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CBSE-XI-Physics

47: The Special Theory of Relativity

with Solutions - page 5

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  • Qstn #23
    An electron and a positron moving at small speeds collide and annihilate each other. Find the energy of the resulting gamma photon.
    Ans : We know,
    Mass of electron = Mass of positron = 9.1 × 10-31 kg
    Both are oppositely charged and annihilate each other to form a gamma photon of rest mass zero. Thus,
    ∆m = me + mp = 2 × 9.1 × 10-31 kg
    This mass will be converted into energy of the resulting γ photon. Thus,
    Eγ= ∆mc2
    Eγ = 2 × 9.1 × 10-31 × 9 × 1016 J
    `` =\frac{2\times 9.1\times 9\times {10}^{-15}}{1.6\times {10}^{-19}}``
    `` =102.37\times {10}^{4}\,\mathrm{\,eV\,}``
    `` =1.02\times {10}^{5}\,\mathrm{\,eV\,}``
    `` =1.02\,\mathrm{\,MeV\,}``
    Page No 458:
  • Qstn #24
    Find the mass, the kinetic energy and the momentum of an electron moving at 0.8c.
    Ans : We know,
    Rest mass of electron, m0 = 9.1 × 10-31 kg
    Velocity of electron, v = 0.8 c
    (a) Mass of electron is given by
    `` m=\frac{{m}_{0}}{\sqrt{1-{v}^{2}/{c}^{2}}}``
    `` \Rightarrow m=\frac{9.1\times {10}^{-31}}{\sqrt{1-0.64{c}^{2}/{c}^{2}}}=\frac{9.1\times {10}^{-31}}{0.6}``
    `` \Rightarrow m=15.16\times {10}^{-31}\,\mathrm{\,kg\,}\approx 15.2\times {10}^{-31}\,\mathrm{\,kg\,}``
    (b) Kinetic energy of electron = mc2 - m0c2
    `` \,\mathrm{\,KE\,}=\left(m-{m}_{0}\right){c}^{2}``
    `` =\left(15.2-9.1\right)\times {10}^{-31}\times 9\times {10}^{16}``
    `` =5.5\times {10}^{-14}\,\mathrm{\,J\,}``
    (c) Momentum of electron, p = mv
    `` p=15.2\times {10}^{-31}\times 0.8\times 3\times {10}^{8}``
    `` =3.65\times {10}^{22}\,\mathrm{\,kgm\,}/\,\mathrm{\,s\,}``
    Page No 458:
  • Qstn #25
    Through what potential difference should an electron be accelerated to give it a speed of
    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)
  • #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:
  • Qstn #26
    Find the speed of an electron with kinetic energy
    Ans : If m0 is the rest mass of an electron and c is the speed of light, then kinetic energy of the electron = mc2 - m0c2 ...(1)
    If `` m=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}``, then
  • #26-a
    1 eV,
    Ans : Kinetic energy of electron = 1 eV = `` 1.6\times {10}^{-19}\,\mathrm{\,J\,}``
    From eq. (1), we get
    `` 1.6\times {10}^{-19}=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}-{m}_{0}{c}^{2}``
    `` \Rightarrow \frac{1.6\times {10}^{-19}}{{m}_{0}{c}^{2}}=\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)``
    `` \Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}-1=\frac{1.6\times {10}^{-19}}{9.1\times {10}^{-31}\times 9\times {10}^{16}}``
    `` \Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}-1=0.019536\times {10}^{-4}``
    `` \Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}=1+0.019536\times {10}^{-4}``
    `` \Rightarrow \frac{1}{\sqrt{1-v/{c}^{2}}}=\frac{1}{1.0000019536}``
    `` \Rightarrow 1-{v}^{2}/{c}^{2}=0.99999613``
    `` \Rightarrow {v}^{2}/{c}^{2}=0.00000387``
    `` \Rightarrow v/c=0.001967231=3\times 0.001967231\times {10}^{8}``
    `` =5.92\times {10}^{5}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
  • #26-b
    10 keV and
    Ans : Kinetic energy of electron = 10 keV`` =1.6\times {10}^{-19}\times 10\times {10}^{3}\,\mathrm{\,J\,}``
    `` {m}_{0}{c}^{2}\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)=1.6\times {10}^{-15}``
    `` \Rightarrow 9.1\times {10}^{-31}\times 9\times {10}^{16}\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)=1.6\times {10}^{-15}``
    `` \Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=\frac{1.6\times {10}^{-15}}{9.1\times 9\times {10}^{-15}}``
    `` \Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=\frac{1.6}{9.1\times 9}``
