When the electron in a Li2+ ion makes a transition from one energy level to another, a photon of wavelength 670.8 nm is emitted. The energy difference between these two energy levels is单项选择题
A
3.37 × 1018 J
B
3.37 × 1027 J
C
4.45 × 10–40 J
D
2.97 × 10–19 J
E
2.97 × 10–28 J

Question

When the electron in a Li2+ ion makes a transition from one energy level to another, a photon of wavelength 670.8 nm is emitted. The energy difference between these two energy levels is

BlackTom AI · Accepted Answer

We are given that a Li2+ ion emits a photon with wavelength 670.8 nm during an electronic transition, and we need the energy difference between the two levels that corresponds to this photon.

First, convert the wavelength to meters: 670.8 nm = 670.8 × 10^-9 m = 6.708 × 10^-7 m.
Second, use the photon energy formula E = h c / λ, where h is Planck's constant (6.626 × 10^-34 J·s) and c is the speed of light (3.00 × 10^8 m/s).
Compute the numerator: h c = (6.626 × 10^-34) × (3.00 × 10^8) ≈ 1.9878 × 10^-25 J·m.
Divide by the wavelength: E ≈ (1.9878 × 10^-25) / (6.708 × 10^-7) ≈ 2.96 × 10^-19 J.
Rounding to appropriate significant figures gives E ≈ 2.97 × 10^-19 J.

Now assess the options:
- Option A: 3.37 × 10^18 J. This is orders of magnitude off (positive mismatch in exponent and magnitude) and does not reflect the tiny energy of a visible photon. It is clearly unreasonable for atomic transitions.
- Option B: 3.37 × 10^27 J. This is far too large by many orders of magnitude; atomic transition energies are typically in the 10^-19 to 10^-18 J range for visible photons, not 10^27 J.
- Option C: 4.45 × 10^-40 J. This is many orders of magnitude smaller than the actual photon energy; it would correspond to an unbelievably tiny energy, not consistent with visible-light photons.
- Option D: 2.97 × 10^-19 J. This matches the calculated value from E = h c / λ for λ = 670.8 nm, and aligns with the physics of a photon emitted during an electronic transition.
- Option E: 2.97 × 10^-28 J. This is still far from the correct energy by about nine orders of magnitude; it's not consistent with the visible photon energy here.

In summary, the correct energy difference for the emitted photon at 670.8 nm is approximately 2.97 × 10^-19 J, while all the other options deviate by several orders of magnitude or misplace the exponent, making them incorrect.

When the electron in a Li2+ ion makes a transition from one energy level to another, a photon of wavelength 670.8 nm is emitted. The energy difference between these two energy levels is单项选择题

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