Physics

Nuclear and Atomic Physics

571 Questions

Nuclear and atomic physics explores the components and properties of the nucleus, including isotopes, radioactive decay, and fundamental forces. These concepts are essential for various competitive exams requiring a strong foundation in physics. The provided questions cover structural properties, mass, energy equivalence, and particle interactions.

Nucleus propertiesIsotopes and mass numberAlpha particle scatteringNuclear forcesMass energy equivalenceBeta particle emission

Nuclear and Atomic Physics Questions

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

A radioactive nucleus has specific binding energy '${E} _{1}$'. It emits an $\alpha$-particle. The resulting nucleus has specific binding energy '${E} _{2}$'. Then

  1. ${E} _{2}={E} _{1}$
  2. ${E} _{2}<{E} _{1}$
  3. ${E} _{2}>{E} _{1}$
  4. ${E} _{2}=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A radioactive nucleus always decays into more stable nucleus and high binding energy corresponds to more stable nucleus. Therefore, after emitting an alpha particle energy $E _2$ will be more than the initial energy $E _1$.

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

The nature of the electrostatic force and nuclear force between a proton and a neutron inside a nucleus are respectively.

  1. Repulsive and attractive

  2. Zero and attractive

  3. Repulsion and repulsive

  4. Attractive and attractive
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since neutron is neutral in nature, the electrostatic force between neutron and proton is zero. They have attractive nuclear force between them which causes the nucleus to bind together. 

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

Range of nuclear force is approximately

  1. $2 \times 10^{-l5} m$
  2. $1.5 \times 10^{-20} m$
  3. $7.2 \times 10^{-4} m$
  4. $1.4 \times 10^{-15} m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The nuclear force is powerfully attractive between nucleons at distance of about 1 femtometre or $1.0\times{10}^{-15}$ metre, but it rapidly decreases to insignificance at distance beyond about 2.5fm
So, $(A)$ is correct.
Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

Which of the following statements is wrong

  1. Strong nuclear forces are the strongest forces

  2. Nuclear forces are very short range forces

  3. nuclear force increases when the number of nucleons is increased

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
There are four fundamental interactions account for all observed forces. These interactions are strong nuclear, electromagnetic, weak nuclear and gravitational
The strongest of these four is the strong nuclear. It is responsible for binding together the fundamental particles of matter to form larger particles.
However this interaction operates at a very short range inside the nucleus (as little as $1$fm- $1$ femo meter or ${ 10 }^{ -15 }$ meter)
Nuclear force is defined as the force exerted between numbers of nucleons. So nuclear force increases when the number of nucleons is increased. All statements are correct.
Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

The force between protons in the nucleus will b

  1. only nuclear

  2. only coulomb

  3. nuclear & coulomb

  4. coulomb & gravitational
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The electrostatic force between an electron and a proton is given by Coulombs Law of force, that is directly proportional; to the product of charges of electron and the proton and inversely proportional to the square of distance between the two particles
Another force on proton is nuclear force as it the force exerted between numbers of nucleons. This force is attractive in nature which binds protons and neutrons in the nucleus together.
Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

If $F _{NN}$, $F _{NP}$, $F _{PP}$ denotes net force between neutron and neutron, neutron and proton, proton and proton then

  1. $F _{NN}$ = $F _{NP}$ = $F _{PP}$
  2. $F _{NN}$ = $F _{NP}$ > $F _{PP}$
  3. $F _{NN}$ = $F _{NP}$ < $F _{PP}$
  4. $F _{NN}$ >$F _{NP}$>$F _{PP}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At separation less than one fermi, hence nuclear force of attraction is strongly active.
Nuclear force is charge independent force.
So, $F _{pp} = F _{pn} = F _{nn}$

Multiple choice chemistry nuclei nuclear force the nuclear force nuclear force and binding energy

Consider an $\alpha$-particle just in contact with a $ _{\;  92}^{238}\textrm{U}$ nucleus. The Coulombic repulsion energy  (i.e, the height of the Coulombic barrier between $^{238}\textrm{U}$ and alpha particle) assuming that the distance between them is equal to the sum of their radii is 

  1. $16.35 \, MeV$
  2. $46.66 \, MeV$
  3. $22.24 \, MeV$
  4. $26.14 \, MeV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The expression for the radius of the nucleus is as shown below.
$r _{nucleus} =1.3\times 10^{-13}(A)^{1/3}$; where $A$ is mass number
Radius of  $ _{92}^{238}\textrm{U}=1.3\times10^{-13}\times (238)^{1/3}$
                           $= 8.06\times 10^{-13}cm$
Radius of $ _{2}^{4}\textrm{He}=1.3\times10^{-13}\times (4)^{1/3}$
                         $=2.06\times10^{-13}cm$
Total distance between uranium and helium nuclei is equal to the sum of their radii. 

