Tag: nuclei

Questions Related to nuclei

Range of weak nuclear force is

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

  1. $10^{16}$ m

  2. $10^{10}$ m

  3. $10^{-12}$ m

  4. $10^{-18}$ m

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Correct Option: D

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 $

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

  1. only in (1)

  2. in(2), (3)

  3. in (1), (3)

  4. in (1), (2) and (3)

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Correct Option: A

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

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

  1. $F _{NN}$ = $F _{NP}$ = $F _{PP}$

  2. $F _{NN}$ = $F _{NP}$ > $F _{PP}$

  3. $F _{NN}$ = $F _{NP}$ < $F _{PP}$

  4. $F _{NN}$ >$F _{NP}$>$F _{PP}$

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Correct Option: A
Explanation:

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

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 

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

  1. $16.35 \, MeV$

  2. $46.66 \, MeV$

  3. $22.24 \, MeV$

  4. $26.14 \, MeV$

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Correct Option: D
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 $26.14\ MeV$.

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)

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

  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

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Correct Option: B,D
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.

Regarding a nucleus, choose the correct options :

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

  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.

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Correct Option: B,D
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).

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

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

  1. 4n

  2. 4n+3

  3. 4n+2

  4. 4n+1

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Correct Option: B

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: 

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

  1. $50$

  2. $250$

  3. $29$

  4. $100$

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Correct Option: B
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.

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

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

  1. is decreased

  2. is increased

  3. remains the same

  4. first (A) then (B)

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Correct Option: B
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.

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

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

  1. $1:2$

  2. $1:4$

  3. $1:5$

  4. $1:10$

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Correct Option: A