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The ratio of the energies of the hydrogen atom in its first to second excited states is (a) \(9 / 4\) (b) \(4 / 1\) (c) \(8 / 1\) (d) \(1 / 8\)

Short Answer

Expert verified
The ratio is (a) 9/4.

Step by step solution

01

Understanding Energy Levels in Hydrogen Atom

The energy of an electron in a hydrogen atom at a particular energy level (n) is given by the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \]The value 13.6 eV is the ionization energy of hydrogen. Each energy level corresponds to a different value of \( n \). The first excited state is \( n = 2 \), and the second excited state is \( n = 3 \).
02

Calculate Energy of the First Excited State (n = 2)

Substitute \( n = 2 \) into the energy formula:\[ E_2 = - \frac{13.6}{2^2} = - \frac{13.6}{4} = - 3.4 \, \text{eV} \]
03

Calculate Energy of the Second Excited State (n = 3)

Substitute \( n = 3 \) into the energy formula:\[ E_3 = - \frac{13.6}{3^2} = - \frac{13.6}{9} \approx - 1.51 \, \text{eV} \]
04

Find the Ratio of Energies

We now find the ratio of the energies from the first excited state to the second excited state:\[ \frac{E_2}{E_3} = \frac{-3.4}{-1.51} \approx 2.25 \]
05

Simplify the Ratio to Match Answer Choices

Convert \( 2.25 \) to a fraction:\[ 2.25 = \frac{9}{4} \]This matches the option (a) \( \frac{9}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Energy
Ionization energy is a fundamental concept in understanding atoms, particularly hydrogen. It represents the amount of energy required to remove an electron from its ground state, or the lowest energy level, in an atom. For hydrogen, this energy is specifically measured as 13.6 electron volts (eV). This is because the hydrogen atom's single electron is bound to the nucleus with this energy at the ground state.
This 13.6 eV is the maximum energy needed to ionize a hydrogen atom. It explores the transition of an electron from the ground state to being completely free of the atom. This concept is crucial for calculating energies at different energy levels, as it provides a comparison for excited and ionization states. Understanding ionization energy helps unravel the complexities of electron transitions and how electrons can be excited from lower to higher energy states. This forms the basis of atomic energy level calculations.
  • Ionization energy is key for understanding electron transitions.
  • Hydrogen's ionization energy is 13.6 eV.
  • Knowing this helps explain how electrons reach higher energy states (excited states).
Excited States
Excited states occur when an electron in an atom gains energy and moves to a higher energy level. In a hydrogen atom, the electron can transition from the ground state (n=1) to higher states, such as n=2 for the first excited state and n=3 for the second excited state. Each level change corresponds to a specific energy that the electron possesses.
Understanding the excited states is crucial for explaining phenomena like emission and absorption spectra. When an electron absorbs energy, it moves to a higher excited state, and when it releases energy, it falls back to a lower energy state, emitting light energy.
In our exercise, the first excited state is determined by n=2 and the second by n=3. As electrons move to these levels, they hold specific energies based on the energy level formula. It becomes vital to comprehend these states accurately for calculations related to energy changes and the behavior of atoms in different energy conditions.
Energy Level Formula
The energy level formula is essential for calculating the energy of electrons at various levels in the hydrogen atom. It is expressed as \( E_n = -\frac{13.6 \, \text{eV}}{n^2} \), where \( n \) is the principal quantum number. This formula indicates that the energy of an electron at a particular energy level is proportional to the square of the quantum number \( n \), but inversely proportional to it.
Using this formula, we determine the negative energy value of an atom's electron, signifying its bound state to the nucleus. The higher the value of \( n \), the closer the energy value is to zero, indicating less energy is required to remove the electron completely from the atom. This is why energy increases significantly as \( n \) increases, as seen in excited states.
  • The formula helps calculate specific energies for different states.
  • Lower \( n \) values mean more bound energy, while higher \( n \) values bring the energy closer to zero.
  • This calculation is vital for understanding ionization and excited states of hydrogen.

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Most popular questions from this chapter

The half-life of radioactive Radon is \(3.8\) days. The time at the end of which ( \(1 / 20\) )th of the Radon sample will remain undecayed is (given \(\log _{10} e=0.4343\) ) (a) \(13.8\) days (b) \(16.5\) days (c) 33 days (d) 76 days

The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is \(B=\frac{m e^{4}}{8 n^{2}{ }_{0}^{2} h^{2}} \cdot(m=\) proton mass \()\) If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be \(B=-\frac{M e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}(M=\) proton mass \() .\) This last expression is not correct because (a) \(n\) would not be integral (b) Bohr-quantisation applies only to electron (c) the frame in which the electron is at rest is not inertial. (d) the motion of the proton would not be in circular orbits, even approximately.

1 Fusion process, line combining two deutrons to form a He nucleus are impossible at ordinary temperatures and pressure. This reasons for this can be traced to the fact [NCERT Exemplar] (a) nuclear forces have short range (b) nuclei are positively charge (c) the original nuclei must be completely ionized before fussion can take place (d) the original nuclei must first break up before combining with each other.

Doubly ionised helium atom and hydrogen ions are accelerated, from rest, through the same potential difference. The ratio of final velocities of helium and hydrogen is (a) \(1: \sqrt{2}\) (b) \(\sqrt{2}: 1\) (c) \(1: 2\) (d) \(2: 1\)

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