This is about the Balmer series of Hydrogen. a) What is the advantage of using a
ID: 1781950 • Letter: T
Question
This is about the Balmer series of Hydrogen.
a) What is the advantage of using a diffraction grating with greater lines/mm over that with a smaller lines/mm?
b) Use Eq. (1) to calculate the corresponding wavelength of the spectral lines with N = 3 and N = 4. Can a 300 lines/mm diffraction grating resolve the two wavelengths? Explain. (The Rydberg constant R = 1.097 x 10^7 m^-1)
uroTOr. Damncr s equaion can be written in the form an integer greater than or equal to 3 and R now known as the RydheroonExplanation / Answer
part a:
as resolution is inversely proportional to slit width and slit width is inversely proportional to number of lines , higher then number of lines, greater the resolution.
part b:
for N=3:
1/lambda=1.097*10^(7)*((1/4)-(1/9))=15.236 *10^5
==>lambda=656.34 nm
for N=4:
1/lambda=1.097*10^(7)*((1/4)-(1/16))=2056875
==>lambda=486.17
resolution power=656.34/(656.34-386.17)=(300)*W
where W is width of the grating
==>W=0.008098 mm
so as naturally produing a grating of this width is not possible, these lines can not be differentiated.
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