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 angstrom `(A^(0))`.Suppose the wavelength of a photon is `lambda` `(A^0)` Then energy of the photon is

Answer

2

Explanation

`E = hv = (hc)/(lambda)`
Here wavelenght =
`lambdaxx10^(-10)m,h=6.62xx10^(-34)js,c = 3xx10^(8)ms^(-1)`
`therefore E = ((6.62xx10^(-34))xx(3xx10^(8)))/(lambdaxx10^(-10))`
`=((6.62xx10^(-34))xx(3xx10^(8)))/((lambdaxx10^(-10))xx(1.6xx10^(-19)))eV=(12400)/(lambda)eV`
`lambda "is taken in" A^(0) and 12400 "in" A^(0) eV`