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I am having problems with these problems. They are review for my test. I have done the rest of the problems but these three seem to be giving me problems. Any help would be greatly appreciated.

1) Bohr’s Model can be used to find the wavelengths of the photons in the absorption or emission spectrum of hydrogen atom.

Using that model, find the wavelengths of the 5→2, 4→2, 3→2 and 2→1 transitions. Specify the colour associated with these wavelengths when possible. To determine the colours, use tables. If the wavelength is not part of the visible spectrum, specify if it is ultraviolet or infrared.

2) A diffraction grating is used to find the visible wavelengths of excited hydrogen gas (wavelengths found in #1). Using a diffraction grating with 8000 lines/cm, find the angle at which these emission lines interfere constructively (m = 1).

3) Why can we use a diffraction grating to identify elements?

1) To predict the wavelength of light emitted transiting from a higher energy state to a lower energy state you calculate the energy of each energy level, subtract these energies and then make this energy equal to the energy of the emitted light photon.

For example consider the wavelength emitted in the transition of hydrogen from n=3 to n=2.

The energy levels for hydrogen gas is given by:

Un=-13.6eV/n^2

For level n=2 this energy will be:

U2=-13.6/2^2=-3.4eV

For level n=3 this energy will be:

U3=-13.6/3^2=-1.51eV

The corresponding energy transition will be equal to the difference in these two levels:

U32=-3.4-1.51=-1.89eV

Since the atom loses 1.89eV of energy the electron must gain this energy. Therefore, the emitted photon will have an energy of:

E=1.89eV=3.01x10^(-19)Joules where 1.6x10^(-19)joules=1eV

The energy of a light photon is given by:

E=h*c/lambda where h=6.63x10^(-34) Joule Sec, c=3.0x10^8 m/s is the speed of light and lambda is the wavelength of the emitted light photon.

Making these two energies equal and solving for the wavelength of the emitted photon:

Lambda=h*c/E=6.61x10^-7m = 661nm which is in the red region of the visible spectrum.

The other calculations are done similarly!

2) The interference pattern produced by a diffraction grating is given by:

m*lambda=d*sin(theta) where m is the order of the antinode, lambda is the wavelength of the light, d is the distance between adjacent slits and sin(theta) is the angle at which the m'th antinode occurs.

In this case m=1, lambda=6.61x10^-7m as calculated above, d=1/800cm=1.25x10^-5m and theta is the answer.

Solving for theta:

theta=inverse sine(m*lambda/d)=inv sin(1*6.61x-7/1.25x10^-5)=3.03degrees.

Again, do similarly for the other wavelengths.

3) Since each element has its own set of energy levels, the emission spectra are unique for each element and can be used like a finger print.

Physics

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.