InterviewSolution
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InterviewSolution
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(a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The specific charge to of the electron, i.e., its `e//m` is given to `1.76xx10^(11)Ckg^(-1)`. (b) Use the same formula you employ in (a) to obtain electron speed for an anode potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified? |
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Answer» (a)Potential difference across the evacuted tube , V=500 V Specific charge of an electron, e/m=`1.76xx10^(11) C kg^(-1)` The speed of each emitted electron is given by the relation for kinetic energy as : `KE=1/2mv^2=eV` `therefore v=((2eV)/m)^(1/2)=(2Vxxe/m)^(1/2)` `=(2xx500xx1.76xx10^11)^(1/2)=1.327xx10^(7)` m/s Therefore, the speed of each emitted electron is `1.327xx10^7` m/s (b)Potential of the anode , V=10 MV =`10xx10^6` V The speed of each electron is given as : `v=(2Ve/m)^(1/2)` `=(2xx10^7 xx 1.76 xx 10^11)^(1/2)` `=1.88xx10^(9)` m/s This result is wrong because nothing can move faster than light. In the above formula, the expression `(mv^2//2)` for energy can only be used in the non-relativistic limit, i.e. , for v lt lt c. For very high speed problems , relativistic equations must be considered for solving them.In the relativistic limit , the total energy is given as : `E=mc^2` Where, m=Relativistic mass `=m_0(1-v^2/c^2)^(1/2)` `m_0` = Mass of the particle at rest Kinetic energy is given as : `K=mc^2-m_0c^2` |
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