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NEET-XII-Physics

29: Electric Field and Potential

with Solutions - page 4
Qstn# iv-15 Prvs-QstnNext-Qstn
  • #15
    Suppose an attractive nuclear force acts between two protons which may be written as F=Ce-kr/r2. (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi-1 and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C. (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi-1 and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.
    Ans : (a) Given, nuclear force of attraction, `` F=C\frac{{e}^{-Kr}}{{r}^{2}}``
    Here e-Kr is just a pure number, i.e. a dimensionless quantity. So,
    `` \left[C\right]=\left[F\right]\times \left[{r}^{2}\right]``
    `` \left[C\right]=\left[{\,\mathrm{\,MLT\,}}^{-2}\right]\times \left[{\,\mathrm{\,L\,}}^{2}\right]``
    `` \left[\,\mathrm{\,C\,}\right]=\left[{\,\mathrm{\,ML\,}}^{3}{\,\mathrm{\,T\,}}^{-2}\right]``
    `` \,\mathrm{\,And\,}\left[K\right]=\frac{1}{\left[r\right]}=\left[{\,\mathrm{\,L\,}}^{-1}\right]`` (b) By Coulomb's Law, electric force,
    `` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{e}^{2}}{{r}^{2}}``
    Taking `` r=5\times {10}^{-15}\,\mathrm{\,m\,}``, we get
    `` F=\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{{\left(5\times {10}^{-15}\right)}^{2}}``
    And nuclear force, F = Ce-kr/r2
    Taking r = 5 × 10-15 m and k = 1 fermi-1, we get
    `` F=\frac{C\times {10}^{-5}}{{\left(5\times {10}^{-15}\right)}^{2}}``
    `` ``
    Comparing both the forces, we get
    `` C=3.4\times {10}^{-26}\,\mathrm{\,N\,}-{\,\mathrm{\,m\,}}^{2}``
    Page No 121: (a) Given, nuclear force of attraction, `` F=C\frac{{e}^{-Kr}}{{r}^{2}}``
    Here e-Kr is just a pure number, i.e. a dimensionless quantity. So,
    `` \left[C\right]=\left[F\right]\times \left[{r}^{2}\right]``
    `` \left[C\right]=\left[{\,\mathrm{\,MLT\,}}^{-2}\right]\times \left[{\,\mathrm{\,L\,}}^{2}\right]``
    `` \left[\,\mathrm{\,C\,}\right]=\left[{\,\mathrm{\,ML\,}}^{3}{\,\mathrm{\,T\,}}^{-2}\right]``
    `` \,\mathrm{\,And\,}\left[K\right]=\frac{1}{\left[r\right]}=\left[{\,\mathrm{\,L\,}}^{-1}\right]`` (b) By Coulomb's Law, electric force,
    `` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{e}^{2}}{{r}^{2}}``
    Taking `` r=5\times {10}^{-15}\,\mathrm{\,m\,}``, we get
    `` F=\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{{\left(5\times {10}^{-15}\right)}^{2}}``
    And nuclear force, F = Ce-kr/r2
    Taking r = 5 × 10-15 m and k = 1 fermi-1, we get
    `` F=\frac{C\times {10}^{-5}}{{\left(5\times {10}^{-15}\right)}^{2}}``
    `` ``
    Comparing both the forces, we get
    `` C=3.4\times {10}^{-26}\,\mathrm{\,N\,}-{\,\mathrm{\,m\,}}^{2}``
    Page No 121:
  • #15-a
    Write down the dimensional formulae and appropriate SI units of C and k.
    Ans : Given, nuclear force of attraction, `` F=C\frac{{e}^{-Kr}}{{r}^{2}}``
    Here e-Kr is just a pure number, i.e. a dimensionless quantity. So,
    `` \left[C\right]=\left[F\right]\times \left[{r}^{2}\right]``
    `` \left[C\right]=\left[{\,\mathrm{\,MLT\,}}^{-2}\right]\times \left[{\,\mathrm{\,L\,}}^{2}\right]``
    `` \left[\,\mathrm{\,C\,}\right]=\left[{\,\mathrm{\,ML\,}}^{3}{\,\mathrm{\,T\,}}^{-2}\right]``
    `` \,\mathrm{\,And\,}\left[K\right]=\frac{1}{\left[r\right]}=\left[{\,\mathrm{\,L\,}}^{-1}\right]``
  • #15-b
    Suppose that k = 1 fermi-1 and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.
    Ans : By Coulomb's Law, electric force,
    `` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{e}^{2}}{{r}^{2}}``
    Taking `` r=5\times {10}^{-15}\,\mathrm{\,m\,}``, we get
    `` F=\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{{\left(5\times {10}^{-15}\right)}^{2}}``
    And nuclear force, F = Ce-kr/r2
    Taking r = 5 × 10-15 m and k = 1 fermi-1, we get
    `` F=\frac{C\times {10}^{-5}}{{\left(5\times {10}^{-15}\right)}^{2}}``
    `` ``
    Comparing both the forces, we get
    `` C=3.4\times {10}^{-26}\,\mathrm{\,N\,}-{\,\mathrm{\,m\,}}^{2}``
    Page No 121: