Question

Maximum frequency of emission obtained in the transition:

& \text{A}\text{. n = 2 to n = 1} \\\ & \text{B}\text{. n = 6 to n = 2} \\\ & \text{C}\text{. n = 1 to n = 2} \\\ & \text{D}\text{. n = 2 to n = 6} \\\ \end{aligned}$$

Answer 1

Hint: For the hydrogen spectrum series, use the Rydberg formula as the expression for the wavelength corresponding to the states of energy level given. From the relationship between the wavelength and the frequency of the emission, derive the formula for frequency and compare the frequency expected for transitions corresponding to each option.

Complete answer:
In 1914, Niels Bohr proposed a theory of hydrogen. The excitement in an atom (in this case Hydrogen, with one electron only) is determined by the position of its electron. The higher the number of orbitals the electron resides in; more is its energy. When the electron falls to an orbital lower than previously, it loses some amount of energy. This energy is emitted in the form of light with a certain wavelength. This is known as the Hydrogen spectrum.
The wavelengths of the hydrogen spectrum could be calculated by the following formula known as the Rydberg formula:
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right)$
Where, $\lambda$ is the wavelength of the light emitted, R is the Rydberg constant for hydrogen and ${{n}_{f}}$, ${{n}_{i}}$ are integers representing the final and initial number of orbital respectively.
We know the relationship between the wavelength $\lambda$and its frequency $\nu$ for an emission. It is given as:
$\nu =\dfrac{c}{\lambda }$, where c is the velocity of light, $c=3\times {{10}^{8}}m/s$.
Therefore, the relation of the frequency with ${{n}_{i}}$ and ${{n}_{f}}$ becomes:

& \Rightarrow c\times \dfrac{1}{\lambda }=c\times R\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right) \\\ & \Rightarrow \nu =Rc\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right) \\\ \end{aligned}$$ Where, R and c are the constants and the value of frequency depends on $${{n}_{i}}$$ and $${{n}_{f}}$$. For option A: $${{n}_{i}}=2$$and $${{n}_{f}}=1$$. Frequency becomes: $$\Rightarrow \nu =Rc\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)=\dfrac{3}{4}Rc$$ For option B: $${{n}_{i}}=6$$and $${{n}_{f}}=2$$. Frequency becomes: $$\Rightarrow \nu =Rc\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{6}^{2}}} \right)=\dfrac{2}{9}Rc$$ For option A: $${{n}_{i}}=1$$and $${{n}_{f}}=2$$. Frequency becomes: $$\Rightarrow \nu =Rc\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{1}^{2}}} \right)=-\dfrac{3}{4}Rc$$ For option A: $${{n}_{i}}=2$$and $${{n}_{f}}=6$$. Frequency becomes: $$\Rightarrow \nu =Rc\left( \dfrac{1}{{{6}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)=-\dfrac{2}{9}Rc$$ Clearly, the correct option is option A. That is the frequency is the highest for the values of $${{n}_{i}}=2$$and $${{n}_{f}}=1$$. Note: Firstly, students should understand what the negative sign for some of the transitions means. The electron is more stable when it lies in a lower energy level. So, when a transition from a higher state to lower state is made, the electron becomes more stable and loses its energy in the form of the emission. That is less the energy, more the electron is stable. Therefore, one should not confuse option A. with C. and option B. with D. Secondly, we have considered the case of the hydrogen spectrum but this concept can be extended to any atom.