    `` \Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}=0.980838``
    `` \Rightarrow 1-{v}^{2}/{c}^{2}=0.962043182``
    `` \Rightarrow {v}^{2}/{c}^{2}=1-0.962043182``
    `` \Rightarrow {v}^{2}=0.341611359\times {10}^{18}``
    `` \Rightarrow v=0.584475285\times {10}^{8}``
    `` \Rightarrow v=5.85\times {10}^{7}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    `` ``
  • #26-c
    10 MeV.
    Ans : Kinetic energy of electron`` =10\,\mathrm{\,MeV\,}={10}^{7}\times 1.6\times {10}^{-19}\,\mathrm{\,J\,}``
    `` \Rightarrow \frac{{m}_{0}{c}^{2}}{2\sqrt{1-{v}^{2}-{c}^{2}}}-{m}_{0}{c}^{2}=1.6\times {10}^{-12}``
    `` \Rightarrow {m}_{0}{c}^{2}\left(\frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1\right)=1.6\times {10}^{-12}``
    `` \Rightarrow \frac{1}{\sqrt{1-{v}^{2}\mathit{/}{c}^{\mathit{2}}}}-1=\frac{1.6\times {10}^{-12}}{9.1\times 9\times {10}^{-31}\times {10}^{16}}``
    `` \Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=1.019536\times {10}^{3}``
    `` \Rightarrow \frac{1}{\sqrt{1-{v}^{2}/{c}^{2}}}-1=1019.536+1``
    `` \Rightarrow \sqrt{1-{v}^{2}/{c}^{2}}=0.000979877``
    `` \Rightarrow 1-{v}^{2}/{c}^{2}=0.99\times {10}^{-6}``
    `` \Rightarrow v=2.999999039\times {10}^{8}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    Page No 458:
  • Qstn #27
    What is the kinetic energy of an electron in electron volts with mass equal to double its rest mass?
    Ans : If m0 is the rest mass of an electron and c is the speed of light, then kinetic energy (E) of the electron = mc2 - m0c2. ...(1)
    According to the question,
    m = 2m0
    E = (2m0 - m0)c2
    E= m0c2 = 9.1 × 10-31 × 9 × 1016 J
    `` E=\frac{9.1\times 9}{1.6}\times \frac{{10}^{-15}}{{10}^{-19}}=51.18\times {10}^{4}\,\mathrm{\,eV\,}=511\,\mathrm{\,keV\,}``
    Page No 458:
  • Qstn #28
    Find the speed at which the kinetic energy of a particle will differ by 1% from its nonrelativistic value
    12 mov2.
    Ans : If m0 is the rest mass of a particle and c is the speed of light, then relativistic kinetic energy of particle = mc2 - m0c2. ...(1)
    If `` m=\frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}``, then nonrelativistic kinetic energy of particle = `` \frac{1}{2}{m}_{o}{v}^{2}``.
    According to the question,
    `` \frac{\left({\displaystyle \frac{{m}_{0}{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}}-{m}_{0}{c}^{2}\right)-{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}{1/2{m}_{0}{v}^{2}}=0.01``
    `` \Rightarrow \left[\frac{{c}^{2}\left(1+{\displaystyle \frac{{v}^{2}}{2{c}^{2}}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}{\displaystyle \frac{{v}^{2}}{{c}^{2}}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}}\times {\displaystyle \frac{5}{6}}{\displaystyle \frac{{v}^{6}}{{c}^{6}}}-1\right)}{{\displaystyle \frac{1}{2}}{v}^{2}}\right]=0.01``
    `` \Rightarrow \frac{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}+{\displaystyle \frac{3}{8}}{m}_{0}{v}^{4}/{c}^{2}+{\displaystyle \frac{15}{96}}{m}_{0}{v}^{6}/{c}^{4}-{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}=0.01``
    `` \Rightarrow \frac{{\displaystyle \frac{3}{8}}{m}_{0}{v}^{2}/{c}^{2}+{\displaystyle \frac{15}{96}}{m}_{0}{\,\mathrm{\,V\,}}^{6}/{c}^{4}}{{\displaystyle \frac{1}{2}}{m}_{0}{v}^{2}}=0.01``
    `` \Rightarrow \frac{3}{4}\frac{{v}^{2}}{{c}^{2}}+\frac{15\times 2{v}^{4}}{96}\frac{{v}^{2}}{{c}^{2}}=0.01``
    Neglecting the v4 term as it is very small, we get
    `` \frac{3}{4}\frac{{v}^{2}}{{c}^{2}}=0.01``
    `` \Rightarrow \frac{{v}^{2}}{{c}^{2}}=\frac{0.04}{3}``
    `` \Rightarrow \frac{v}{c}=\frac{0.2}{\sqrt{3}}``
    `` \Rightarrow v=\frac{2}{2\sqrt{3}}``
    `` c=\frac{0.02}{1.732}\times 3\times {10}^{8}=\frac{0.6}{1.722}\times {10}^{8}``
    `` =0.345\times 10\,\mathrm{\,m\,}/\,\mathrm{\,s\,}=3.46\times {10}^{7}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
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