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Problems · Problem 2.10

Q.What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?

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✓ Free question

When an electron drops from n=5n=5 to n=2n=2 in hydrogen, it emits a photon whose energy equals the difference between these two levels. Using the Rydberg formula, we find wavelength λ=434\lambda = 434 nm and frequency ν=6.91×1014\nu = 6.91 \times 10^{14} Hz (visible blue-violet light).

The hydrogen atom's electron exists in discrete energy levels labeled by the principal quantum number nn. When an electron transitions from a higher level to a lower one, it must shed the exact energy difference by emitting a photon. That photon's frequency and wavelength are locked to this energy through Planck's relation E=hνE = h\nu and the wave equation c=λνc = \lambda\nu.

The energy of the nn-th level in hydrogen is given by:

En=−13.6 eVn2E_n = -\frac{13.6 \text{ eV}}{n^2}

The negative sign reflects that the electron is bound; zero energy corresponds to ionization.

Step-by-step calculation

  1. Find the initial and final energies

    Using the direct-Joule form of the Rydberg energy expression, En=−2.18×10−18n2E_n = -\dfrac{2.18 \times 10^{-18}}{n^2} J:

    For n=5n = 5:

E5=−2.18×10−1825=−8.72×10−20 JE_5 = -\frac{2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20} \text{ J}

For n=2n = 2:

E2=−2.18×10−184=−5.45×10−19 JE_2 = -\frac{2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19} \text{ J}

  1. Calculate the energy of the emitted photon

    The photon carries away the energy difference:

ΔE=E5−E2=−8.72×10−20−(−5.45×10−19)=4.58×10−19 J\Delta E = E_5 - E_2 = -8.72 \times 10^{-20} - (-5.45 \times 10^{-19}) = 4.58 \times 10^{-19} \text{ J}

  1. Determine the frequency using Planck's relation

    From E=hνE = h\nu, where h=6.626×10−34h = 6.626 \times 10^{-34} J·s:

ν=ΔEh=4.58×10−196.626×10−34=6.91×1014 Hz\nu = \frac{\Delta E}{h} = \frac{4.58 \times 10^{-19}}{6.626 \times 10^{-34}} = 6.91 \times 10^{14} \text{ Hz}

  1. Find the wavelength from the wave equation

    Using c=λνc = \lambda\nu, where c=3×108c = 3 \times 10^8 m/s:

λ=cν=3×1086.91×1014=4.34×10−7 m=434 nm\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{6.91 \times 10^{14}} = 4.34 \times 10^{-7} \text{ m} = 434 \text{ nm}

Note

This transition is part of the Balmer series (all transitions ending at n=2n=2), which produces visible light. The 434 nm line appears as blue-violet and is one of the prominent lines in hydrogen's emission spectrum.

Rydberg formula (alternative direct approach):

1λ=RH(1nf2−1ni2)\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

where RH=1.097×107R_H = 1.097 \times 10^7 m−1^{-1} is the Rydberg constant. For ni=5n_i=5, nf=2n_f=2:

1λ=1.097×107(14−125)=1.097×107×0.21=2.30×106 m−1\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{25}\right) = 1.097 \times 10^7 \times 0.21 = 2.30 \times 10^6 \text{ m}^{-1}

λ=434 nm\lambda = 434 \text{ nm}

Watch out

Always subtract energies in the correct order: Einitial−EfinalE_{\text{initial}} - E_{\text{final}}. Since both are negative and E5>E2E_5 > E_2 (less negative = higher), the difference is positive, as it must be for an emitted photon.

✓Final answer

The photon has wavelength λ=434\lambda = 434 nm and frequency ν=6.91×1014\nu = 6.91 \times 10^{14} Hz.

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