If an X-ray tube operates at the voltage of 10kV, find the ratio of the de-broglie wavelength of the incident electrons to the shortest wavelength of X-ray producted. The specific charge of electron is `1.8xx10^11 C/(kg)`.

Options

A. 1
B. 0.1
C. 1.8
D. 1.2

Answer

B

Explanation

Kinetic energy gained by a charge `q` after being accelerated through a potential difference `V` volt, is given by
`q V = (1)/(2) m v^(2)`
`m V = sqrt(2 M q V)`
As `lambda_(b) = (h)/(m V) = (h)/(sqrt (2 q m V)`
for cut- off wavelength of X-ray , we have `q V = (h c)/(lambda_(m))`
or `lambda_(m) = (hc)/(q V)`
Now, `(lambda_(b))/(lambda_(m)) = sqrt ((qV)/(2 m))/(c)`
As `(q)/(m) xx 1.8 xx10^(11) C kg^(-1)` for electron,
we have `(lambda_(b))/(lambda_(m)) = sqrt(1.8 xx 10^(11) xx 10 xx 10^(3)//2)/(3 xx 10^(8)) = 0.1`