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The $\mathrm{H}_{\alpha}$ line in hydrogen has a wavelength of 656.20 $\mathrm{nm} .$ This line differs in wavelength from the corresponding spectral line in deuterium (the heavy stable isotope of hydrogen) by 0.18 $\mathrm{nm}$ . (a) Determine the minimum number of lines a grating must have to resolve these two wavelengths in the first order. (b) Repeat part (a) for the second order.

a. $3.6 \times 10^{3}\mathrm{lines}$

b. $1.8 \times 10^{3}\mathrm{lines}$

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So in this problem, it is given that the web link of the H off a hydrogen line is Lambda calls 6 56.2 nanometers and the difference in Web link between the age Alfa hydrogen line and off do Torrey, um, line is Delta Lambda equals 0.18 nanometer. So for part A to find the minimum number of lance needed in the creating elements to be able to resolve these two lines in the first order, we used the fact that the resulting power of the diffraction creating is for part a The resolving power is our coast Lambda over Delta Lambda. Okay, which is an times M. And here is number of flights, and M is the order off diffraction pattern. All right, so from here, we can write down and because are over em, which is going to be cool. D'oh! Lambda Over. Delta Lambda! Over M. Okay, so this is going to give me Lambda is 6 56.2 nanometer delta Lambda, 0.18 nanometer. And remember, this is the first order. So am is going to be go to one. So 6 56.2 over 0.18 Divide by one and this is going to give me 3.6 times tenderly, three lies. So for part B, everything is basically the same, but the order is to now. So what we're gonna do is for part B. M is going to be replaced by too. So and is going to be Lambda over Delta Lambda. Like I said, it was going to be too now. And this is very straightforward. We will get 1.8 times 10 33 less.

University of Wisconsin - Milwaukee