An electron inside an excited atom drops from an energy state of -1.20\text{ eV} to a lowe… — JAMB UTME

An electron inside an excited atom drops from an energy state of $-1.20\text{ eV}$ to a lower energy state of $-4.50\text{ eV}$ . What is the frequency of the emitted electromagnetic radiation, in units of $10^{14}\text{ Hz}$ ? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Cevap: 8 10^14 Hz

Cevap

The frequency of the emitted electromagnetic radiation is

8.0 \times 10^{14}\text{ Hz}

(or 8.0 in units of

10^{14}\text{ Hz}

The energy of the photon emitted during a downward transition between discrete atomic energy levels is equal to the energy difference between the initial and final states. Calculating

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

, converting to Joules gives

3.30 \times 1.6 \times 10^{-19}\text{ J} = 5.28 \times 10^{-19}\text{ J}

. Dividing this energy by Planck's constant

6.6 \times 10^{-34}\text{ J s}

yields a frequency of

8.0 \times 10^{14}\text{ Hz}

Adım Adım Çözüm

Determine the energy of the emitted photon in electron-volts

\Delta E = 3.30\text{ eV}

The energy of the emitted photon equals the difference between the upper and lower atomic energy levels:

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

Convert photon energy from electron-volts to Joules

\Delta E = 5.28 \times 10^{-19}\text{ J}

Since Planck's constant is given in SI units (J s), energy must be converted to Joules using

1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

Calculate photon frequency using Planck's relation

f = 8.0 \times 10^{14}\text{ Hz}

Applying

f = \frac{\Delta E}{h} = \frac{5.28 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J s}}

yields

8.0 \times 10^{14}\text{ Hz}

Anahtar Kavram

Photon Emission and Energy Level Transitions

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