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The ionization energy of hydrogen atom is \(13.6 \mathrm{eV}\). Following Bohr's theory, the energy corresponding to a transition between 3 rd and 4 th orbit is (a) \(3.40 \mathrm{eV}\) (b) \(1.51 \mathrm{eV}\) [c) \(0.85 \mathrm{eV}\) (d) \(0.66 \mathrm{eV}\)

Short Answer

Expert verified
The energy corresponding to the transition is \(0.66 \mathrm{eV}\) (Option d).

Step by step solution

01

Understand Energy Levels in a Hydrogen Atom

According to Bohr's theory, the energy of an electron in a hydrogen atom at a particular orbit (n) is given by the formula \(E_n = - \frac{13.6 \, \mathrm{eV}}{n^2}\), where \(13.6 \, \mathrm{eV}\) is the ionization energy of the hydrogen atom.
02

Calculate Energy for n=3

For the orbit with \( n = 3 \), the energy is \( E_3 = - \frac{13.6 \, \mathrm{eV}}{3^2} = - \frac{13.6 \, \mathrm{eV}}{9} = -1.51 \, \mathrm{eV} \).
03

Calculate Energy for n=4

For the orbit with \( n = 4 \), the energy is \( E_4 = - \frac{13.6 \, \mathrm{eV}}{4^2} = - \frac{13.6 \, \mathrm{eV}}{16} = -0.85 \, \mathrm{eV} \).
04

Determine the Transition Energy

The energy corresponding to a transition from the 4th orbit to the 3rd orbit is \( \Delta E = E_3 - E_4 = (-1.51 \, \mathrm{eV}) - (-0.85 \, \mathrm{eV}) \).
05

Calculate the Energy Difference

Calculate the difference \( \Delta E = -1.51 \, \mathrm{eV} + 0.85 \, \mathrm{eV} = -0.66 \, \mathrm{eV} \). Thus, the energy released during the transition is \( 0.66 \, \mathrm{eV} \) because it is a transition to a lower energy state (emission).

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Key Concepts

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

Hydrogen Atom Energy Levels
In Bohr's model of the hydrogen atom, the energy level of an electron is quantized, meaning it can only occupy certain defined "rungs" on an energy ladder. Each rung corresponds to a specific orbit labeled by the quantum number \( n \). The energy levels are negative because the electron is bound to the nucleus. The closer an electron is to the nucleus (lower \( n \)), the lower its energy. The energy of an electron in the \( n^{th} \) orbit is described by the formula:
  • \( E_n = -\frac{13.6 \, \mathrm{eV}}{n^2} \)
This formula tells us that as \( n \) increases, the absolute value of this negative energy increases, showing that the electron is less tightly bound. In other words, higher energy levels are less stable and the electron requires less energy to transition to a higher state or to escape entirely from the atom.
Electron Transition
An electron transition occurs when an electron moves from one energy level to another, either absorbing or releasing energy in the process. This is crucial in explaining atomic spectra, where such transitions emit or absorb specific wavelengths of light. When an electron drops to a lower energy level, it releases energy, often in the form of a photon.For transitions in the hydrogen atom:
  • The energy change \( \Delta E \) associated with a transition between two levels \( n_i \) and \( n_f \) is given by \( \Delta E = E_i -E_f \).
  • Here, \( E_i \) and \( E_f \) are the initial and final energy levels computed using the formula \( E_n = -\frac{13.6 \, \mathrm{eV}}{n^2} \).
For example, the transition from the 4th to the 3rd orbit in hydrogen results in energy release of \( 0.66 \, \mathrm{eV} \). This energy difference corresponds to the emission of light with a specific frequency.
Ionization Energy
Ionization energy is an important concept when discussing the binding of electrons in an atom. In the case of hydrogen, the ionization energy is the energy required to remove its single electron from the ground state to infinity, effectively freeing it from the nucleus. For hydrogen, this energy is precisely \( 13.6 \, \mathrm{eV} \), which reflects the energy of the lowest orbit (\( n = 1 \)).Here’s what happens during ionization:
  • The electron absorbs exactly \( 13.6 \, \mathrm{eV} \) of energy, increasing its energy to zero as it moves from being bound to free.
  • This signifies that the electron has overcome the attractive force of the proton, illustrating the concept that zero energy marks the threshold of being unbound.
Understanding ionization energy helps in comprehending how atoms absorb energy from external sources to trigger electron liberation or excite them to higher energy levels. In any heavier atom, this energy quantification also aids in explaining why multi-electron atoms require additional energy compared to a hydrogen atom for ionization.

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Most popular questions from this chapter

\(M_{x}\) and \(M_{y}\) denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The \(Q\)-value of a \(\beta^{-}\)decay is \(Q_{1}\) and that for a \(\beta^{+}\) decay is \(Q_{2}\). If \(m_{e}\) denotes the mass of an electron, then which of the following statements is correct? [NCERT Exemplar] (a) \(Q_{1}=\left(M_{x}-M_{y}\right) c^{2}\) and \(Q_{2}=\left(M_{x}-M_{y}-2 m_{e}\right) c^{2}\) (b) \(Q_{1}=\left(M_{x}-M_{y}\right) c^{2}\) and \(Q_{2}=\left(M_{x}-M_{y}\right) c^{2}\) (c) \(Q_{1}=\left(M_{x}-M_{y}-2 m_{e}\right) c^{2}\) and \(Q_{2}=\left(M_{x}-M_{y}+2 m_{e}\right) c^{2}\) (d) \(Q_{1}=\left(M_{x}-M_{y}+2 m_{e}\right) c^{2}\) and \(Q_{2}=\left(M_{x}-M_{y}+2 m_{e}\right) c^{2}\)

Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 year. After 1 year [NCERT Exemplar] (a) all the containers will have 5000 atoms of the materials. (b) all the containers will contain the same number of atoms of the material but that number will only be approximately \(5000 .\) (c) the containers will in general have different numbers of the the atoms of the material but their average will be close to 5000 . (d) None of the containers can have more than 5000 atoms.

Hydrogen atom excites energy level from fundamental state to \(n=3 .\) Number of spectrum lines, according to Bohr, is (a) 4 (b) 3 (c) 1 (d) 2

The energy equivalent of one atomic mass unit is (a) \(1.6 \times 10^{-19} \mathrm{~J}\) (b) \(6.02 \times 10^{-23} \mathrm{~J}\) (c) \(931 \mathrm{~J}\) (d) \(931 \mathrm{MeV}\)

For light of wavelength \(5000 \AA\), photon energy is nearly \(2.5 \mathrm{eV}\). For \(X\)-rays of wavelength \(1 \AA\), the photon energy will be close to [a) \([2.5 \div 5000] \mathrm{eV}\) (b) \(\left[2.5 \div(5000)^{2}\right] \mathrm{eV}\) (c) \([2.5 \times 5000] \mathrm{eV}\) (d) \(\left[2.5 \times(5000)^{2}\right] \mathrm{eV}\)

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