Question

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

Answer 1

Answer:

The energy of photon, $E=8\times 10^{-15}\ J$

Explanation:

It is given that,

Voltage of anode, $V=50\ kV=50\times 10^3\ V=5\times 10^4\ V$

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

$E=1.6\times 10^{-19}\times 5\times 10^4$

$E=8\times 10^{-15}\ J$

So, the maximum energy of the x- ray radiation is $8\times 10^{-15}\ J$. Hence, this is the required solution.