Question

A photon with a wavelength of 2.29 × 10^–7 meter strikes a mercury atom in the ground state.

Calculate the energy, in joules, of this photon. [Show all work, including the equation and
substitution with units.]

Answer 1

detailed solution is attached

Answer 2

Energy of photon,
`E=8.64* 10^(-19)\ J`

Step-by-step explanation:

It is given that,

Wavelength of photon,
`\lambda=2.29* 10^(-7)\ m`

It strikes a mercury atom in the ground state. We have to find the energy of this photon. It can be calculated using below relation as :

`E=(hc)/(\lambda)`

Where

h is the Planck's constant

c is the speed of light

So,
`E=(6.6* 10^(-34)\ J-s* 3* 10^8\ m/s)/(2.29* 10^(-7)\ m)`

`E=8.64* 10^(-19)\ J`

Hence, the above value is the energy of this photon.