Question

A certain atom has an energy level 2.58 eV above the ground level. Once excited to this level, the atom remains at this level for 1.64E-7 s (on average) before emitting a photon and returning to the ground level. a) What is the energy of the photon (in electron volts)? What is its wavelength (in nanometers)? b) What is the smallest possible uncertainty in energy of the photon? Give your answer in electron volts. c) Show that|?E/E|=|??/?|if |??/?|?1. Use this to calculate the magnitude of the smallest possible uncertainty in the wavelength of the photon. Give your answer in nanometers. (39.56)

Answer 1

a)

E = energy of emitted photon = difference in the energy levels = 2.58 eV

In Joules

E =  2.58 (1.6 x 10^-19) J

wavelength is given as

= hc/E = (6.63 x 10^-34) (3 x 10⁸)/(2.58 (1.6 x 10^-19))

= 482 nm

b)

Uncertainty in energy is given as

E = h/(2t) = (6.63 x 10^-34)/(2 (3.14) (1.64 x 10^-7)) = 6.44 x 10^-7