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

Answer:

$\lambda=3451*10^{10}m$

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

$\lambda=\frac{hc}{E}$

$\lambda=\frac{6.626*10^{-34}*3*10^8}{5.76*10^{-19}}$

$\lambda=3451*10^{10}m$