let w = wavelength (I can't type the symbol).
E = hc/w; So the slope is E/(1/w) = E*w.
You can see that E*w = slope = hc so slope/c = h.
Yes, pick a couple of points on the y axis and determine the slope.
In the cchemistry lab we conducted an experiment called atomic spectroscopy.
We had to find the color, energy, wavelength, and 1/lamda for Hydrogen, neon, and mercury.
For hydrogen:
red 1.90ev 650nm 1.54e^-3
red 2.02ev 610nm 1.64e^-3
green 2.30ev 545nm 1.83e^-3
for neon:
green 2.30ev 550nm 1.82e^-3
red 2.05ev 610nm 1.64e^-3
red 1.80ev 690 1.45e^-3
yellow 2.12 580 1.72e^-3
blue 2.65 470 2.13e^-3
for mercury:
green 2.30ev 548 1.823^-3
red 2.05 590 1.69e^-3
violet 2.85 440 2.27e^-3
blue 2.55 470 2.13e^-3
yellow 2.15 570 1.75e^-3
for the first question we had to derive the equation that would allow us to find Planck's constant from your graph using the formulas
c= lamda X v and E=hv
The teacher said that the equation is h=slope / c
The next question says to show the calculations of Planck's constant, including the unit conversions, for mercury, neon, and hydrogen using the equation created above.
I am not sure how to do this. I do not know where to start.
We had to create graphs for neon and hydrogen on the computer so would i use the slope from there?
on the x axis was 1/wavelength (nm-1) and the y axis was energy(ev). The line for both graphs of neon and hydrogen was positive. So would I pick two points on the graph to find the slope and then do h=slope/c?
2 answers
In the cchemistry lab we conducted an experiment called atomic spectroscopy.
We had to find the color, energy, wavelength, and 1/lamda for Hydrogen, neon, and mercury.
For hydrogen:
red 1.90ev 650nm 1.54e^-3
red 2.02ev 610nm 1.64e^-3
green 2.30ev 545nm 1.83e^-3
for neon:
green 2.30ev 550nm 1.82e^-3
red 2.05ev 610nm 1.64e^-3
red 1.80ev 690 1.45e^-3
yellow 2.12 580 1.72e^-3
blue 2.65 470 2.13e^-3
for mercury:
green 2.30ev 548 1.823^-3
red 2.05 590 1.69e^-3
violet 2.85 440 2.27e^-3
blue 2.55 470 2.13e^-3
yellow 2.15 570 1.75e^-3
for the first question we had to derive the equation that would allow us to find Planck's constant from your graph using the formulas
c= lamda X v and E=hv
The teacher said that the equation is h=slope / c
The next question says to show the calculations of Planck's constant, including the unit conversions, for mercury, neon, and hydrogen using the equation created above.
I am not sure how to do this. I do not know where to start.
We had to create graphs for neon and hydrogen on the computer so would i use the slope from there?
on the x axis was 1/wavelength (nm-1) and the y axis was energy(ev). The line for both graphs of neon and hydrogen was positive. So would I pick two points on the graph to find the slope and then do h=slope/c?
Chemistry - DrBob222, Tuesday, November 1, 2011 at 7:50pm
let w = wavelength (I can't type the symbol).
E = hc/w; So the slope is E/(1/w) = E*w.
You can see that E*w = slope = hc so slope/c = h.
Yes, pick a couple of points on the y axis and determine the slope.
to calculate the slope i picked points (2.85, 0.00227) and (2.05, 0.00169). So i did 2.85-2.05 / .00227 -.00169 and ended up with .80/.00058. Is this correct.
We had to find the color, energy, wavelength, and 1/lamda for Hydrogen, neon, and mercury.
For hydrogen:
red 1.90ev 650nm 1.54e^-3
red 2.02ev 610nm 1.64e^-3
green 2.30ev 545nm 1.83e^-3
for neon:
green 2.30ev 550nm 1.82e^-3
red 2.05ev 610nm 1.64e^-3
red 1.80ev 690 1.45e^-3
yellow 2.12 580 1.72e^-3
blue 2.65 470 2.13e^-3
for mercury:
green 2.30ev 548 1.823^-3
red 2.05 590 1.69e^-3
violet 2.85 440 2.27e^-3
blue 2.55 470 2.13e^-3
yellow 2.15 570 1.75e^-3
for the first question we had to derive the equation that would allow us to find Planck's constant from your graph using the formulas
c= lamda X v and E=hv
The teacher said that the equation is h=slope / c
The next question says to show the calculations of Planck's constant, including the unit conversions, for mercury, neon, and hydrogen using the equation created above.
I am not sure how to do this. I do not know where to start.
We had to create graphs for neon and hydrogen on the computer so would i use the slope from there?
on the x axis was 1/wavelength (nm-1) and the y axis was energy(ev). The line for both graphs of neon and hydrogen was positive. So would I pick two points on the graph to find the slope and then do h=slope/c?
Chemistry - DrBob222, Tuesday, November 1, 2011 at 7:50pm
let w = wavelength (I can't type the symbol).
E = hc/w; So the slope is E/(1/w) = E*w.
You can see that E*w = slope = hc so slope/c = h.
Yes, pick a couple of points on the y axis and determine the slope.
to calculate the slope i picked points (2.85, 0.00227) and (2.05, 0.00169). So i did 2.85-2.05 / .00227 -.00169 and ended up with .80/.00058. Is this correct.