Answer:
The energy of the photon is
.
Step-by-step explanation:
Given:
The wavelength of the photon is given as:
![\lambda =8.6* 10^3\ nm\\1\ nm = 10^(-9)\ m\\\therefore \lambda = 8.6* 10^(3)* 10^(-9)=8.6* 10^(-6)\ m](https://img.qammunity.org/2020/formulas/chemistry/middle-school/mzrpb6l8txo0hhm58tylmrqifh1rkfua07.png)
The energy of a photon in terms of its wavelength is given as:
![E_p=(hc)/(\lambda)\\Where,\ h\rightarrow \textrm{Planck's constant}=6.626* 10^(-34)\ Js\\c\rightarrow \textrm{velocity of light}=3* 10^8\ m/s](https://img.qammunity.org/2020/formulas/chemistry/middle-school/x041chlxr700furtmpj4xin5bn50l585fk.png)
Plug in all the given values and calculate energy of the photon,
. This gives,
![E_p=(6.626* 10^(-34) * 3* 10^8)/(8.6* 10^(-6))\\E_p=(19.878* 10^(-26))/(8.6* 10^(-6))\\E_p=2.311* 10^(-20)\ J](https://img.qammunity.org/2020/formulas/chemistry/middle-school/za0na98ncsqsxb4hzzqwuadkgo0m8re2wc.png)
Therefore, the energy of the photon is
.