in the ground state.

energy equivalent of neutrinos or electron is $0.511 MeV$

$1 amu = 1.67 \times 10^{- 24} g$ and $1 amu$ is equivalent to $931 MeV$ of energy, speed of light $3 \times 10^{8} {ms}^{- 1}$ .

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>NUCLEAR REACTIONS>Q 6.249.

Аn alpha particle with kinetic energy,

T_{α} = 7.0 MeV

, is scattered elastically by an initially stationary

{Li}^{6}

nucleus. Find the kinetic energy of the recoil nucleus if the angle of divergence of the two particles is

θ = 60 ° .

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>NUCLEAR REACTIONS>Q 6.250.

A neutron collides elastically with an initially stationary deuteron. Find the fraction of the kinetic energy lost by the neutron,
(a) in a head-on collision.
(b) in scattering at right angles.

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 6 - ATOMIC AND NUCLEAR PHYSICS>NUCLEAR REACTIONS>Q 6.251.

Find the greatest possible angle through which a deuteron is scattered as a result of elastic collision with an initially stationary proton.

Find the energy Q liberated in β- and β+ decays and in K capture, if the masses of the parent atom Mp, the daughter atom Md and an electron m are known.( c is speed of light)

Important Questions on ATOMIC AND NUCLEAR PHYSICS

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Find the energy $Q$ liberated in $β^{-}$ and $β^{+}$ decays and in $K$ capture, if the masses of the parent atom $M_{p},$ the daughter atom $M_{d}$ and an electron $m$ are known.( $c$ is speed of light)