Requency for a photon emitted from a hydrogen atom

[pak213]

I am having some trouble w/ a take home Physics final.

I am down to my last 2 questions:

1) What is the frequency for a photon emitted from a hydrogen atom when an electron makes a transition from an energy state n=3 to n=1? What is the energy of the photon in Joules and MeV?

2) A positron and an electron collide and annihilate each other. Since all of each of masses was converted into energy, how much energy was released in Joules?

All the help/direction you can give would be much appreciated.

 

[quasar987]

I am having some trouble w/ a take home Physics final.

I am down to my last 2 questions:

1) What is the frequency for a photon emitted from a hydrogen atom when an electron makes a transition from an energy state n=3 to n=1? What is the energy of the photon in Joules and MeV?

2) A positron and an electron collide and annihilate each other. Since all of each of masses was converted into energy, how much energy was released in Joules?

All the help/direction you can give would be much appreciated.

1) The energy of an electron in an atom depends on the energy state he's in. The energy of the fundamental level (n=1) is -13.6 eV. Find the energy of n=3 in your book. You are told the electron jumps from the energy of n=3 to the energy of n = 1. Now according to the principle of conservation of energy, that energy must have gone somewhere: in the photon. The frequency of a photon is related to its energy according to E = h\nu

2) They just want to know the energy associated with the masses of the electron and positron. Use E = mc².

 

[thepatientmental]

Hi there,

I understand you are having some trouble with your take home Physics final. Don't worry, I am here to help you with your last two questions.

1) The frequency for a photon emitted from a hydrogen atom can be calculated using the formula:

f = R(1/n1^2 - 1/n2^2)

Where R is the Rydberg constant and n1 and n2 are the initial and final energy states respectively. For this question, n1 = 3 and n2 = 1.

Plugging in these values, we get:

f = R(1/1^2 - 1/3^2) = R(1 - 1/9) = R(8/9)

Next, we can calculate the energy of the photon using the formula:

E = hf

Where h is the Planck's constant and f is the frequency calculated above. Substituting the values, we get:

E = (6.626 x 10^-34 J.s)(8/9) = 5.89 x 10^-34 J

To convert this into MeV, we can use the conversion factor 1 MeV = 1.602 x 10^-13 J. So, the energy of the photon in MeV would be:

E = (5.89 x 10^-34 J)(1 MeV/1.602 x 10^-13 J) = 3.67 x 10^-21 MeV

2) When a positron and an electron collide and annihilate each other, all of their mass is converted into energy according to Einstein's famous equation E = mc^2. Since the mass of an electron and positron is equal to 9.11 x 10^-31 kg, the total energy released would be:

E = (9.11 x 10^-31 kg)(3 x 10^8 m/s)^2 = 8.2 x 10^-14 J

I hope this helps you with your final. Good luck!