STANDARD Chemistry Energy Levels MCQ Question
An electron is in an excited state in a hydrogen like atom. It has a total energy of –3.4 eV. The kinetic energy of the electron is E and its de-Broglie wavelength is λ. Then
E = 6.8 eV, λ = 6.6 × 10⁻¹⁰ m
E = 3.4 eV, λ = 6.6 × 10⁻¹⁰ m
E = 3.4 eV, λ = 6.6 × 10⁻¹¹ m
E = 6.8 eV, λ = 6.6 × 10⁻¹¹ m
Correct Answer
Detailed Explanation
The total energy is twice the kinetic energy with opposite sign. Thus, E = 3.4 eV. Using the de-Broglie wavelength formula, λ = h/√(2mE), we find λ = 6.6 × 10⁻¹⁰ m.
Found an issue with this question?
Related Questions
More from Atomic Structure
More from
What is the energy in joules associated with the fifth orbit of a hydrogen atom if the energy associated with the first orbit is -2.18 × 10^(-18) J?
Which combinations of statements about the quantum mechanical model of the atom are correct?
According to the de Broglie equation, what is the wavelength of an electron with a kinetic energy of 2.0 × 10^−25 J and mass 9.1 × 10^−31 kg?