Color and Spectra

In the experiment, we view the spectrum of colors and found in white light, and observed the effect colored filters have on the spectrum of white light. The purpose of this experiment is to find the wavelength of different color for the wavelength spectrum by applying the concept of constructive interference.

Equation for wavelength is:

λ= yd/R

λ is the wavelengh of the color on the spectruem, y is the width of the slit, d is the distance from the center to the first dark fridge, and R is the distance between the light source and the slits. In the experiment R^2=D^2+L^2 and y=D.

therefore, the equation for wavelength in the experiment is :

λ= Dd/sqr(D^2+L^2)




Part 1 White Light

Obtaining Wavelengh Uncertainty by plotting and get the slope.

Part 2 measure spectra of unknown#4 tube.

Distance of the diffraction grating d=2 μ meter

Distance from the bulb to to the grating D= 1.98 m

Y(m)         sinθ           λ                      calibrated

0.542        0.264024 5.28048E-07    5.33578E-07

0.606        0.29266   5.8532E-07       5.92534E-07

0.634        0.30495   6.099E-07         6.17837E-07

0.664        0.317951 6.35902E-07    6.44604E-07

0.736        0.348424 6.96848E-07    7.07342E-07

The spectrum of the #4 tube is not continuous and only some waves have strength range.

http://ioannis.virtualcomposer2000.com/spectroscope/elements

Professor announces the #4 is neno(Ne), and the spectrum range is similiar. However we confuse about the source color, for the photograph shows the source color is orange to red but the website give us white to gray. The white to gray color source color is Neptunium, but Np is radioactive element. We are not sure the problem coming from the bad camera to make the photo color change or something else.

Part 3. measure spectra of hydrogen.

Distance of the diffraction grating d=2 μ meter

Distance from the bulb to to the grating D= 1.98 m

Y(m)         sinθ           λ                      calibrated

0.4036      0.199731 3.99462E-07    4.0121E-07

0.487        0.238841 4.77682E-07    4.81731E-07

0.672        0.321388 6.42776E-07    6.5168E-07

The spectrum is not continuous and only few strength light.

By the formula of hydrogen, 1/λ= R(1/n^2-1/n'^2)

 R=1.0974*10^7

wave length        1/(λ*R)            n       n'      1/n^2-1/n'^2      error%

4.0121E-07       0.227123974   2       5       0.21                 8.154273

4.81731E-07    0.189160619   2       4       0.1875               0.885663

6.5168E-07       0.139830071   2       3       0.13888889      0.677651

The n'=5 part, if it change to n'=6, the error percent will reduce from 8 to 2%.

In the theroy, If the n=1, n'=infinite, wave length will be 91.127*10^-8 m

E= hc/wavelength= 2.18267*10^-18 J= 13.64eV

By our data, R as unknow n=2, n'=4, wavelength=4.81731E-07m

so we get R=11071185.7

If the n=1, n'=infinite, wave length will be 9.03246E-8

E=hc/wavelength= 2.20206*10^-18 J=13.7629eV

error= (13.7629-13.64)/13.64=0.9%

Conclusion

The spectra of the hydrogen approach the formula of hydrogen, 1/λ= R(1/n^2-1/n'^2).

The electron drop from different n and release the photon with different wave length.

The #4 tube is not hydrogen, so we may not use this formula of the hydrogen. Otherwise, the source color we photo is different from the website.