Physics, 30.05.2020 14:57 lusan4u

Explain why a 5eV photon of light cannot be absorbed or emitted by the hydrogen atom

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None of the transitions in the hydrogen atom corresponds to a photon energy of 5eV hence no photon of this energy is absorbed or emitted by the hydrogen atom.

Explanation:

Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy. If it is in the second energy level, it must have -3.4 eV of energy and so on.

If the electron wants to jump from the first energy level, n = 1, to the second energy level n = 2. The second energy level has higher energy than the first, so to move from n = 1 to n = 2, the electron needs to gain energy. It needs to gain (-3.4) - (-13.6) = 10.2 eV of energy to be excited to the second energy level.

The step from the second energy level to the third is much smaller. It takes only 1.89 eV of energy for this excitation to take place. It takes even less energy to excite electrons in hydrogen from the third energy level to the fourth, and even less from the fourth to the fifth.

None of these transitions in the hydrogen atom corresponds to a photon energy of 5eV hence no photon of this energy is absorbed or emitted by the hydrogen atom.

Answer from: Quest

Which of the following are risk factors for emphysema ? correct answer : a and and c good luck and have a nice benevolentpuppy day! : )

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Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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Explain why a 5eV photon of light cannot be absorbed or emitted by the hydrogen atom...

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