# 6.E: The Quantum-Mechanical Model of the Atom (Exercises)

These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

## Q7.45

What wavelength of light in meters produces this amount of energy? Round to two significant figures.

Formula for Energy is E=hc/lambda

h is Planck's constant it is equivalent to 6.626 x 10-34

c is the speed of light constant it is equivalent to 2.998 x 108

lambda is the wavelength, which in this problem you are trying to find

E is the energy which is what you will be given in this problem

Examples:

a) 1.33 x 10-19J

1.33 x 10-19J =hc/lambda

1.33 x 10-19 = (6.626 x 10-34)(2.998 x 108)/lambda

1.33 x 10-19= 1.98 x 10-25/lambda

1.33 x 10-19(lambda)= 1.986 x 10-25

lambda =1.98 x 10-25/1.33 x 10-19

lambda= 1.4936 x 10-6

lambda = 1.5 x 10-6

lambda = 1500 x 10 -9 m

b) 3.98 x 10-19 J

3.98 x 10-19J = hc/lambda

3.98 x 10 -19 = (6.626 x 10-34)(2.998 x 108)/lambda

3.98 x 10-19 = 1.986 x 10-25/lambda

3.98 x 10-19(lambda) = 1.986 x 10-25

lambda = 1.986 x 10-25/3.98 x 10-19

lambda = 4.989 x 10-7

lambda = 5.0 x 10-7

lambda = 500 x 10-9 m

c) 1.32 x 10-18J

1.32 x 10-18J = hc/lambda

1.32 x 10-18 = (6.626 x 10-34)(2.998 x 108)/lambda

1.32 x 10-18 = 1.986 x 10-25/lambda

1.32 x 10-18(lambda) = 1.986 x 10-25

lambda = 1.5045 x 10-7

lambda = 1.5 x 10-7

lambda = 150 x 10-9 m

## Q7.57a

How many electrons are in a s orbital verses a p orbital?

Notation to know:

• e- is the notation for an electron
• Orbital- the energy state in an atom where an electron will be found
1. Figure out where an s orbital and p orbital would be located
2. Determine which level each one is
3. Use which level they are to determine how many boxes electrons can occupy
4. Realize that two opposite spinning electrons can occupy one box

1. Using the number of boxes and the knowledge that two electrons can occupy a box, multiply the number of boxes by the two orbitals

Solution

The s orbital has one box that could have electrons in it since s is the lowest orbital. p is the next level up, which has three boxes electrons can occupy. Each box has the capacity for two electrons to occupy the space. Since s has one box, and p has three boxes, the number of electrons that could possibly be in an s orbital is two, while a p orbital has 6.

## Q7.57b

If an electron is excited from a 3p orbital to a 2p orbital, does it give off energy or gain energy? Why?

Strategy

1. Identify the location of each orbital and how this relates to their energy levels
2. Explain why your answer is correct, referring to the movement of photons

The 3p orbital is further away from the nucleus than the 2p orbital.The closer an electron is to the nucleus the less amount of energy it has.Thus, the electron will give off energy due to its movement from a higher level orbital to a lower level orbital releasing a photon of energy in the process. When the electron gets excited it will jump back up to a higher level orbital.

## Q7.59

When each value of n is given, what are the possible values for l?

1. 2
2. 4
3. 6
4. 8

Strategy:

• The relationship between the quantum number (n) and the angular momentum quantum number (l).
• The principle quantum number is an integer that determines the overall size and energy of an orbital. The possible values for n are 1, 2, 3, … and so on.
• The angular momentum quantum number is an integer that determines the shape of the orbital. Every value of n has a certain l value. In other words, for a given value of n, l can be any integer up to n-1.

Example:

If n=3, what are the possible values of l?

Since l = n-1, and n=3. Substitute 3 with n (l=3-1), which l equals to 2 maximum values or 0, 1, 2.

Solution:

1. n= 2

l= n-1

l= 0, 1

1. n= 4

l= n-1

l= 0, 1, 2, 3

1. n= 6

l= n-1

l= 0, 1, 2, 3, 4, 5

1. n= 8

l= n-1

l= 0, 1, 2, 3, 4, 5, 6, 7

## Q7.78a

A photon of wavelength 0.787 nm strikes a surface. The surface has a binding energy of 1.35 *1010 10^-19 kJ/mol. What is the kinetic energy of the emitted electron in eV?

Strategy

Solution

1. Convert kJ/mol à J to keep units consistent.
2. Solve For energy of the photon
3. Subtract binding energy from the energy of the photon
4. Convert J --> eV
1. 1.35*10^-19 kJ/mol = 1.35*10-16 J/mol
2. Energy=(Planck's Constant)(Speed of light)/ wavelength
1. h= 6.626* 10^-34 (Planck’s Constant)
2. c=2.99*1010 10^-16 m/s (speed of light)
3. E=Energy
4. wavelength (convert to meters)
3. Energy of photon= binding energy + kinetic energy
1. 1.67*r10^-16 J=Kinetic energy
4. 1.167*10^-16 J* * 1 eV/1.603*10^-19 J
1. =728.5 eV

## Q7.78b

A photon which has a wave length of 0.875 nm strikes a surface. The emitted electron has a kinetic energy of 900 eV. Calculate the binding energy of the electron in kJ/mol.

Solution

First we must change our value in units of eV to joules using the conversion 1 eV=1.602x10-19 J. This gives us a value of 1.4418x10-16J