The allowed energies of a simple atom are 000 eV 421 eV and

The allowed energies of a simple atom are 0.00 eV, 4.21 eV , and 6.54 eV . What wavelengths appear in the atom\'s emission spectrum?

Part A

What wavelengths appear in the atom\'s emission spectrum?

Part B

What wavelengths appear in the atom\'s absorption spectrum?

Solution

n = 3 n = 2

lambda = hc / E = (4.136 * 10^-15eV * 3 * 10⁸) / (2.33 eV)

lambda = 533 nm

n = 3 n = 1

lambda = hc / E = (4.136 * 10^-15 eV * 3 * 10⁸) / (6.54 eV)

lambda = 190 nm

n = 2 n = 1

lambda = hc / E = (4.136 * 10^-15 eV * 3 * 10⁸) / (4.21 eV)

lambda = 295 nm

------------------------------------------------------

The same as part (a)