What Factors Affect the Emission of a Photon by an Electron?

[NATSALANE]

an electron travels at 1 x 10⁶ m/s. when it hits a target, its speed decrease in half and a photo is released. find the phton's wavelength.

so far i know E _photon = (hc) / lamda

what about the speed? does the photons' energy same as kinetic energy of the electron which is just 1/2 mv ^2? if so, do i use the halved speed or the initial?

 

[nicksauce]

Apply conservation of energy:

E_electron_inital = E_electron_final + E_phton

 

[baba89]

I would like to clarify a few things regarding the emission of a photon by an electron. First, the speed of an electron is usually measured in units of meters per second (m/s), not meters per second squared (m/s^2). So, the given speed of 1 x 10^6 m/s is correct.

Secondly, when an electron hits a target, its speed does not decrease in half. The decrease in speed depends on the nature of the target and the surrounding environment. In some cases, the electron may lose all of its kinetic energy and come to a complete stop, while in other cases it may only lose a fraction of its speed.

Now, to answer your question about the energy of the photon released by the electron. Yes, the energy of the photon is equal to the kinetic energy of the electron before it hits the target. This is known as the photoelectric effect, where the energy of the photon is equal to the work function of the material it is hitting, which in turn is equal to the kinetic energy of the electron.

To calculate the wavelength of the photon, you can use the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength. You can use the kinetic energy of the electron to calculate the energy of the photon, and then solve for the wavelength.

However, it is important to note that the energy of the photon released may not always be equal to the kinetic energy of the electron. This depends on the specific conditions and materials involved in the emission process.

In summary, the emission of a photon by an electron is a complex process that involves many factors. The speed of the electron, the material it hits, and the energy of the photon are all important considerations in understanding this phenomenon. I hope this helps clarify your understanding.