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PhyHardAdv 2013 • Paper 2

Emitted Wavelengths of De-exciting Hydrogen Like Atom

Consider a hydrogen-like atom where the electron's orbital radius is 4.5a04.5 a_0, with a0a_0 representing the Bohr radius. The orbital angular momentum of the electron is 3h2π\frac{3h}{2\pi}, where hh is Planck's constant and RR is the Rydberg constant. Identify all possible wavelength(s) of radiation emitted when the atom de-excites:

Options

A

932R\frac{9}{32R}

Correct
B

916R\frac{9}{16R}

C

95R\frac{9}{5R}

Correct
D

43R\frac{4}{3R}

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To determine the possible wavelengths emitted when the atom de-excites, we use Bohr's model of the hydrogen-like atom.

1. Determine the principal quantum number nn: According to Bohr's quantization condition, the orbital angular momentum LL of an electron is given by: L=nh2πL = \frac{n h}{2\pi}

Given that the orbital angular momentum is 3h2π\frac{3h}{2\pi}: nh2π=3h2π  ⟹  n=3\frac{n h}{2\pi} = \frac{3h}{2\pi} \implies n = 3

2. Determine the atomic number ZZ: The radius of the nthn^{\text{th}} orbit in a hydrogen-like atom is given by: rn=n2Za0r_n = \frac{n^2}{Z} a_0

Given that rn=4.5a0=92a0r_n = 4.5 a_0 = \frac{9}{2} a_0 and n=3n = 3: 32Za0=92a0  ⟹  9Z=92  ⟹  Z=2\frac{3^2}{Z} a_0 = \frac{9}{2} a_0 \implies \frac{9}{Z} = \frac{9}{2} \implies Z = 2

3. Calculate the emitted wavelengths during de-excitation: The Rydberg formula for the wavelength λ\lambda of radiation emitted in a transition from an initial state n2n_2 to a final state n1n_1 is: 1λ=RZ2(1n12−1n22)\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Substituting Z=2Z = 2: 1λ=4R(1n12−1n22)\frac{1}{\lambda} = 4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Since the electron starts in the state n=3n = 3, the possible transitions are:

  • Transition 3→13 \to 1: 1λ3→1=4R(112−132)=4R(1−19)=4R(89)=32R9\frac{1}{\lambda_{3 \to 1}} = 4R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 4R \left( 1 - \frac{1}{9} \right) = 4R \left( \frac{8}{9} \right) = \frac{32R}{9} λ3→1=932R\lambda_{3 \to 1} = \frac{9}{32R}

  • Transition 3→23 \to 2: 1λ3→2=4R(122−132)=4R(14−19)=4R(536)=5R9\frac{1}{\lambda_{3 \to 2}} = 4R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9} λ3→2=95R\lambda_{3 \to 2} = \frac{9}{5R}

  • Transition 2→12 \to 1: 1λ2→1=4R(112−122)=4R(1−14)=4R(34)=3R\frac{1}{\lambda_{2 \to 1}} = 4R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R λ2→1=13R\lambda_{2 \to 1} = \frac{1}{3R}

Matching these results with the given options, we find that:

  • 932R\frac{9}{32R} corresponds to Option (A).
  • 95R\frac{9}{5R} corresponds to Option (C).

Therefore, the correct options are A and C.

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