Tag: nuclear physics

Questions Related to nuclear physics

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

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

  1. $\alpha-decay$

  2. $\beta^--decay$

  3. $\beta^+-decay$

  4. k-capture

Show answer
Correct Option: B
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.

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

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

  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 $\gamma-decay$

Show answer
Correct Option: C
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.

The electron emitted in beta radiation originates from

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

  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

Show answer
Correct Option: C
Explanation:

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

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

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

  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 $\beta-decay$

Show answer
Correct Option: D
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.

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

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

  1. neutrino

  2. photon

  3. antineutrino

  4. graviton

Show answer
Correct Option: C
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.

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

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

  1. $146$

  2. $144$

  3. $140$

  4. $142$

Show answer
Correct Option: A
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.

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

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

  1. Positron

  2. $\beta$-particle

  3. $\alpha $-particle

  4. None of these

Show answer
Correct Option: A
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.

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

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

  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

Show answer
Correct Option: B
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.

Which of the following statement is correct?

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

  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.

Show answer
Correct Option: A

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

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

  1. Proton

  2. Neutron

  3. Electron

  4. Positron

  5. Deuteron

Show answer
Correct Option: B
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.