
The ratio of minimum to maximum wavelength of radiation emitted by transition of an electron to ground state of Bohr's hydrogen atom is:
A. $\dfrac{3}{4} \\ $
B. $\dfrac{1}{4} \\ $
C. $\dfrac{1}{8} \\ $
D. $\dfrac{3}{8}$
Answer
162.9k+ views
Hint:As the transition is from $n = \infty $ to $n = 1$. So it is considered in the Lyman series as the final state is $n=1$. The energy of the radiation produced increases with the difference between the degrees of transition, resulting in a decrease in wavelength. To solve the question, we need to use the energy formula, which is expressed in terms of the wavelength, Planck's constant, and the speed of light in a vacuum.
Formula used: The expression of the energy of the radiation corresponds to the transition between two energy levels is given by:
$E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV$
And, the energy which is necessary for the transition of an electron is given by:
$E = \dfrac{{hc}}{\lambda }$
Here, $h$ is the Planck constant and $c$ is the speed of light.
Complete step by step solution:
As we know that the Energy of the radiation corresponds to the transition between two energy levels ${n_1}$ and ${n_2}$ is given by:
$E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV \\ $
Now, the transition of the electron must occur in a way that releases the least amount of energy in order for the maximal wavelength to be emitted. The ground state of Bohr's hydrogen atom is ${n_1} = 1$, the just-next-energy state, which is ${n_2} = 2$, would be the nearest energy state for the minimum energy, so we can substitute the values in the above formula, then we have:
${E_{\min }} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)eV \\$
$\Rightarrow {E_{\min }} = 13.6\left( {1 - \dfrac{1}{4}} \right)eV \\$
$\Rightarrow {E_{\min }} = 13.6 \times \dfrac{3}{4}eV \\$
Similarly, when the atom is ionised then the ground state of Bohr's hydrogen atom is ${n_1} = 1$, the just-next-energy state, which is ${n_2} = \infty $, would be the nearest energy state for the maximum energy, so we can substitute the values in the above formula, then we have:
${E_{\max }} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)eV \\$
$\Rightarrow {E_{\max }} = 13.6\left( {1 - \dfrac{1}{\infty }} \right)eV \\$
$\Rightarrow {E_{\max }} = 13.6eV \\$
As per the question, the proportion of an electron's transition to the ground state of the hydrogen atom proposed by Bohr's by its minimum to maximum wavelength of radiation. The energy released on each of the transitions mentioned in the question must now be compared to this value, so we obtain:
$\dfrac{{{E_{\min }}}}{{{E_{\max }}}} = \dfrac{{13.6 \times \dfrac{3}{4}eV}}{{13.6eV}}\,\,\,\,\,...(i)$
To identify the wavelength, we can disregard the negative sign, which suggests that energy was released during the transition. Then, we can keep in mind that the energy and wavelength are related by:
$E = \dfrac{{hc}}{\lambda }\,\,\,\,\,....(ii)$
Now, substitute the equation $(ii)$ in the equation $(i)$, we obtain:
$\dfrac{{{{\left( {\dfrac{{hc}}{\lambda }} \right)}_{\min }}}}{{{{\left( {\dfrac{{hc}}{\lambda }} \right)}_{\max }}}} = \dfrac{{13.6 \times \dfrac{3}{4}eV}}{{13.6eV}} \\$
$\therefore \dfrac{{{\lambda _{\min }}}}{{{\lambda _{\max }}}} = \dfrac{3}{4} $
Therefore, the radiation's emission's wavelength ratio is $\dfrac{3}{4}$.
Thus, the correct option is A.
Note: It should be noted that an EMR transition's energy is inversely proportional to wavelength and directly proportional to frequency. Therefore, we may conclude that the energy of the EMR will increase with greater frequency and decrease with a higher wavelength of the EMR.
Formula used: The expression of the energy of the radiation corresponds to the transition between two energy levels is given by:
$E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV$
And, the energy which is necessary for the transition of an electron is given by:
$E = \dfrac{{hc}}{\lambda }$
Here, $h$ is the Planck constant and $c$ is the speed of light.
Complete step by step solution:
As we know that the Energy of the radiation corresponds to the transition between two energy levels ${n_1}$ and ${n_2}$ is given by:
$E = 13.6\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)eV \\ $
Now, the transition of the electron must occur in a way that releases the least amount of energy in order for the maximal wavelength to be emitted. The ground state of Bohr's hydrogen atom is ${n_1} = 1$, the just-next-energy state, which is ${n_2} = 2$, would be the nearest energy state for the minimum energy, so we can substitute the values in the above formula, then we have:
${E_{\min }} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)eV \\$
$\Rightarrow {E_{\min }} = 13.6\left( {1 - \dfrac{1}{4}} \right)eV \\$
$\Rightarrow {E_{\min }} = 13.6 \times \dfrac{3}{4}eV \\$
Similarly, when the atom is ionised then the ground state of Bohr's hydrogen atom is ${n_1} = 1$, the just-next-energy state, which is ${n_2} = \infty $, would be the nearest energy state for the maximum energy, so we can substitute the values in the above formula, then we have:
${E_{\max }} = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)eV \\$
$\Rightarrow {E_{\max }} = 13.6\left( {1 - \dfrac{1}{\infty }} \right)eV \\$
$\Rightarrow {E_{\max }} = 13.6eV \\$
As per the question, the proportion of an electron's transition to the ground state of the hydrogen atom proposed by Bohr's by its minimum to maximum wavelength of radiation. The energy released on each of the transitions mentioned in the question must now be compared to this value, so we obtain:
$\dfrac{{{E_{\min }}}}{{{E_{\max }}}} = \dfrac{{13.6 \times \dfrac{3}{4}eV}}{{13.6eV}}\,\,\,\,\,...(i)$
To identify the wavelength, we can disregard the negative sign, which suggests that energy was released during the transition. Then, we can keep in mind that the energy and wavelength are related by:
$E = \dfrac{{hc}}{\lambda }\,\,\,\,\,....(ii)$
Now, substitute the equation $(ii)$ in the equation $(i)$, we obtain:
$\dfrac{{{{\left( {\dfrac{{hc}}{\lambda }} \right)}_{\min }}}}{{{{\left( {\dfrac{{hc}}{\lambda }} \right)}_{\max }}}} = \dfrac{{13.6 \times \dfrac{3}{4}eV}}{{13.6eV}} \\$
$\therefore \dfrac{{{\lambda _{\min }}}}{{{\lambda _{\max }}}} = \dfrac{3}{4} $
Therefore, the radiation's emission's wavelength ratio is $\dfrac{3}{4}$.
Thus, the correct option is A.
Note: It should be noted that an EMR transition's energy is inversely proportional to wavelength and directly proportional to frequency. Therefore, we may conclude that the energy of the EMR will increase with greater frequency and decrease with a higher wavelength of the EMR.
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
