Calculating the Kinetic Energy and De Broglie Wavelength of an Ejected Electron

[thursdaytbs]

If 15eV photon interacts with a Hydrogen atom at groundstate [-13.6eV], and all 15eV is transferred to the atom. How would the KE of the ejected electron be found? and what is the de Broglie wavelength of the electron?

For the Kinetic Energy, i said 15eV = KE + 13.6eV, KE = 1.4eV

for the wavelength, i have no clue. any help?

 

[theCandyman]

The de Broglie wavelength is found by dividing Planck's constant by the momentum of the moving object. You can find the momentum using the kinetic energy.

 

[thepatientmental]

To find the kinetic energy of the ejected electron, we can use the equation KE = hf - BE, where KE is the kinetic energy, hf is the energy of the incident photon (15eV in this case), and BE is the binding energy of the electron in the ground state (-13.6eV for Hydrogen). Plugging in the values, we get KE = 15eV - (-13.6eV) = 1.4eV. This means that the ejected electron will have a kinetic energy of 1.4 electron volts.

To find the de Broglie wavelength of the electron, we can use the equation λ = h/p, where λ is the de Broglie wavelength, h is Planck's constant (6.626 x 10^-34 joule seconds), and p is the momentum of the electron. The momentum of the electron can be calculated using the equation p = √(2mKE), where m is the mass of the electron (9.109 x 10^-31 kg) and KE is the kinetic energy we calculated earlier (1.4eV).

Plugging in the values, we get p = √(2(9.109 x 10^-31 kg)(1.4eV)(1.602 x 10^-19 J/eV)) = 1.48 x 10^-24 kg m/s. Now, we can use this value to calculate the de Broglie wavelength: λ = (6.626 x 10^-34 joule seconds)/(1.48 x 10^-24 kg m/s) = 4.47 x 10^-10 meters. Therefore, the de Broglie wavelength of the ejected electron is 4.47 x 10^-10 meters.