Answer:
![E=1.32* 10^(-17)\ J](https://img.qammunity.org/2020/formulas/physics/high-school/85o5eyrvyhxnezyiqf37xvkffn0zjmu7go.png)
Step-by-step explanation:
Given that,
The wavelength of x ray,
![\lambda=15\ nm=15* 10^(-9)\ m](https://img.qammunity.org/2020/formulas/physics/high-school/msx2ipfd20gdvn4a8fr0nb9pppf5aegm9u.png)
We need to find the energy if a photon on the x-rays. The formula for the energy of the photon is given by :
![E=(hc)/(\lambda)](https://img.qammunity.org/2020/formulas/physics/middle-school/qsv4fkctsmfst9aln3wy2xtnfdwqn1eov2.png)
![E=(6.63* 10^(-34)* 3* 10^8)/(15* 10^(-9))](https://img.qammunity.org/2020/formulas/physics/high-school/h2gmayw1udaxhsyeaeipv2dauz6c1ws4t3.png)
![E=1.32* 10^(-17)\ J](https://img.qammunity.org/2020/formulas/physics/high-school/85o5eyrvyhxnezyiqf37xvkffn0zjmu7go.png)
So, the energy of a photon of the X-rays is
. Hence, this is the required solution.