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Question: Energy level of a hypothetical atom are given as shown. Which transition will give photon of wavelen...

Energy level of a hypothetical atom are given as shown. Which transition will give photon of wavelength 275 nm ?

A

A

B

B

C

C

D

D

Answer

B

Explanation

Solution

  1. Calculate Photon Energy: The energy (EE) of a photon is related to its wavelength (λ\lambda) by the formula:

    E=hcλE = \frac{hc}{\lambda}

    where hh is Planck's constant (6.626×10−34 J s6.626 \times 10^{-34} \text{ J s}), and cc is the speed of light (3×108 m/s3 \times 10^8 \text{ m/s}). Given λ=275 nm=275×10−9 m\lambda = 275 \text{ nm} = 275 \times 10^{-9} \text{ m}.

    E=(6.626×10−34 J s)×(3×108 m/s)275×10−9 m≈7.228×10−19 JE = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{275 \times 10^{-9} \text{ m}} \approx 7.228 \times 10^{-19} \text{ J}

  2. Convert Energy to Electron Volts (eV): Convert the photon energy from Joules to eV using the conversion factor 1 eV=1.602×10−19 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}.

    EeV=7.228×10−19 J1.602×10−19 J/eV≈4.51 eVE_{\text{eV}} = \frac{7.228 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 4.51 \text{ eV}

    (Alternatively, use the shortcut E(eV)=1240λ(nm)=1240275≈4.509 eVE(\text{eV}) = \frac{1240}{\lambda(\text{nm})} = \frac{1240}{275} \approx 4.509 \text{ eV})

    So, the photon has an energy of approximately 4.5 eV4.5 \text{ eV}.

  3. Calculate Energy Differences for Transitions: For each given transition, calculate the energy difference between the initial and final energy levels.

    • Transition A: From 0 eV0 \text{ eV} to −2 eV-2 \text{ eV}

      ΔEA=0−(−2 eV)=2 eV\Delta E_A = 0 - (-2 \text{ eV}) = 2 \text{ eV}

    • Transition B: From 0 eV0 \text{ eV} to −4.5 eV-4.5 \text{ eV}

      ΔEB=0−(−4.5 eV)=4.5 eV\Delta E_B = 0 - (-4.5 \text{ eV}) = 4.5 \text{ eV}

    • Transition C: From −2 eV-2 \text{ eV} to −4.5 eV-4.5 \text{ eV}

      ΔEC=−2−(−4.5 eV)=2.5 eV\Delta E_C = -2 - (-4.5 \text{ eV}) = 2.5 \text{ eV}

    • Transition D: From −4.5 eV-4.5 \text{ eV} to −10 eV-10 \text{ eV}

      ΔED=−4.5−(−10 eV)=5.5 eV\Delta E_D = -4.5 - (-10 \text{ eV}) = 5.5 \text{ eV}

  4. Match the Energies: Compare the calculated photon energy (4.5 eV4.5 \text{ eV}) with the energy differences for each transition. Transition B has an energy difference of 4.5 eV4.5 \text{ eV}, which exactly matches the energy of the photon with a wavelength of 275 nm.