Experiment 2: The Wave Model vs. The Quantum Model of light: AP Physics B

Experiment 2

Page Name:

Rich Text Content

FIRST ORDER:

Color	Wavelength (nm)	Frequency (Hz x10^14)	Stopping potential (V)
Yellow	576	5.2	.67
Green	545	5.5	.79
Blue	441	6.9	1.35
Violet	411	7.3	1.55
Ultraviolet	366	8.2	1.84

SECOND ORDER:

Color	Wavelength (nm)	Frequency (Hz x10^14)	Stopping potential(V)
Yellow	576	5.2	.7
Green	545	5.5	.82.
Blue	441	6.9	1.37
Violet	411	7.3	1.56
Ultraviolet	366	8.2	1.86

Image/File Upload

Rich Text Content

The key to this graph is seeing that the equation

V=mf+b

for the slope of this line is very similar to

$KE_{\max}=hf-w_o$

Where we need to multiply the entire equation by the charge of an electron to make sure we have an energy unit obtained on the left (in this case electron volts)

Because m=.3937 , and m is the ratio $\frac{h}{e}$

First, we multiply m by e, and remember that my Hz values include an extra $10^{14}$ .

by first multiplying the slope by the electron, we obtain $h=6.299\cdot\frac{10^{-20}\:}{10^{14}}$ for planks constant. But then, with the $10^{14}$ in the denominator of the slope, we get $h=6.992\cdot10^{-34}$

The Work function can simply be found as the y-intercept of the graph. Meaning, electrons are being knocked off the metal, but not getting any kinetic energy. It is therefore 1.356 eV

Allow Comments on this Page

Make Comments Public

HTML Editor

[{:section_type=>"rich_text", :content=>"FIRST ORDER:\n \n \n<table border=\"0\" cellspacing=\"2\" cellpadding=\"5\"><tbody>\n<tr>\n<td>Color</td>\n<td>Wavelength (nm)</td>\n<td>Frequency (Hz x10^14)</td>\n<td>Stopping potential (V)</td>\n</tr>\n<tr>\n<td>Yellow</td>\n<td>576</td>\n<td>5.2</td>\n<td>.67</td>\n</tr>\n<tr>\n<td>Green</td>\n<td>545</td>\n<td>5.5</td>\n<td>.79</td>\n</tr>\n<tr>\n<td>Blue</td>\n<td>441</td>\n<td>6.9</td>\n<td>1.35</td>\n</tr>\n<tr>\n<td>Violet</td>\n<td>411</td>\n<td>7.3</td>\n<td>1.55</td>\n</tr>\n<tr>\n<td>Ultraviolet</td>\n<td>366</td>\n<td>8.2</td>\n<td>1.84</td>\n</tr>\n</tbody></table> \nSECOND ORDER:\n \n<table border=\"0\" cellspacing=\"2\" cellpadding=\"5\"><tbody>\n<tr>\n<td>Color</td>\n<td>Wavelength (nm)</td>\n<td>Frequency (Hz x10^14)</td>\n<td>Stopping potential(V)</td>\n</tr>\n<tr>\n<td>Yellow</td>\n<td>576</td>\n<td>5.2</td>\n<td>.7</td>\n</tr>\n<tr>\n<td>Green</td>\n<td>545</td>\n<td>5.5</td>\n<td>.82.</td>\n</tr>\n<tr>\n<td>Blue</td>\n<td>441</td>\n<td>6.9</td>\n<td>1.37</td>\n</tr>\n<tr>\n<td>Violet</td>\n<td>411</td>\n<td>7.3</td>\n<td>1.56</td>\n</tr>\n<tr>\n<td>Ultraviolet</td>\n<td>366</td>\n<td>8.2</td>\n<td>1.86</td>\n</tr>\n</tbody></table> \n "}, {:section_type=>"attachment", :attachment_id=>34484184}, {:section_type=>"rich_text", :content=>" \nThe key to this graph is seeing that the equation\n<img class=\"equation_image\" title=\"V=mf+b\" src=\"/equation_images/V%253Dmf%2Bb\" alt=\"V=mf+b\">\nfor the slope of this line is very similar to\n <img class=\"equation_image\" title=\"KE_{\\max}=hf-w_o\" src=\"/equation_images/KE_%257B%255Cmax%257D%253Dhf-w_o\" alt=\"KE_{\\max}=hf-w_o\">\nWhere we need to multiply the entire equation by the charge of an electron to make sure we have an energy unit obtained on the left (in this case electron volts)\n \nBecause <img class=\"equation_image\" title=\"m=.3937\" src=\"/equation_images/m%253D.3937\" alt=\"m=.3937\"> , and m is the ratio <img class=\"equation_image\" title=\"\\frac{h}{e}\" src=\"/equation_images/%255Cfrac%257Bh%257D%257Be%257D\" alt=\"\\frac{h}{e}\">\nFirst, we multiply m by e, and remember that my Hz values include an extra <img class=\"equation_image\" title=\"10^{14}\" src=\"/equation_images/10%255E%257B14%257D\" alt=\"10^{14}\">.\nby first multiplying the slope by the electron, we obtain <img class=\"equation_image\" title=\"h=6.299\\cdot\\frac{10^{-20}\\:}{10^{14}}\" src=\"/equation_images/h%253D6.299%255Ccdot%255Cfrac%257B10%255E%257B-20%257D%255C%253A%257D%257B10%255E%257B14%257D%257D\" alt=\"h=6.299\\cdot\\frac{10^{-20}\\:}{10^{14}}\"> for planks constant. But then, with the <img class=\"equation_image\" title=\"10^{14}\" src=\"/equation_images/10%255E%257B14%257D\" alt=\"10^{14}\">in the denominator of the slope, we get <img class=\"equation_image\" title=\"h=6.992\\cdot10^{-34}\" src=\"/equation_images/h%253D6.992%255Ccdot10%255E%257B-34%257D\" alt=\"h=6.992\\cdot10^{-34}\">\n \nThe Work function can simply be found as the y-intercept of the graph. Meaning, electrons are being knocked off the metal, but not getting any kinetic energy. It is therefore 1.356 eV\n "}]

Rich Text Content

AP Physics B

Experiment 2

Page Comments

Using an ePortfolio

Step