Electron Transitions

  • Thread starter scorpa
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Hello Everyone,

I've been doing some physics homework and have gotten all but two of the questions. They are on electron transitions.

#1) In a hypothetical hyrogen like atom the energy of the second energy level is -13.6eV. Calculate the momentum of the electron in the third energy level of this atom.

En = (1/3^2)(13.6) = 1.51 ev
1.51 eV = 2.42 x 10^-19 J
Ek = (1/2)mv^2
square root of (2.42x10^-19J/(0.5 x 9.11 x 10^-31) )
v = 7.3 x 10^5 m/s
p = mv
p = (9.11x10^-31kg)(7.3x10^5m/s) = 6.7 x 10^-25 kg*m/s

The answer should be 1.34 x 10^-24 kg*m/s

I have no idea what I've done wrong



#2) A spectra line in the hydrogen bright line spectra (NOTE: visible region) passes perpendicularily through a diffraction grating (d= 5.00x10^-6m). This spectra line produces a first order maximum at an angle of 4.97 degrees. What is the transition of the elctron that produces this spectra line?

wavelength = (5.00 x 10^-6m)(sin4.97)/1 = 4.33 x 10^-7m

Using Rydberg's equation I subsituted and manipulated and got:

(1/4) - ((1/4.33x10^-7)/1.1x 10^7) then I took the square root of that and got 0.2 which definitely isn't right.

The answer should be 5 to 2. Once again I don't know where I've gone wrong.
Any help is greatly appreciated, thanks in advance!
 

Answers and Replies

  • #2
OlderDan
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scorpa said:
#1) In a hypothetical hyrogen like atom the energy of the second energy level is -13.6eV. Calculate the momentum of the electron in the third energy level of this atom.

En = (1/3^2)(13.6) = 1.51 ev
1.51 eV = 2.42 x 10^-19 J
Ek = (1/2)mv^2
square root of (2.42x10^-19J/(0.5 x 9.11 x 10^-31) )
v = 7.3 x 10^5 m/s
p = mv
p = (9.11x10^-31kg)(7.3x10^5m/s) = 6.7 x 10^-25 kg*m/s

The answer should be 1.34 x 10^-24 kg*m/s

I have no idea what I've done wrong
I will look at these one at a time.

Two things you need to look at. 1) Your equation for En is not giving you the correct result for the initial energy level that was stated to be the SECOND energy level, not the first. 2) What are you doing about the potential energy in your calculations?
 
  • #3
OlderDan
Science Advisor
Homework Helper
3,021
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scorpa said:
#2) A spectra line in the hydrogen bright line spectra (NOTE: visible region) passes perpendicularily through a diffraction grating (d= 5.00x10^-6m). This spectra line produces a first order maximum at an angle of 4.97 degrees. What is the transition of the elctron that produces this spectra line?

wavelength = (5.00 x 10^-6m)(sin4.97)/1 = 4.33 x 10^-7m

Using Rydberg's equation I subsituted and manipulated and got:

(1/4) - ((1/4.33x10^-7)/1.1x 10^7) then I took the square root of that and got 0.2 which definitely isn't right.

The answer should be 5 to 2. Once again I don't know where I've gone wrong.
Any help is greatly appreciated, thanks in advance!
I think your wavelength is OK, but I don't follow your application of the Rydberg. Take a look here and find a common denominator for the difference of 1/n^2 and combine fractions so that you can solve for the reciprocal of the fraction. You will have n1^2n2^2 in the denominator. You have the information to find the value of this fraction. Figure out which combination of small integer values of n give you the right ratio.

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html
 

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