# Requency for a photon emitted from a hydrogen atom

• pak213
In summary, the conversation is about two physics questions, the first one involving the frequency and energy of a photon emitted from a hydrogen atom during an electron's transition between energy states, and the second one involving the energy released from the annihilation of an electron and positron. The solution to the first question involves finding the energy of the electron in different energy states and using the principle of conservation of energy to find the frequency of the photon. The solution to the second question involves using the famous equation E=mc² to calculate the energy released from the masses of the particles.
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.

pak213 said:
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.

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!

## 1. What is the equation for calculating the frequency of a photon emitted from a hydrogen atom?

The frequency of a photon emitted from a hydrogen atom can be calculated using the equation E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 joule seconds), and f is the frequency in hertz.

## 2. How does the energy level of a hydrogen atom affect the frequency of the emitted photon?

The energy level of a hydrogen atom directly affects the frequency of the emitted photon. As the energy level increases, the frequency of the emitted photon also increases. This relationship is described by the Rydberg formula: 1/λ = R(1/n1^2 - 1/n2^2), where λ is the wavelength of the emitted photon, R is the Rydberg constant (1.097 x 10^7 m^-1), and n1 and n2 are the initial and final energy levels, respectively.

## 3. What is the relationship between frequency and wavelength for a photon emitted from a hydrogen atom?

The frequency and wavelength of a photon emitted from a hydrogen atom are inversely proportional. This means that as the frequency increases, the wavelength decreases, and vice versa. This relationship is described by the equation c = fλ, where c is the speed of light in a vacuum (3.00 x 10^8 m/s), f is the frequency, and λ is the wavelength.

## 4. How does the mass of a hydrogen atom affect the frequency of the emitted photon?

The mass of a hydrogen atom does not directly affect the frequency of the emitted photon. However, the mass of the atom does affect its energy level, which in turn affects the frequency of the emitted photon. This is because the energy level is determined by the mass and charge of the electron in the atom.

## 5. What is the significance of the frequency of a photon emitted from a hydrogen atom?

The frequency of a photon emitted from a hydrogen atom is significant because it determines the type of electromagnetic radiation it belongs to. Higher frequencies correspond to higher energy photons, which can have more damaging effects on living organisms. Additionally, the frequency of a photon can be used to identify the element from which it was emitted, as each element has a unique emission spectrum.

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