Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

What is the frequency of the emitted radiation in Rydberg’s unit when electrons in a hydrogen atom jumps from third to the first orbit?

seo-qna
Last updated date: 14th Jul 2024
Total views: 347.1k
Views today: 7.47k
Answer
VerifiedVerified
347.1k+ views
Hint:To solve this problem, we must first understand the Rydberg formula, its application, and the interpretation of each term specified in the formula. We'll also use Rydberg's equation to solve the problem.

Complete step by step answer:
The Rydberg formula is a mathematical formula that predicts the wavelength of light generated by an electron moving between atomic energy levels. The charge of an electron varies as it moves from one atomic orbital to another. A photon of light is produced when an electron moves from a high-energy orbital to a lower-energy orbital. When an electron passes from a low-energy to a higher-energy state, the atom absorbs a photon of light.

This formula was created by combining his findings with Bohr's atomic model:
\[1/\lambda {\text{ }} = {\text{ }}RZ{\;^2}\left( {1/n{\;_1}{\;^2}\; - {\text{ }}1/n{\;_2}{\;^2}} \right)\]
Where, \[\lambda \]= is the wavelength of the photon (wavenumber = 1/wavelength), \[R\] = Rydberg's constant, \[\;Z\] = atomic number of the atom and \[n{\;_1}\;\] and are integers where \[n{\;_2}\; > {\text{ }}n{\;_1}\].

So, for the given question,wavelength is represented by $\lambda $ and is given by following expression
$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right] \\
\Rightarrow \dfrac{1}{\lambda } = \left( {1.097 \times {{10}^7}{m^{ - 1}}} \right)\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{3^2}}}} \right] \\
\Rightarrow \lambda = \,\dfrac{9}{{8 \times 1.097 \times {{10}^7}}} \\
\Rightarrow \lambda= \dfrac{9}{{8.776 \times {{10}^7}}} = 1.025 \times {10^{ - 7}} \\
\Rightarrow \text{frequency is given by (v)} = \dfrac{c}{\lambda } \\
\Rightarrow \text{frequency is given by (v)}= \dfrac{{3 \times {{10}^8}}}{{1.025 \times {{10}^{ - 7}}}} \\
\therefore \text{frequency is given by (v)}= 2.926 \times {10^{15}}\,Hz$

Hence,the frequency of the emitted radiation is $2.926 \times {10^{15}}\,Hz$.

Note:The question now is, why can't the Rydberg equation be applied to all atoms?Since the Bohr model of the atom breaks down when there are more than two electrons, the Rydberg equation only works for Hydrogen and Hydrogen-like (species with only one electron). As a result, the Schrodinger equation yields the following mathematical model of the atom: $\,H{\text{ }} = {\text{ }}E$.