A physics student is using a diffraction grating to separate the different wavel

Question

A physics student is using a diffraction grating to separate the different wavelengths of visible light in the Balmer series of atomic Hydrogen. The wavelengths of the Balmer series can be determined from the following equation: 1/lambda = R(1/2^2 - 1/n^2) n = 3, 4, 5... a) Using the formula given above, find the longest wavelength in the Balmer series. b) Using the formula given above, find the shortest wavelength in the Balmer series. c) Find the energy and momentum of a photon with the wavelength you found in b). d) Find the speed of an electron that has a de Broglie wavelength equal to the wavelength you found in b). e) To do this experiment the student shines a Hydrogen gas discharge tube light on to a diffraction grating. The separated spectral lines appear on a screen that is a distance of 0.870 meters away from the grating. The first order bright lines for the longest and shortest wavelengths in the Balmer series are separated by 2.00 cm on the screen. How many lines per centimeter does the diffraction grating have?

Explanation / Answer

a) For longest wavelength,n=3

Wavelength=36/(5R)

b) for shortest wavelength, n=infinity

Wavelength=4/R

c)Rydberg constant R=1.1*10^7

Wavelength of part b = (4/1.1)*10^-7 =3.636*10^-7m =363.6Nm

Energy = hc/wavelength=(6.626*10^-34)*(3*10^8)/(3.636*10^-7 )=5.47*10^-19 Joules

Momentum = h/wavelength =( 6.626*10^-34)/3.636*10^-7

=1.822*10^-27 kgm/s

d)velocity = p/2m where p=momentum and m= mass of electron

velocity =(1.822*10^-27)/(2*9.1*10^-34)=10^6 m/s

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