It is $=(8.06 + 2.06)\times10^{-13}=10.12\times10^{-13}cm$ 

The Coulombic repulsion energy is: 
$\displaystyle \frac{Q _1Q _2}{r}$ $\displaystyle =\frac{92\times 4.8\times 10^{-10}\times 2\times 4.8\times 10^{-10}}{10.12\times 10^{-13}}erg$                (because $Q _1$ and  $Q _2$  in  esu and r in cm)     
                                      
            $=418.9\times 10^{-7}erg= 418.9\times 10^{-14}$J

            $=418.9\times 10^{-14}/1.602\times 10^{-19}\ eV$
              
            $\displaystyle =\frac{26.14\times 10^6}{10^6}\ MeV$

            $=26.14 \, MeV$

Hence, the coulombic repulsion energy is $26.14\ MeV$.

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

Regarding a nucleus, choose the correct options :

  1. Density of a nucleus is directly proportional to mass number A.
  2. Nucleus radius $ \propto {{A}^{1/3}}$

  3. Nuclear forces are dependent on the nature of nucleons.

  4. Nuclear forces are short range forces.
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

Density of nucleus is: $\rho=\dfrac{A}{\dfrac{4}{3}\pi R^3}$
The radius of a nucleus, $R=r _0A^{1/3},$ so density of nucleus is independent of A and $R\propto {^3\sqrt{A}}$
The nuclear force is a short-range force because the distance between the nucleon is less than $0.7$ fermi (then the force is repulsive) and if greater than $10.7$ fermi (the force is attractive).

Multiple choice physics motion and measurement of distances motion around us motion and rest moving things around us

The distance of closest approach of an $\alpha $-particle projected towards a nucleus with momentum p is r. What will be the distance of closest approach, when the momentum of projected $\alpha $-particle is 2p?

  1. 2r

  2. 4r
  3. $\dfrac { r }{ 2 } $
  4. $\dfrac { r }{ 4 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At the distance of closest approach:$KE=PE$

$\dfrac{p^2}{2m}=\dfrac{k 2 eq}{r}$
When the momentum chnages to 2p:
$\dfrac{(2p)^2}{2m}=k\dfrac{2eq}{r'}$
$\dfrac{4p^2}{2m}=k\dfrac{2eq}{r'}$
$\dfrac{2p^2}{m}=k\dfrac{2eq}{r'}$
$r'=k\dfrac{2eq}{p^2} \times \dfrac{m}{2}$
$=\dfrac{r}{2m}\times \dfrac{m}{2}$
$r'=\dfrac{r}{4}$

Multiple choice detection and recording of x-ray images option c: imaging physics

From the $\alpha$-particle scattering experiment, Rutherford concluded that

  1. $\alpha$- particles can come within a distance of the order of $10^{-14}$ m of the nucleus.
  2. The radius of the nucleus is less than $10^{-14}$ m

  3. Scattering follows coulomb's law

  4. The positively charged parts of the atom move with extremely high velocities
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rutherford's alpha-particle scattering experiment allowed him to estimate the distance of closest approach, which provided an upper bound for the radius of the nucleus, concluding it is less than 10^-14 m.

Multiple choice detection and recording of x-ray images option c: imaging physics

In Compton effect, if the incident x-rays have low energy and the scattering atom has high atomic number then the electrons appear as

  1. bound with no measurable Compton shift

  2. free with measurable Compton shift

  3. bound with measurable Compton shift

  4. free with no measurable Compton shift
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If atomic number of an atom is high- suggests that more energy is required to eject the electron or electron is more tightly bound . If energy of x rays is low- suggests that x-ray photon possesses insufficient energy to cause any measurable effect on electron. Thus, considering both these factors, A is the correct option

Multiple choice physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

An $\alpha$-particle is the nucleus of a helium atom. It has a mass $m=6.64 \times 10^{-27}$kg and a charge $q = + 2e =  3.2 \times 10^{-19}$ C. Compare the force of the electric repulsion between two a-particles with the force of gravitational attraction between them.

  1. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{32}$
  2. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{35}$
  3. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{38}$
  4. $\displaystyle \dfrac{F _e}{F _g} = 1 \times 10^{32}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Electrostatic repulsion = $F _e=\dfrac{1}{4 \pi \epsilon } \dfrac{q^2}{r^2 }=\dfrac{1}{4 \pi \epsilon } \dfrac{4e^2}{r^2 } $

gravitational attraction = $ F _g = G\dfrac{m^2}{r^2} $

$\dfrac{F _e}{F _g} = \dfrac{1}{4 \pi \epsilon } \dfrac{4 e^2}{G m^2} = 3.1 \times 10^{35} $