While working with light and `X`-rays, there is a useful relation between the energy of a photon in electron volts `(eV)` and the wavelength of the photon in angstom `(A^(0))`.Suppose the wavelength of aphoton is `lambdaA`. Then energy of the photon is

Explanation

`E=hv=(hc)/lambda`
Here wavelength=`lambdaxx10^(-10)m,h=6.62xx10^(-34)Js,c=3xx10^(8)ms^(-1)`
`:.E=((6.62xx10^(-34))xx(3xx10^(8)))/(lambdaxx10^(-10)`
`((6.62xx10^(-34))xx(3xx10^(8)))/((lambdaxx10^(-10))xx(1.6xx10^(-19)))eV=12400/lambdaeV`
`:.E=12400/lambdaeV`