Question

Question #1427c

Answer 1

`E_(gamma)=4.31*10^(-33)J=26.94` `feV`

Explanation:

`E_(gamma)=hv`
where:

• `E_(gamma)` is the photon energy (in `J`).
• `h` is the Planck constant (`6.63*10^(-34)J*s`).
• `v` is the frequency (in `Hz=1/s`).

So if you substitute with the values,
`E_(gamma)=6.63*10^(-34)*6.5`
`E_(gamma)=4.31*10^(-33)J`

if you want it in `eV` and not in `J`, you use the equivalence:
`1` `eV=1.6*10^(-19)` `J`

so,
`E_(gamma)=2.694*10^(-14)eV`

or,
`E_(gamma)=26.94` `feV`