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# Working with Wavelentgths and Quantum numbers

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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 This solution is FREE courtesy of BrainMass!

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 = [5^2/25-1](1/R) = 25/24*R = 1.041/R

and for our n=1 to n=3, lambda3 = (9/8R) = 1.125/R (given value)

from this, lambda3 > lambda1 > lambda2

wavelengths corresponding to the transition from n=1 to n=4 and n=5 are with lesser wavelength than 102.6nm (which is our lambda3, or l3)

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a) l = 2 corresponds to the d orbital.
Hence the orbital type is 4d.
But m = -1. Hence it correspond to the 4dy orbital (orientation in y direction).

b) For the principal quantum number n, there will be n^2 orbitals in the atom.
Therefore, for n = 5, there will be 5^2 = 25 = 25 orbitals.

If n = 5, l can take values, l = 4,3,2,1,0

Therefore, five subshells will be there for n = 5. They are 5g, 5f, 5d, 5p, 5s.

The letter labels are s, p, d, f, g corresponding to l = 0,1,2,3,4 respectively.

c) For a given l value there will be (2l + 1) orbitals.
In an f subshell there are 7 orbitals (l = 3) which can accomodate a total of 2*7 = 14 electrons.

The possible values of m for a given value of l range fron +l through 0 to -l.
l = 3 for the f subshell.
Therefore, the values of m are, m = 3, 2, 1, 0, -1, -2, -3

Number of orbitals

a) n =4, l=3
Here the orbital type is 4f.

Now, there are (2l + 1) orbitals associated with a given l value. That is 2*3 + 1 = 7 orbitals with m values ranging from -3 to 0 to +3.

So, we can identify a maximum of 7 orbitals.

b)
n=2, l=2
Here the orbital type is 2d.
There are 2l+1 = 5 orbitals associated with the d subshell with m values ranging from -2 to 0 to 2.

So, we can identify a maximum of 5 orbitals.

c) n = 5

In this case we can identify a maximum of 25 orbitals. the reason is that for n = 5, l can take any value from 0 to 4.

That is we can have 5s, 5p, 5d, 5f and 5g orbitals for l= 0,1,2,3,4.

Now, the s subshell has 1 orbital, p has 3 , d has 5, f has 7 and g has 9 orbiatla (2l+1).

Therefore, we have a total of 1+3+5+7+9 = 25 orbitals (n^2).

d)
n=3, l=1 and m=-1
Here we can identify only one orbital with the label 3p with orientation in the x direction (m=-1). That is 3px orbital.

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