Question

For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?

Answer 1

The energy of photon,
`E=8* 10^(-15)\ J`

Step-by-step explanation:

It is given that,

Voltage of anode,
`V=50\ kV=50* 10^3\ V=5* 10^4\ V`

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

`E=eV`

e is charge of electron

`E=1.6* 10^(-19)* 5* 10^4`

`E=8* 10^(-15)\ J`

So, the maximum energy of the x- ray radiation is
`8* 10^(-15)\ J`. Hence, this is the required solution.