
The wavelength of radiation emitted is ${\lambda _0}$ when an electron jumps from the third to second orbit of a hydrogen atom. For the electron jumping from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be:
A. $\dfrac{{16}}{{25}}{\lambda _0} \\ $
B. $\dfrac{{20}}{{27}}{\lambda _0} \\ $
C. $\dfrac{{27}}{{20}}{\lambda _0} \\ $
D. $\dfrac{{25}}{{16}}{\lambda _0}$
Answer
161.1k+ views
Hint:As the transition is from ${n_2} = 3$ to ${n_1} = 2$ and from ${n_2} = 4$ to ${n_1} = 2$. So in both the cases Balmer series is considered as the final state corresponds to n=2. The energy of the radiation produced increases with the difference between the degrees of transition, resulting in a decrease in wavelength.
Formula used:
Rydberg formula is given by:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Here, ${R_H}$ is the Rydberg constant, ${n_1}$ is the final orbit, ${n_2}$ is the beginning orbit, and $\lambda $ is the wavelength of the radiation
Complete step by step solution:
In order to know that when an electron transitions from a high to a low energy level in hydrogen gas, the emission spectrum manifests as discrete lines of various wavelengths. These lines, which include some of the Balmer series and are seen in the visible spectrum. The Rydberg formula can be used to determine the wavelength of light that is emitted whenever an electron moves from a higher orbit to a lower orbit:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
The Balmer series refers to a group of visible lines in the spectrum of hydrogen atoms. In both cases, when an electron transitions from a high energy level $(n = 3,4,5....)$ to a lower energy level ${n_1} = 2$, it enters the second orbit, so consider $\lambda = {\lambda _0}$ for the first case. Substitute ${n_1} = 2$and ${n_2} = 3$ in the above formula, then we have:
$\dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(3)}^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{5}{{36}}} \right)\,\,\,\,...(i) \\$
Similarly, consider $\lambda = {\lambda _1}$for the second case,
Substitute ${n_1} = 2$and ${n_2} = 4$ in the above formula, then we have:
$\dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(4)}^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{3}{{16}}} \right)\,\,\,\,...(ii) \\$
Now, from equation $(i)$ and $(ii)$, we obtain:
$\dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{{R_H}\left( {\dfrac{5}{{36}}} \right)\,}}{{{R_H}\left( {\dfrac{3}{{16}}} \right)\,}} \\$
$\Rightarrow \dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{5{R_H}}}{{36}} \times \dfrac{{16}}{{3{R_H}}} \\$
$\Rightarrow \dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{20}}{{27}} \\$
$\therefore {\lambda _1} = \dfrac{{20}}{{27}}{\lambda _0} \\$
Thus, the correct option is B.
Note: It should be noted that the wavelength of the radiation released as an electron in a hydrogen atom moves from the stationary state to infinity. Also Rydberg acquired the improvised technique for determining the wavelength in the hydrogen emission spectrum after the Balmer series. The formulae used to determine the wavelength of the hydrogen spectral lines are directly generalised by the formula provided by Rydberg, according to the analysis.
Formula used:
Rydberg formula is given by:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Here, ${R_H}$ is the Rydberg constant, ${n_1}$ is the final orbit, ${n_2}$ is the beginning orbit, and $\lambda $ is the wavelength of the radiation
Complete step by step solution:
In order to know that when an electron transitions from a high to a low energy level in hydrogen gas, the emission spectrum manifests as discrete lines of various wavelengths. These lines, which include some of the Balmer series and are seen in the visible spectrum. The Rydberg formula can be used to determine the wavelength of light that is emitted whenever an electron moves from a higher orbit to a lower orbit:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
The Balmer series refers to a group of visible lines in the spectrum of hydrogen atoms. In both cases, when an electron transitions from a high energy level $(n = 3,4,5....)$ to a lower energy level ${n_1} = 2$, it enters the second orbit, so consider $\lambda = {\lambda _0}$ for the first case. Substitute ${n_1} = 2$and ${n_2} = 3$ in the above formula, then we have:
$\dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(3)}^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{5}{{36}}} \right)\,\,\,\,...(i) \\$
Similarly, consider $\lambda = {\lambda _1}$for the second case,
Substitute ${n_1} = 2$and ${n_2} = 4$ in the above formula, then we have:
$\dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(4)}^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{3}{{16}}} \right)\,\,\,\,...(ii) \\$
Now, from equation $(i)$ and $(ii)$, we obtain:
$\dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{{R_H}\left( {\dfrac{5}{{36}}} \right)\,}}{{{R_H}\left( {\dfrac{3}{{16}}} \right)\,}} \\$
$\Rightarrow \dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{5{R_H}}}{{36}} \times \dfrac{{16}}{{3{R_H}}} \\$
$\Rightarrow \dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{20}}{{27}} \\$
$\therefore {\lambda _1} = \dfrac{{20}}{{27}}{\lambda _0} \\$
Thus, the correct option is B.
Note: It should be noted that the wavelength of the radiation released as an electron in a hydrogen atom moves from the stationary state to infinity. Also Rydberg acquired the improvised technique for determining the wavelength in the hydrogen emission spectrum after the Balmer series. The formulae used to determine the wavelength of the hydrogen spectral lines are directly generalised by the formula provided by Rydberg, according to the analysis.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

If a wire of resistance R is stretched to double of class 12 physics JEE_Main
