Examkin

Soru

Zorluk: OrtaEnergy Levels and Atomic Spectra

In a mercury vapor tube, an excited atom transitions from an upper energy level of −3.71 eV-3.71\text{ eV} to a lower energy level of −5.54 eV-5.54\text{ eV}. What is the frequency of the emitted photon? (Take h=6.6×10−34 J sh = 6.6 \times 10^{-34}\text{ J s} and 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

  1. 4.44×1014 Hz4.44 \times 10^{14}\text{ Hz}Cevap
  2. B
    2.24×1015 Hz2.24 \times 10^{15}\text{ Hz}
  3. C
    1.34×1015 Hz1.34 \times 10^{15}\text{ Hz}
  4. D
    2.77×1033 Hz2.77 \times 10^{33}\text{ Hz}

Cevap

The frequency of the emitted photon is 4.44×1014 Hz4.44 \times 10^{14}\text{ Hz}.
The energy lost by the atom during transition is ΔE=−3.71 eV−(−5.54 eV)=1.83 eV\Delta E = -3.71\text{ eV} - (-5.54\text{ eV}) = 1.83\text{ eV}. Converting to Joules gives 1.83×1.6×10−19 J=2.928×10−19 J1.83 \times 1.6 \times 10^{-19}\text{ J} = 2.928 \times 10^{-19}\text{ J}. Dividing this energy by Planck's constant (6.6×10−34 J s6.6 \times 10^{-34}\text{ J s}) yields a photon frequency of 4.44×1014 Hz4.44 \times 10^{14}\text{ Hz}.

Adım Adım Çözüm

1
Calculate the energy difference (ΔE\Delta E) between the two atomic energy levels.
ΔE=Einitial−Efinal=−3.71 eV−(−5.54 eV)=1.83 eV\Delta E = E_{\text{initial}} - E_{\text{final}} = -3.71\text{ eV} - (-5.54\text{ eV}) = 1.83\text{ eV}
The energy of the emitted photon equals the difference in energy between the initial and final states.
2
Convert the energy difference from electron-volts (eV) to Joules (J).
ΔE=1.83×1.6×10−19 J=2.928×10−19 J\Delta E = 1.83 \times 1.6 \times 10^{-19}\text{ J} = 2.928 \times 10^{-19}\text{ J}
Planck's constant is given in SI units (J s), so the energy must be in Joules.
3
Apply Planck's equation E=hfE = hf to find the photon frequency ff.
f=ΔEh=2.928×10−19 J6.6×10−34 J s≈4.44×1014 Hzf = \frac{\Delta E}{h} = \frac{2.928 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J s}} \approx 4.44 \times 10^{14}\text{ Hz}
The frequency of an emitted photon is directly proportional to its energy difference.

Anahtar Kavram

Atomic transition photon frequency calculation (E=hf=E2−E1E = hf = E_2 - E_1).
Tahmini Süre:1m 30s
Bu soruyu puanla

Konular

Energy Levels and Atomic Spectra

Bu konudan daha fazla soru