Tag: the nuclear force

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Questions Related to the nuclear force

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 the ionization energy of hydrogen atom is $13.6 eV$ then the wavelength of the radiation required to excite the electron in $L{ i }^{ ++ }$ from first to third Bohr orbit is approximately

  1. $1140 A$
  2. $914 A$
  3. $11.4 A$
  4. $134 A$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ionisation energy is given by

$E = 13.6\ eV$ (given)
or $E = 13.6 \times 1.6 \times 10^{-19} J ....(1)$
also $E = hv$
$E = \dfrac{hc}{\lambda} ....(2)$
equation $(1)$ and $(2)$
$\dfrac{hc}{\lambda} = 13.6 \times 1.6 \times 10^{-19}$
$\lambda = \dfrac{h \times c}{13.6 \times 1.6 \times 10^{-19}}$
$\lambda = \dfrac{3 \times 10^{8} \times 6.63 \times 10^{-34}}{13.6 \times 1.6 \times 10^{-19}}$
$\lambda = 914 \times 10^{-10} m$
$\lambda = 914 A^o$

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).