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An electron, in a hydrogen-hydrogen-like atom, is in an excited state. It has a total energy of – 3.4 eV. Calculate (i) the kinetic energy and (ii) the de Broglie wavelength of the electron

An electron, in a hydrogen-hydrogen-like atom, is in an excited state. It has a total energy of – 3.4 eV. Calculate (i) the kinetic energy and (ii) the de Broglie wavelength of the electron 

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Hello Student,
Please find the answer to your question
(i) En = - 3.4 eV
The kinetic energy is equal to the magnitude of total energy in this case.
∴ K.E. = + 3.4 eV
(ii) The de Broglie wavelength of electron
Λ = h / √ 2mK = 6.64 x 10-34 / √ 2 x 9.1 x 10-31 x 3.4 x 1.6 x 10-19 eV
= 0.66 x 10-9 m
Thanks
Navjot Kalra
askIITians Faculty

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