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(a) Find the maximum frequency of the X-rays emitted by an X-ray tube operating at 30 kV. (b) An X-ray tube operates at 20 kV. A particular electron loses 5% of its kinetic energy to emit an X-ray photon at the first collision. Find the wavelength corresponding to this photon. (c ) An X-ray tube is operated at 20 kV and the current through the tube is 0.5 mA. Find (i) the number of electrons hitting the target per second, (ii) the energy falling on the target per second as the kinetic anergy of the electrons and (iii) the cut-off wavelength of the X-rays emitted. |
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Answer» Solution :(a) `lambda_(min)=(hc)/(EV), V=30 kV` `lambda_(min)=1242/(E(eV))=1242/(30xx10^(3))=41.4xx10^(-3) nm` `=41.4xx10^(-12)m` `v_(max)=c/lambda(min)=(3xx10^(8))/(41.4xx10^(-12))=7.2xx10^(18) Hz` (B) KINETIC energy of electron `=20 keV` `5%` of this converted to photon `5/100xx20=1 ke V` `lambda=1242/(1xx10^(3)) nm=1.242 nm` (c ) `V=20 kV, i=0.5 mA` (i) `i=n eimpliesn=i/e=(0.5xx10^(-3))/(1.6xx10^(-19))=3.125xx10^(15)` (ii) `P=Vi=20xx0.5==10W` (iii) `lambda_(min)=1242/(20xx10^(3))=0.062 nm` |
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