NOTE: No Chem 1E03-01 students will have a formal tutorial during the week of October 11- 15. Solutions will be posted.

**1. The McMaster student FM radio station CFMU broadcasts at a frequency
of 93.3 MHz (MHz = 1 megaherz = 10 ^{6} sec^{- 1}).**

**a) What is the wavelength of this signal in meters?**

nu = c / lambda

lambda = c / nu = (3.00 x 10^{8} m s^{-1})/(96.3 x
10^{6} s^{-1})

lambda = 3.22 m

**b) What is the energy of one photon of this frequency?**

E = h x nu = 6.62 x 10^{-34} J s x 9.33 x 10^{7} s^{-1}

E = 6.18 x 10^{-26} J

**c) Compare the photon energy of part b with the energy of a photon
of red light with wavelength 685 nm.**

A red photon (its nu is about 10 ^{15} s^{-1}) has an
energy of about E = h x nu = 6.62 x 10^{-34} J s x 10^{15}
s^{-1} = 6.62 x 10^{-19} J. Its energy is about ten million
times larger than the FM radio frequency of part a.

**2. A cup containing 250 cm ^{3} of water, initially at 25
^{o}C,
is warmed to 95 ^{o}C in a microwave oven that operates at a frequency
of 2.45 GHz (GHz = 1 gigaherz = 10^{9} sec^{-1}). How many
microwave photons of this frequency must be absorbed to warm the water?
Assume that the heat capacity of the cup is negligible, compared to the
heat capacity of the water.**

Density of H_{2}O = 1.00 g/cm^{3}; thus, 250 cm^{3}
H_{2}O has a mass of 250 g

heat flow q = mass of H_{2}O x C_{sp,H2O} x delta T
= 7.32 x 10^{4} J

E of one microwave photon = h x nu = 6.62 x 10^{-34} J s x 2.45
x 10^{9} s^{-1}

E = 1.62 x 10^{-24} J/photon

# of photons required = 7.32 x 10^{4} J/1.62 x 10^{-24}
J/photon

= 4.52 x 10^{28} photons

**3. The longest-wavelength light that causes an electron to be emitted
from gaseous lithium atom is 520 nm. Gaseous lithium atoms are irradiated
with light of wavelength 400 nm. What is the kinetic energy of the emitted
electrons, in kJ/mol?**

KE_{electron} = E_{400 nm} - E_{520 nm}, where
520 nm corresponds to the minimum energy needed to remove the electron

KE = h x c x (1/lambda of 400 nm - 1/lambda of 520 nm)

KE_{electron} = [6.62 x 10^{-34} J s x 3.00 x 10^{8}
m/s]/[1.00 x 10^{-7} m x (1/4.00 - 1/5.20)]

KE_{electron} = 1.15 x 10^{-19} J/electron.

For one mole of electrons, KE = 1.15 x 10^{-19} J/electron x
6.022 x 10^{23} electrons/mole x 10^{-3} kJ/J

KE = 69.3 kJ/mol

**4. Assume that you have graduated with a McMaster Engineering degree,
and that you are designing a space probe to land on a distant planet. You
wish to use a switch that works by the photoelectric effect. The metal
that you wish to use in your device requires 6.7 x 10 ^{-19} J/atom
to eject an electron. You know that the atmosphere of the planet on which
your device must work filters out all light of wavelengths shorter than
540 nm. Will your device work on the planet? Why or why not?**

The energy associated with the highest energy possible photon on the planet is

E = h x c / lambda

= [6.62 x 10^{-34} J s x 3.00 x 10^{8} m/s]/[5.40 x
10^{-7} m]

= 3.69 x 10^{-19} J

This photon is not energetic enough to eject an electron from the chosen metal. So the device will not work on the planet. You must choose another metal!

