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PhyHardAdv 2019 • Paper 2

Hydrogen Atom Photon Absorption Emission and Momentum Change

A stationary free hydrogen atom absorbs a photon with wavelength λa\lambda_a, transitioning from energy level n=1n = 1 to n=4n = 4. Following this, the electron drops to a level n=mn = m by releasing a photon of wavelength λe\lambda_e. The changes in momentum of the atom corresponding to the absorption and emission processes are Δpa\Delta p_a and Δpe\Delta p_e, respectively. Given that λa/λe=15\lambda_a / \lambda_e = \frac{1}{5}, select the correct option(s): [Take hc=1242 eV nmhc = 1242\text{ eV nm}; 1 nm=10−9 m1\text{ nm} = 10^{-9}\text{ m}; hh and cc represent Planck's constant and light speed]

Options

A

m=2m = 2

Correct
B

λe=418 nm\lambda_e = 418\text{ nm}

C

Δpa/Δpe=12\Delta p_a / \Delta p_e = \frac{1}{2}

D

The ratio of the kinetic energy of the electron in state n=mn = m to that in state n=1n = 1 equals 14\frac{1}{4}

Correct

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To determine the correct options, we analyze the energy level transitions of the hydrogen atom according to Bohr's model.

1. Energy of Absorbed and Emitted Photons

The energy levels of a hydrogen atom are given by: En=−E0n2E_n = -\frac{E_0}{n^2} where E0=13.6 eVE_0 = 13.6\text{ eV}.

For the absorption process (n=1→n=4n = 1 \to n = 4), the energy of the absorbed photon is: Ea=E4−E1=E0(1−142)=1516E0=1516×13.6 eV=12.75 eVE_a = E_4 - E_1 = E_0 \left( 1 - \frac{1}{4^2} \right) = \frac{15}{16} E_0 = \frac{15}{16} \times 13.6\text{ eV} = 12.75\text{ eV}

Since the photon energy is inversely proportional to its wavelength (E=hcλE = \frac{hc}{\lambda}), the ratio of energies is given by: EeEa=λaλe=15\frac{E_e}{E_a} = \frac{\lambda_a}{\lambda_e} = \frac{1}{5}

Thus, the energy of the emitted photon during the transition (n=4→n=mn = 4 \to n = m) is: Ee=Ea5=12.75 eV5=2.55 eVE_e = \frac{E_a}{5} = \frac{12.75\text{ eV}}{5} = 2.55\text{ eV}

2. Determination of mm (Option A)

The energy emitted during the transition n=4→n=mn = 4 \to n = m is also expressed as: Ee=E4−Em=13.6(1m2−116) eVE_e = E_4 - E_m = 13.6 \left( \frac{1}{m^2} - \frac{1}{16} \right)\text{ eV}

Equating this to 2.55 eV2.55\text{ eV}: 13.6(1m2−116)=2.5513.6 \left( \frac{1}{m^2} - \frac{1}{16} \right) = 2.55

1m2−116=2.5513.6=316\frac{1}{m^2} - \frac{1}{16} = \frac{2.55}{13.6} = \frac{3}{16}

1m2=316+116=416=14\frac{1}{m^2} = \frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}

m2=4  ⟹  m=2m^2 = 4 \implies m = 2

Thus, Option A is correct.

3. Calculation of Emission Wavelength λe\lambda_e (Option B)

Using hc=1242 eV nmhc = 1242\text{ eV nm}: λe=hcEe=1242 eV nm2.55 eV≈487.06 nm\lambda_e = \frac{hc}{E_e} = \frac{1242\text{ eV nm}}{2.55\text{ eV}} \approx 487.06\text{ nm}

Since 487.06 nm≠418 nm487.06\text{ nm} \neq 418\text{ nm}, Option B is incorrect.

4. Ratio of Momentum Changes (Option C)

The magnitude of the change in momentum of the atom due to absorption or emission of a photon is equal to the momentum of the corresponding photon: Δpa=hλaandΔpe=hλe\Delta p_a = \frac{h}{\lambda_a} \quad \text{and} \quad \Delta p_e = \frac{h}{\lambda_e}

The ratio of the momentum changes is: ΔpaΔpe=λeλa=5\frac{\Delta p_a}{\Delta p_e} = \frac{\lambda_e}{\lambda_a} = 5

Since 5≠125 \neq \frac{1}{2}, Option C is incorrect.

5. Ratio of Kinetic Energies (Option D)

The kinetic energy of an electron in the nn-th orbit is: Kn=E0n2K_n = \frac{E_0}{n^2}

For n=m=2n = m = 2 and n=1n = 1: KmK1=K2K1=1/221/12=14\frac{K_m}{K_1} = \frac{K_2}{K_1} = \frac{1/2^2}{1/1^2} = \frac{1}{4}

Thus, Option D is correct.

Conclusion

The correct options are A and D.

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