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Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

In which of the following processes, the number of protons in the nucleus increase?

  1. $\alpha-decay$
  2. $\beta^--decay$
  3. $\beta^+-decay$

  4. k-capture
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ \beta  }^{ - }\quad decay$ :
Example: $ _{ 6 }^{ 14 }{ C }\longrightarrow _{ 7 }^{ 14 }{ N }+{ e }^{ - }+\bar { { v } _{ e } }$
Atomic number increases by 1 unit, which implies increase in proton.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Atomic masses of two isobars $ _{29}^{63}Cu$ and $ _{30}^{64}Zn$ are $63.9298 u$ and $63.9292 u$, respectively. It can be concluded from this data that

  1. both the isobars are stable
  2. $^{64}Zn$ is radioactive, decaying to $^{64}Cu$ through $\beta-decay$
  3. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\beta-decay$
  4. $^{64}Cu$ is radioactive, decaying to $^{64}Zn$ through $\gamma-decay$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Zn$ with higher no.of nucleons has a lower mass than $Cu$, which means that the binding energy/nucleon is higher in $Zn$.
Which means that $Zn$ is more stable than $Cu$.
Hence, $Cu$ will have a tendency to convert to $Zn$ by radioactive decay
$\beta$ decay to change the atomic number.
$\gamma$ decay can't help in changing the no. of protons in the nucleus.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The electron emitted in beta radiation originates from

  1. inner orbits of atoms

  2. free electrons existing in nuclei

  3. decay of a neutron in a nucleus

  4. photon escaping from the nucleus
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The electron emitted in beta radiation may originates from neutron and it increases the atomic number $1$.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Masses of two isobars $ _{29}Cu^{64}$ and $ _{30}Zn^{64}$ are $63.9298\ u$ and $63.9292\ u$, respectively. It can be conclude from these data that

  1. Both the isobars are stable
  2. $Zn^{64}$ is radioactive, decaying to $Cu^{64}$ through $\beta-decay$
  3. $Cu^{64}$ is radioactive, decaying to $Zn^{64}$ through $\gamma-decay$
  4. $Cu^{64}$ is radioactive, decaying to $Zn^{64}$ through $\beta-decay$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Same no. of nucleons for $Cu^{64}$ and $Zn^{64}$
However, $M _{Cu}> M _{Zn}$
which indicates that the mass defect/ nuclear binding energy per nucleon is lesser of $Cu$, hence it will have a tendency to get to more stable form by $\beta $ decay.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Neutron decay in free space is given as follows
$ _{ 0 }{ n }^{ 1 }\longrightarrow _{ 1 }{ H }^{ 1 }+ _{ 1 }{ e }^{ 0 }+$[  ]
Then the parenthesis [  ] represents a

  1. neutrino

  2. photon

  3. antineutrino

  4. graviton
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the charge is already conserved in the decay process, proton cannot be ejected.

To conserve the spin angular momentum of the initial neutron particle, an antineutrino (chargeless) is ejected.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The number of neutrons in the element L in the following nuclear changes is 
$^{238} _{92}M\, \rightarrow\, ^x _y\, N\, +\, ^4 _2\, He$
$^X _YN\, \rightarrow\, ^A _BL\, +\, 2\beta^+$

  1. $146$
  2. $144$
  3. $140$
  4. $142$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ _{ 92 }^{ 238 }{ M }\rightarrow _{ y }^{ x }{ N }+ _{ 2 }^{ 4 }{ He }$
$y=92-2=90 \\ x=238-4=234$
$ _{ 90 }^{ 234 }{ N }\rightarrow _{ B }^{ A }L+2 _{ +1 }^{ o }{ \beta  }\\ B=90-2=88\\ A=234$
no. of neutrons $=234-88 \\ =146$
So, $(A)$ is correct option.
Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

$^{11} _{6}C\, \rightarrow\,  ^{11} _{5}B$ decay produces -

  1. Positron
  2. $\beta$-particle
  3. $\alpha $-particle

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The positron or antielectron is the antiparticle or the antimatter counterpart of the electron. The positron has an electric charge of $+1e$, a spin of $\cfrac{1}{2}$ and has the same mass as an electron.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

In radioactive decay process, the emitted negatively charged $\beta$ - particles are :

  1. the electrons present inside the nucleus

  2. the electrons produced as a result of the decay of neutrons inside the nucleus

  3. the electrons produced as a result of collisions between atoms

  4. the electrons orbiting around the nucleus
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$ - decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.
In negative $\beta$- decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted $\beta$- particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Which of the following statement is correct?

  1. The rest mass of a stable nucleus is less than the sum of the rest masses of ts separated nucleons.

  2. The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons

  3. In nuclear fission, energy is released by fusion two nuclei of medium mass (approximately 100 amu).

  4. In nuclear fission, energy is released by fragmentation of a very low nucleus.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Stable nucleus has mass deficit: binding energy means M_nucleus < Σ(m_nucleons). This mass defect equals binding energy/c². Option B is wrong. Option C mentions fusion, not fission. Option D has 'very low nucleus' which is garbled; should say 'very heavy nucleus'.

Multiple choice physics nuclear physics beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Find out the missing particle in the following nuclear reaction?
$^2 _1H+^{63} _{29}Cu \rightarrow ^{64} _{30}Zn+(?)$

  1. Proton

  2. Neutron

  3. Electron

  4. Positron

  5. Deuteron
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given reaction :    $^2 _1 H$  $+$  $^{63} _{29} Cu  \rightarrow      ^{64} _{30} Zn  $  $+$  $^A _Z X$

In a nuclear reaction, mass number and the atomic number are conserved.
$\therefore$    $2+ 63  = 64 + A$   $\implies A = 1$
Also,  $1 + 29  = 30+Z$    $\implies Z = 0$
Thus, the missing particle is $^1 _0 n$  i.e  neutron.