Question

A certain atom has an energy state 3.50 eV above the ground state. When excited to this state, the atom remains for 2.0 ms, on average, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon

Answer 1

`\lambda=3451*10^(10)m`

Step-by-step explanation:

From the question we are told that:

Energy state
`e=3.50 eV`

Time
`t=2ms`

Generally the equation for energy of Photon is mathematically given by

`E=e-e_0`

`E=3.6*10^(-19)J`

`E=5.7*10^(-19)J`

Generally the equation for Wave-length of Photon is mathematically given by

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

`\lambda=(6.626*10^(-34)*3*10^8)/(5.76*10^(-19))`

`\lambda=3451*10^(10)m`