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In photosynthesis in plants, light is absorbed in light-harvesting complexes that consist of protein and pigment molecules. The absorbed energy is then transported to a specialized complex called the reaction center.

Quantum-mechanical effects may play an important role in this energy transfer. In a recent experiment, researchers cooled rubidium atoms to a very low temperature to study a similar energy transfer process in the lab. Laser light was used to excite an electron in each atom to a state with large n. This highly excited electron behaves much like the single electron in a hydrogen atom, with an effective (screened) atomic number${\mathbf{Z}}_{\mathbf{eff}}\mathbf{=}\mathbf{1}$. Because n is so large, though, the excited electron is quite far from the atomic nucleus, with an orbital radius of approximately 1 mm, and is weakly bound. Using these so-called Rydberg atoms,the researchers were able to study the way energy is transported from one atom to the next. This process may be a model for understanding energy transport in photosynthesis. (Source: 鈥淥bserving the

Dynamics of Dipole-Mediated Energy Transport by Interaction Enhanced Imaging,鈥� by G. G眉nter et al., Science342(6161): 954鈥�956, Nov. 2013.)

Assume that the researchers place an atom in a state with,. What is the magnitude of the orbital angular momentum LS associated with this state? (a)$\sqrt{\mathbf{2}}\mathit{魔}$; (b)$\sqrt{\mathbf{6}}\mathit{魔}$; (c)$\sqrt{\mathbf{200}}\mathit{魔}$; (d)$\sqrt{\mathbf{10100}}\mathit{魔}$.

Step by step solution

01

(a) Identification of the concept

The Bohr model of Hydrogen atom has three postulates it is based on:

(i) There are circular orbits around the nucleus where the electron revolves

(ii) The orbiting electron鈥檚 angular momentum is quantized and discrete depending on the orbit and associated principal quantum number.

(iii) There is a release of photon when there is a transition of the electron from one orbit to another.

02

(b) Determination of the orbital angular momentum. 

The total orbital angular momentum is,

$L=\sqrt{l\left(l+1\right)}魔$

For, n=100 and l=2,

$$\begin{gathered} L=\sqrt{2\left(2+1\right)}魔 \\ =\sqrt{6}魔 \end{gathered}$$

Hence option (b) is correct.

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