User:Chad A McCoy/Notebook/Jr. Lab/2008/10/26

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Balmer Lab: 10/13-26/2008
$$m=1.0034$$ and $$m=1.0021$$.
 * Calibration of the spectrometer was done using Microsoft Excel, via fitting a least squares regression line to the know and measured values. In doing so, I found values of:
 * I fit the data with the y-intercept fixed at zero to get a direct regression for the wavelength, and found the coefficient of determination to be greater than 99.99% for both calibration sets
 * That proved that it is valid to have a linear regression for calculating the known and measured values from each other.
 * Steve Koch: I don't think this proves it, because in this particular lab your precision is at the 0.01% level!
 * The two calibration lines are shown below.

$$\frac{1}{lambda}=R×(\frac{1}{2^2}-\frac{1}{n^2})$$ where R is the Rydberg Constant, Lambda is the wavelength and n is the quantum number. $$R=\frac{1}{lambda×(\frac{1}{2^2}-\frac{1}{n^2})}$$ $$R=1.09556*10^7\frac{1}{m}$$ $$Error-R=\sqrt{2834^2+(-145)^2}=2838m^{-1}$$ $$R=1.09556(28)*10^7\frac{1}{m}$$
 * My full calculations can be found in the Excel file located at [[Media:Balmer Calculations.xls|here]]
 * Doing so resulted in adjusted values for the spectral lines at:
 * I got these values by multiplying the slope of my regression line with their measured value.
 * From here I was able to calculate the Rydberg Constant using each wavelength and the formula:
 * Solving the equation for R leads to:
 * From this I was able to plug in the values of lambda and n and came out with 8 values of R, such that after averaging the values:
 * The standard error in the value R, strictly from the average, I found to be $$SE-R=2834m^{-1}$$
 * Using the Standard error of the projected Y values, as given by the LINEST function, along with the standard error of the data set, I calculated the error from the measurement and calibration and found a value $$Cal-Error=-145m^{-1}$$
 * By summing the error in quadrature, I was able to find the overall error as
 * Therefore my final value for the Rydberg Constant can be written as:
 * Comparing my value to the known value of the Rydberg constant: $$R=1.0967758*10^7m^{-1}$$ shows that the actual value does not lie within 3 standard deviations of the mean as that would result a 99.7% confidence interval of $$R=1.09556(85)*10^7m^{-1}$$, demonstrating that my answer is significantly lower than the actual value.

$$R=(\frac{2pi^2m·e^4}{h^3c})(\frac{mu_0*c^2}{4*pi})^2=1.0973733*10^7m^{-1}$$ $$lambda=\frac{1}{R×(\frac{1}{2^2}-\frac{1}{3^2})}=\frac{1}{1.0973733*10^7*(.13889)}=656.1nm$$
 * Rydberg constant of Deuterium:
 * From this the alpha line can be found as:
 * Comparing the alpha line of deuterium to the known value of hydrogen gives that the difference of .2 nanometers as the space differential of the lines.


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