What should be the approzimate kinetic energy of an electron so that its de-Broglie wavelength is equal to the wavelength of X-ray of maximum energy, produced in an X-ray tube operating at 24800 V? (given that `h=6.6xx10^(-34)` joule-sec, mass of electron `=9.1xx10^(-31)kg`)

Options

A. 600 eV
B. 365 eV
C. 120 eV
D. 300 eV

Answer

A

Explanation

The wavelength `lamda_(min)` emitted by X-ray tgube operating at a voltage V is given by `eV=(hc)/(lamda_(min))`
Kinetic energy of the electron `=(1)/(2)mv^(2)=(1)/(2)((m^(2)v^(2))/(m))=(h^(2))/(2mlamda_(1)^(2))`
Now the wavelength `lamda` of an electron moving with the velocity v is given by
`lamda=h//mvimplieslamda=(h)/(sqrt(2meV))`
`thereforeK.E.=600eV`