Tag: nuclei

Quizzes (7) Questions (160)

Questions Related to nuclei

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

Mass numbers of the elements A, B, C and D are 30, 60, 90 and 120 respectively. the  specific binding energy of them are 5 MeV, 8.5 MeV, 8 MeV and 7 MeV respectively. then, in which of the following reaction/s energy is released?
(1) $ D \rightarrow 2B $
(2) $ C \rightarrow B+A $
(3) $ B \rightarrow 2A $

  1. only in (1)

  2. in(2), (3)

  3. in (1), (3)

  4. in (1), (2) and (3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Energy is released when the product has a higher specific binding energy than the reactants. (1) D(7 MeV) -> 2B(8.5 MeV): 8.5 > 7, so energy is released. (2) C(8 MeV) -> B(8.5 MeV) + A(5 MeV): Average BE is (8.5+5)/2 = 6.75 < 8, so energy is absorbed. (3) B(8.5 MeV) -> 2A(5 MeV): 5 < 8.5, so energy is absorbed.

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

A hydrogen atom having kinetic energy $E$ collides with a stationary hydrogen atom. Assume all motions are taking place along the line of motion of the moving hydrogen atom. For this situation, mark out the correct statement(s)

  1. For $E\ge20.4\space eV$ only, collision would be elastic
  2. For $E\ge20.4\space eV$ only, collision would be inelastic
  3. For $E = 2.4\space eV$, collision would be perfectly inelastic
  4. For $E = 18\space eV$, the $KE$ of initially moving hydrogen atom after collision is zero
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

K.E=2P.E
For electron in hydrogen to excite, a minimum of 10.2eV energy is required. Therefore, minimum 20.4eV K.E is required for inelastic collision otherwise, electron would not accept energy. And if E=20.4eV, collision would be perfectly inelastic.
If E is less than 20.4eV, collision is elastic and the two hydrogen atoms exchange velocities.
Therefore, B,D are the correct answers.

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

The mass number of an element in a radioactive series is 223. Then the radioactive series is ................

  1. 4n

  2. 4n+3

  3. 4n+2

  4. 4n+1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Radioactive series are classified by their mass number modulo 4. For a mass number 223, 223 / 4 = 55 with a remainder of 3. Thus, it belongs to the 4n+3 series (Actinium series).

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

A radio isotope X has a half life of $10s$. Find the number of active nuclei in the sample (if initally there are $1000$ isotopes which are falling from rest from a height of $3000m$) when it is at a height of $1000m$ from the reference plane: 

  1. $50$
  2. $250$
  3. $29$
  4. $100$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time taken in falling a height $h=3000-1000=2000m$ 

is given as $t=\sqrt[2]{\dfrac{2h}{g}}$
putting $g=10,h=2000$ we get $t=20second$
number of half life in this time period is $n=20/10=2$
So number of active nuclei$ = initial/2^n=initial/2^2=inital/4=1000/4=250$
Option B is correct.

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

When a $\beta^-$ particle is emitted from a nucleus, the neutron-proton ratio:

  1. is decreased

  2. is increased

  3. remains the same

  4. first (A) then (B)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ _{A}^{Z}\textrm{X}$ $\rightarrow  _{A-1}^{Z}\textrm{Y} $  $+  \beta^{-1}$


So,  the neutron-proton ratio before emission $ = \dfrac{Z-A}{A}$

And, the neutron-proton ratio after emission $ = \dfrac{Z-A+1}{A-1}$
Since, $ \dfrac{Z-A+1}{A-1}$  $ >\dfrac{Z-A}{A}$
Therefore, B is correct option.

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

A certain mass of an ideal diatomic gas contained in a closed vessel is heated. It is observed that half the amount of gets dissociated, but the temperature remains constant. The ratio of the heat supplied to the gas to the initial internal energy of the gas will be

  1. $1:2$
  2. $1:4$
  3. $1:5$
  4. $1:10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For an ideal diatomic gas, the internal energy is U = (f/2)nRT, where f = 5. When half the gas dissociates into monoatomic gas, the total number of moles of atoms changes, but since temperature remains constant, the internal energy depends on the total degrees of freedom. Let initial moles be n. Initial internal energy U_i = (5/2)nRT. When half dissociates, let's analyze carefully: a diatomic molecule has 5 degrees of freedom, becoming 2 monoatomic atoms each with 3 degrees of freedom. Using energy conservation and heat supplied Q = Delta U + W, at constant temperature for dissociation, the heat supplied goes into bond dissociation energy and internal energy changes. With standard ideal gas dissociation problems where T is constant, Q equals the dissociation energy, and the ratio of heat supplied to initial internal energy simplifies to 1:10 based on standard derivation for diatomic dissociation.