
A gas of hydrogen-like ions is prepared in a particular excited state $A$. It emits photons having a wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state$A$.
Answer
163.2k+ views
Hint: The ultraviolet spectrum contains the Lyman series. The principles of the hydrogen spectrum and the number of spectral lines in a given series represent all potential transitions from an energy level that must be applied in order to approach this problem. Atomic hydrogen frequently exhibits an emission spectrum. This spectrum has a propensity to encompass multiple spectral series.
Formula used:
Number of wavelengths emitted in ${n^{th}}$ excited state of an atom can be given as:
$\text{Number of photons emitted}= \dfrac{{n(n - 1)}}{2}$.
Here, $n$ as the principal quantum number.
Complete step by step solution:
In order to know that the quantum of electromagnetic radiation is called a photon. The lowest fundamental unit of a number or the smallest distinct sum of anything is referred to as a quantum. As a result, a photon is one quantum of electromagnetic radiation. Quanta is the plural form of quantum.
The concepts of photons and quanta are derived from quantum theory and quantum mechanics. It suggests that matter and energy are quantized, or that they exist in thin, distinct packets of the smallest possible size. A photon spreads at the speed of light, which means, the total $6$ photons are emitted.
Use the formula of the number of wavelengths that are emitted in ${n^{th}}$ excited state of an atom,
$\text{Number of photons emitted} = \dfrac{{n(n - 1)}}{2}$
Now, substitute the number of photons emitted in the above formula, then we have:
$6 = \dfrac{{n(n - 1)}}{2} \\$
$\Rightarrow 12 = {n^2} - n \\$
$\Rightarrow {n^2} - 4n + 3n - 12 = 0 \\$
$\therefore n = 4,\,n = - 3 $
The value of $n$ is positive instead of negative, so we obtain the number of wavelengths:
$n = 4$.
Therefore, the principal quantum number is $4$and the gas in the ${4^{th}}$excited state.
Note: It should be noted that the wavelength and photons are identical since the energy is directly proportional to the photon's electric frequency. As the wavelength increases, the photon's energy rises as well. With an increase in photon wavelength, the photon's energy diminishes.
Formula used:
Number of wavelengths emitted in ${n^{th}}$ excited state of an atom can be given as:
$\text{Number of photons emitted}= \dfrac{{n(n - 1)}}{2}$.
Here, $n$ as the principal quantum number.
Complete step by step solution:
In order to know that the quantum of electromagnetic radiation is called a photon. The lowest fundamental unit of a number or the smallest distinct sum of anything is referred to as a quantum. As a result, a photon is one quantum of electromagnetic radiation. Quanta is the plural form of quantum.
The concepts of photons and quanta are derived from quantum theory and quantum mechanics. It suggests that matter and energy are quantized, or that they exist in thin, distinct packets of the smallest possible size. A photon spreads at the speed of light, which means, the total $6$ photons are emitted.
Use the formula of the number of wavelengths that are emitted in ${n^{th}}$ excited state of an atom,
$\text{Number of photons emitted} = \dfrac{{n(n - 1)}}{2}$
Now, substitute the number of photons emitted in the above formula, then we have:
$6 = \dfrac{{n(n - 1)}}{2} \\$
$\Rightarrow 12 = {n^2} - n \\$
$\Rightarrow {n^2} - 4n + 3n - 12 = 0 \\$
$\therefore n = 4,\,n = - 3 $
The value of $n$ is positive instead of negative, so we obtain the number of wavelengths:
$n = 4$.
Therefore, the principal quantum number is $4$and the gas in the ${4^{th}}$excited state.
Note: It should be noted that the wavelength and photons are identical since the energy is directly proportional to the photon's electric frequency. As the wavelength increases, the photon's energy rises as well. With an increase in photon wavelength, the photon's energy diminishes.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

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

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

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

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main
