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What should be the approzimate kinetic energy of an electron so that its de-Broglie wavelength is equal to the wavelength of X-ray of maximum energy, produced in an X-ray tube operating at 24800 V? (given that `h=6.6xx10^(-34)` joule-sec, mass of electron `=9.1xx10^(-31)kg`)A. `600 eV`B. `36.5 eV`C. `6000 eV`D. `300 eV` |
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Answer» The wavelength `lambda_(min)` emitted by the `X-` ray tube operating at a voltage `V` is given by `eV=(hc)/(lambda_(min))` Kinetic energy of the electron `=(1)/(2)mv^(2)=(1)/(2)((m^(2)v^(2))/(m))=(h^(2))/(2mlambda_(1)^(2))` Now the wavelength `lambda` of an electron moving with the velocity `v` is given by `lambda=h//mvrArrlambda=(h)/(sqrt(2meV))` `:. K.E.=600eV` |
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