Eletrons with de- Broglie wavwlength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays is
A
`lambda_(0) = (2mclambda^(2))/(h)`
B
`lambda_(0) = (2h)/(mc) `
C
`lambda_(0) = (2m^(2) c^(2) lambda^(3))/(h^(2))`
D
`lambda_(0) = lambda`
Text Solution
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The correct Answer is:
A
The cut off wavelength is given by `lambda_(0) = (hc)/(eV)` …..(i) according to de broglie equation `lambda = (h)/(p) = (h)/(sqrt(2 MeV))` `rArr lambda^(2) = (h^(2))/(2meV) rArr V = ( (h^(2))/(2me lambda^(2))` ….(ii) From (i) and (ii) , `lambda _(0) = (hc xx 2me lambda^(2))/(eh^(2)) = (2mclambda^(2))/(h)`
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