Electrons with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

Options

A. `lamda_(0)=(2mclamda^(2))/(h)`
B. `lamda_(0)=(2h)/(mc)`
C. `lamda_(0)=(2m^(2)c^(2)lamda^(3))/(h^(2))`
D. `lamda_(0)=lamda`

Answer

B

Explanation

We know that , de-Broglie wavelength of photon `E=hv`
but we know that also,
`E=(1)/(2)mv^(2) implies 2V=(p^(2))/(m)` (for proton)
`becausep=mv`
`(p^(2))/(m)=mv^(2)`
`implies p=sqrt(2mE)` . .. (i)
According to the question, `lamda_(1)=(h)/(p)` . .. (ii)
and `lamda_(2)=(hc)/(E)` . . . (iii)
From the Eqs. (ii) and (iii), we get
`(lamda_(1))/(lamda_(2))=(h//p)/(hc//E)=(E)/(Pc)=(E)/(sqrt(2mE)*c)=(sqrt(E))/(sqrt(2m)*c)`
So, `(lamda_(1))/(lamda_(2))=(sqrt(E))/(sqrt(2m)*c) implies (lamda_(1))/(lamda_(2))propsqrt(E)`
`implies lamda_(1):lamda_(2)=E^(1//2)`