# Working with Wavelentgths and Quantum numbers

If energy is absorbed by a hydrogen atom in its general state, the atom is excited to a higher energy state. For example, the excitation of an electron from the level with n=1 to the level n=3 requires radiation with a wavelength of 102.6 nm. Which of the following transitions would require radiation of longer wavelength then this?

a) n=2 to n=4

b) n=1 to n=4

c) n=1 to n=5

d) n=3 to n=5

Quantum numbers

a) When n=4, l=2, and m= -1, to what orbital type does this refer? (Give the orbital label, such as 1s.)

b) How many orbitals occur in the n=5 electron shell? How many subshells? What are the letter labels of the subshells?

c) How many orbitals occur in an f subshell? What are the values of m?

What is the maximum number of orbitals that can be identified by each of the following sets of quantum numbers? When "none" is the correct answer, explain your reasoning.

a) n=4, l=3

b) n=2, l=2

n=5

n=3,l=1, m=-1

https://brainmass.com/chemistry/modern-quantum-theory/working-wavelengths-quantum-numbers-16044

#### Solution Preview

Answer is a and d for the first question

explanation:

n = 1 to n = 3 falls in the lyman series (UV region: 102.6 nm)

n = 2 to n = N falls in the Balmer series(visible)

n = 3 to n = N falls in the Paschen series(IR)

n = 4 to n = N falls in the Bracket series(IR)

n = 5 to N = N falls in the Pfund series (IR)

UV has the lowest wavelength (high energy)

Hence immediately we can say that n = 3 to 5 and n=2 to 4 are of longer wavelength(low energy) than the given value (102.6nm)

to check the other two, we will use the formula,

lambda = [n^2/(n^2 - 1)]*(1/R) where R is the rydberg constant

for n=1 to 4, lambda1 = [4^2/(4^2 - 1)]*(1/R) = (16/15*R) = 1.07/R

for n=1 to 5, lambda2 = ...

#### Solution Summary

A wavelength and quantum numbers are analyzed. A very detailed answer with more that 30 steps. A good reference.