| 
>>
View the other Key Equations and Concepts in this
chapter
Energy of Light
>> Parts of this equation/concept include:
E = hc/ (Equation 1.3)
As with all equations, it is important that the proper units be
used. In this example, E is in J, h is 6.626 x 10–34
J • s, c is 2.998 x 108 m/s, and 
is m.
E represents the energy required to remove an electron from
a metal. In this case it is the work function of the metal,  .
The energy is also a threshold energy, so more energy is acceptable.
Remember that higher energies are shorter wavelengths.
>> Example 1
What is the work function for a metal that exhibits the photoelectric
effect at 500 nm?
Solution:
|
= |
500 nm |
|
= 5.00 x 10–7 m |
| E |
= |
| hc |
|
|
|
| |
= |
| (6.626 x 10–34 J • s)(2.998 x 108 m/s) |
|
| 5.00 x 10–7 m |
|
| |
= |
3.97 x 10–19 J |
>> Example 2
At what will a metal with a work
function of = 8.20 x 10–19
J exhibit the photoelectric effect?
Solution:
| E |
= |
|
= |
8.20 x 10–19 J |
= |
| (6.626 x 10–34 J • s)(2.998 x 108 m/s) |
|
|
|
| (8.20 x 10–19) |
= |
1.9865 x 10–25 |
|
= |
2.42 x 10–7 m |
|
|
= |
242 nm |
It will exhibit the photoelectric effect at wavelengths shorter
than 242 nm.
>> back
to the Top of the Page
| B. Energy of Electronic Transitions, Equation 3.4 |
| 1 |
|
|
|
= |
1.097 x 10–2 nm–1 |
( |
|
– |
|
) |
This equation is used to determine the transition associated with
a wavelength in the line spectrum. The n refers to the principal
quantum number. Note that in this
equation is in nanometers rather than meters.
Wavelength can be converted to energy using the equation E
= hc/, if the question asks
for energy rather than wavelength.
>> Example 3
What is the wavelength of the transition from n = 2 to
n = 4?
Solution:
| 1 |
|
|
|
= |
0.01097 |
( |
|
– |
|
) |
| |
= |
0.01097 |
(0.025 – 0.0625) |
| |
= |
0.01097 |
(0.1875) |
| |
= |
2.056 x 10–3
|
|
= |
|
|
= |
486.2 nm |
>> Example 4
What wavelength of light is required to remove an electron from
the n = 3 energy level.
Solution:
Since larger values of n represent electrons further from
the nucleus, the electron is removed when it is furthest from
the nucleus or n =  . Note
that 2 =
and 1/ = 0. The initial value
is n = 3. Therefore
| 1 |
|
|
|
= |
0.01097 |
( |
|
– |
| 1 |
|
| 2 |
|
) |
| |
= |
0.01097 |
( |
|
) |
| |
= |
1.219 x 10–3
|
|
= |
|
|
= |
820.4 nm |
>> back
to the Top of the Page
| C. De Broglie Equation, Equation 3.14 |
= h/mv
where = wavelength (m), h
= 6.626 x 10–34 J • s, m = mass (kg),
and v = velocity (m/s). The key to using this equation is
to make sure each unit is in the correct units.
>> Example 5
What is the wavelength of an object weighing 20.0 lb and moving
at 4.0 mph?
Solution:
= h/mv
| mass |
= |
20.0 lb |
|
|
|
|
= 9.08 kg |
|
= |
| 6.626 x 10–34 |
|
| (9.08 kg)(1.8 m/s) |
|
= 4.1 x 10–35 m |
>> Example 6
At what velocity must a neutron (mass = 1.67 x 10–24
g) be moving to have a wavelength of 1.73 cm?
Solution:
= h/mv
| mass |
= |
1.67 x 10–24 g |
|
|
= 1.67 x 10–27 kg |
|
= |
1.73 cm |
|
|
= 0.0173 m |
| 0.0173 |
= |
| 6.626 x 10–34 |
|
| (1.67 x 10–27)v |
|
| |
| (0.0173)(1.67 x 10–27)v |
= |
6.626 x 10–34 |
| |
| v |
= |
| 6.626 x 10–34 |
|
| (1.67 x 10–27)(0.0173) |
|
| |
| v |
= |
2.29 x 10–5 m/s |
>> View
the other Key Equations and Concepts in this chapter
|
