A free electron is captured by an alpha particle to form an excited state of a h

Question

A free electron is captured by an alpha particle to form an excited state of a helium ion. During this process, a photon of energy E is emitted followed by another one of 48.4 eV. No further photons are emitted. The ionization energy for the helium ion is 54.4 eV.

a) Determine the frequency of the 48.4 eV photon.

b) For the first photon emitted:

i. Determine its energy E

ii. Determine its wavelength

c) The ion is in its ground state when it is struck by a 70 eV photon. All the photons energy is transferred to the electron, freeing it. Determine the de Broglie wavelength of the emitted electron.

Explanation / Answer

a) 48.4 = h*frequency/ (1.6*10^-19)

=> frequency = 1.17*10^16 Hz

b) E = 54.4 - 48.4 = 6 eV

6 = h*c/(wavelenrth *1.6*10^-19)

=> wavelength = 2*10^-7 m

c) kinetic energy of electron = 70 -54.4 = 15.6 eV =2.5*10^-18 J

=> de Broglie wavelength = h/sqrt(2*m*kinetic energy in joules) = 3 angstrom

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