**5. In the Bohr model for hydrogen atom,**

**E _{n} = -2.178 x 10^{-18} J/n^{2}**

**Calculate whether a photon of green light, of wavelength 500 nm,
has enough energy to excite the electron in the hydrogen atom from n =
1 to n = 2.**

Here we are concerned with the transition E_{1} --> E_{2}

E_{photon required} = difference in energy between E_{2}
and E_{1}

So we calculate what the photon energy must be to bring this about:

E_{photon} = E_{2} - E_{1}

= [(-2.178 x 10^{-18}/4) - (-2.178 x 10^{-18}/1)] J

E_{photon} = 1.634 x 10^{-19} J

The photon of green light under consideration has an energy of

E = h x c / lambda

= [6.62 x 10^{-34} J s x 3.00 x 10^{8} m/s]/[5.00 x
10^{-7} m]

= 3.97 x 10^{-19} J

So the green photon has too little energy to excite the electron from
E_{1} to E_{2} (need a UV photon)

**6. When the hydrogen atom is at relatively low temperatures, its
electron is in the lowest energy level, n = 1. This is called the ground
state.**

**a) What is the longest wavelength of radiation that can be absorbed
by such an atom?**

The longest wavelength of light that can be absorbed corresponds to
the smallest difference in energy levels. Since n_{initial} = 1,
n_{final} must be 2, so as to have the smallest delta E.

In problem 5 (above), we have calculated that this light must have an
energy of 1.634 x 10^{-19} J

Since E = h x nu = h x c / lambda, lambda = hc/E

lambda = [6.62 x 10^{-34} J s x 3.00 x 10^{8} m/s]/[1.634
x 10^{-18} J]

lambda = 1.21 x 10^{-7} m (or 121 nm)

**b) What is the minimum amount of energy required to remove electrons
from a mole of ground state hydrogen atoms?**

To REMOVE an electron from H atom, we must consider n = 1 --> n = infinity. For one electron, this energy is h x nu. To remove an electron from N hydrogen atoms requires an energy of N x h x nu.

E = N x h x nu = N x (E_{infinity} - E_{1})

= 6.022 x 10^{23} x 2.178 x 10^{-18}) J x [0 - (- 1)]

E = 1.31 x 10^{6} J = 1.33 MJ (see p 333 in Kotz and Treichel)

**7. Calculate the wavelength in nm for the longest wavelength emission
transition seen in the Brackett series of hydrogen, namely where the final
state corresponds to n = 4.**

The Brackett series of lines are emission lines, all of which end in n = 4. The longest wavelength line in this series corresponds to the smallest energy gap, i.e., n = 5 --> n = 4.

E = h x nu = h x c / lambda; lambda = h x c / E_{photon}

lambda = h x c /(E_{n = 5} - E_{n = 4})

6.62 x 10^{-34} J s x 3.00 x 10^{8} m/s

= ---------------------------------------------------

[(-2.178 x 10^{-18})/25 - (-2.178 x 10^{-18})/16)]J

lambda = 4.05 x 10^{-6} m (4050 nm)

**8. The Brackett series of emission lines from atomic hydrogen occurs
in the far infrared region. One of the lines has a wavelength of 2625 nm.
Determine the values for the quantum number n for the two energy levels
involved in the transition. Show your calculations.**

E_{photon} = h x c / lambda

6.62 x 10^{-34} J s x 3.00 x 10^{8} m/s

= -----------------------------------------

2.625 x 10^{-6}m

E_{photon} = 7.568 x 10^{-20}J

This is a Brackett series line, n_{lower} = 4.

Therefore Delta E_{hydrogen} = 7.568 x 10^{-20}J = -2.178
x 10^{-18} (1/n^{2} - 1/16)J

7.568 x 10^{-20} J - 13.61 x 10^{-20} J = -2.178 x 10^{-18}
J/(n^{2})

n^{2} = 2.178 x 10^{-18}/6.04 x 10^{-20} = 36.06

Therefore n = 6 (6 --> 4 transition)

**9. The n = 3 to n = 1 emission line for atomic hydrogen occurs in
the UV region (it is a member of the Lyman series). Without doing any calculations,
decide which of the following emission lines for atomic hydrogen occur
at longer wavelengths than this line.**

**a) n = 4 --> n = 2;**

**b) n = 4 --> n = 1;**

**c) n = 5 --> n = 1;**

**d) n = 5 --> n = 3.**

We need to determine those transitons that involve LESS energy than the 3 --> 1 transition. Therefore the 4 --> 1 and 5--> 1 transitions are eliminated, since they require more energy. The 4 - -> 2 transition is a VISIBLE transition (Balmer series) and requires less energy than the UV transition. Likewise the 5 --> 3 transition in the infrared region requires less energy than the UV transition.