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CBSE-XI-Physics

47: The Special Theory of Relativity

with Solutions - page 5
Qstn# iv-25 Prvs-QstnNext-Qstn
  • #25
    Through what potential difference should an electron be accelerated to give it a speed of (a) 0.6c, (b) 0.9c and (c) 0.99c?
    Ans : Kinetic energy of electron = mc2 - m0c2 ...(1)
    Suppose the electron is accelerated through a potential difference of V. Then,
    KE of electron = eV
    `` m=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}``
    Putting the values of m and KE in eq. (1), we get
    `` eV=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}-{m}_{0}{c}^{2}`` ...(2) (a) Velocity of electron, v = 0.6c
    Rest mass of electron, m0 = `` 9.1\times {10}^{-31}\,\mathrm{\,kg\,}``
    Charge of electron, e = `` 1.6\times {10}^{-19}\,\mathrm{\,C\,}``
    Putting the values of m0, v and e in eq. (2), we get
    `` e\,\mathrm{\,V\,}=\frac{9.1\times {10}^{-31}\times 9\times {10}^{16}}{\sqrt{1-{\displaystyle \frac{0.36{c}^{2}}{{c}^{2}}}}}-9.1\times {10}^{-31}\times 9\times {10}^{16}``
    `` \Rightarrow V=\frac{9.1\times {10}^{-31}\times 9\times {10}^{16}\left(1.25-1\right)}{1.6\times {10}^{-19}}``
    `` \Rightarrow V=0.51\times 0.25\,\mathrm{\,MV\,}``
    `` \Rightarrow V=0.1275\,\mathrm{\,Mev\,}=127.5\,\mathrm{\,kV\,}`` (b) Putting v = 0.9c in eq. (2), we get
    V = 661 kV
    (b) Putting v = 0.99c in eq. (2), we get
    V = 3.1 MV
    Page No 458:
  • #25-a
    0.6c,
    Ans : Velocity of electron, v = 0.6c
    Rest mass of electron, m0 = `` 9.1\times {10}^{-31}\,\mathrm{\,kg\,}``
    Charge of electron, e = `` 1.6\times {10}^{-19}\,\mathrm{\,C\,}``
    Putting the values of m0, v and e in eq. (2), we get
    `` e\,\mathrm{\,V\,}=\frac{9.1\times {10}^{-31}\times 9\times {10}^{16}}{\sqrt{1-{\displaystyle \frac{0.36{c}^{2}}{{c}^{2}}}}}-9.1\times {10}^{-31}\times 9\times {10}^{16}``
    `` \Rightarrow V=\frac{9.1\times {10}^{-31}\times 9\times {10}^{16}\left(1.25-1\right)}{1.6\times {10}^{-19}}``
    `` \Rightarrow V=0.51\times 0.25\,\mathrm{\,MV\,}``
    `` \Rightarrow V=0.1275\,\mathrm{\,Mev\,}=127.5\,\mathrm{\,kV\,}``
  • #25-b
    0.9c and (c) 0.99c?
    Ans : Putting v = 0.9c in eq. (2), we get
    V = 661 kV
    (b) Putting v = 0.99c in eq. (2), we get
    V = 3.1 MV
    Page No 458: