1.

Through what minimum potential must an electron in an X-ray tube be accelerated so that it can produce X- rays with wavelength of 0.050 nm ?

Answer»

12.62KV
24.847KV
32.52KV
40.8KV

Solution :We know, `V=(hc)/(e lambda)`
Given `lambda=0.050 nm=0.050 XX 10^(-9)m`
`h=6.626 xx 10^(-34) JS, c=3 xx 10^(8)m`
and `e=1.6 xx 10^(-19)C`
`V=(6.626 xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) xx 0.050 xx 10^(-9))`
`=2.48475 xx 10^(4) V=24.847kV`


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