An X- ray tube produces a continuous spectrum of raditation with its shortest wavelength of `45xx10^(-2)`A. The maximum energy of a photon in the radiation of in eV is `(h=6.62xx10^(-34)J-s,c =3xx10^(8)ms^(-1))`

Options

A. 27500
B. 22500
C. 17500
D. 12500

Answer

A

Explanation

For an X-ray tube ,
Wavelength ,`lamda=45xx10^(-2)Å`
Planck's constant, `h=6.62xx10^(-34)J-s`
Speed of light ,`c=3xx10^(8)ms^(-1)`
The expression for maximum energy of a photon is given as
`E=(hc)/(lamda)`
=`(6.62xx10^(-34)(3xx10^(8)))/(45xx10^(-12))=(6.62xx10^(-34)(3xx10^(8)))/(45xx10^(-12))=(0.44xx10^(-4))/(1.6xx10^(-19))`
`0.275xx10^(5)eV`
`therefore E=27,500`