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91Ó°ÊÓ

Ionisation energy of an electron present in the second Bohr's orbit of hydrogen is (a) \(54.4 \mathrm{eV}\) (b) \(13.6 \mathrm{eV}\) (c) \(1.5 \mathrm{eV}\) (d) \(3.4 \mathrm{eV}\)

Short Answer

Expert verified
The ionization energy is \(3.4 \, \text{eV}\), so the correct answer is (d).

Step by step solution

01

Understanding the Problem

The problem requires us to find the ionization energy of an electron in the second Bohr's orbit of hydrogen. Ionization energy is the energy needed to completely remove an electron from an atom.
02

Bohr's Ionization Energy Formula

Bohr's model gives the formula for ionization energy as \( E_n = - \frac{13.6 \, \text{eV}}{n^2} \), where \( n \) is the principal quantum number corresponding to the orbit.
03

Determining the Principal Quantum Number

For the second Bohr's orbit, \( n = 2 \). Substitute \( n = 2 \) into the ionization energy formula.
04

Calculate Ionization Energy

Substitute \( n = 2 \) into the formula: \[ E_2 = - \frac{13.6 \, \text{eV}}{2^2} = - \frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV}. \] Since ionization energy is a magnitude, we take the positive value as \( 3.4 \, \text{eV} \).
05

Conclusion

The ionization energy of an electron in the second Bohr's orbit of hydrogen is positive \( 3.4 \, \text{eV} \), which matches option (d).

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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 atomic physics. It is the energy required to completely remove an electron from an atom or ion in its gaseous state.
This energy is a measure of how strongly an atom's nucleus holds onto its electrons. When we discuss ionization energy within the context of Bohr's model, we specifically look at hydrogen-like atoms where only one electron orbits the nucleus. In Bohr's model, the ionization energy can be calculated using the formula \[ E_n = - \frac{13.6 \text{ eV}}{n^2} \]where - \( E_n \) represents the ionization energy at principal quantum number \( n \). - \( 13.6 \text{ eV} \) is the ionization energy of a hydrogen atom at the ground state.The negative sign indicates the energy is being released when the electron is added to the atom. However, in our calculations, we are often concerned with the magnitude, which is why we may take a positive value at the end. In exercises similar to this, correctly identifying the orbit or shell of the electron and using the appropriate \( n \) will directly lead to finding the correct ionization energy.
Principal Quantum Number
The principal quantum number, denoted by \( n \), plays a critical role in both the Bohr model and quantum mechanics.
This number essentially determines the size of the electron's orbit and its energy level. Bohr's model maps the electron in discrete orbits around the nucleus, each corresponding to a unique principal quantum number. - As \( n \) increases, the electron is at a higher energy level, farther from the nucleus. - The principal quantum number is also an integer, starting from 1. For example, \( n = 1 \) is the first energy level (closest to the nucleus), \( n = 2 \) is the second, and so on.In the solution to the problem, \( n = 2 \) corresponds to the second orbit of hydrogen. By substituting \( n = 2 \) into the ionization energy formula, we are able to determine the specific energy required to ionize an electron at this energy level. Understanding the principal quantum number is crucial for conceptualizing electron transitions between orbits and calculating ionization energies, as it directly relates to the electron's energy within an atom.
Electron Orbits
In the Bohr model, electrons revolve around the atomic nucleus in defined orbits or shells, reminiscent of how planets orbit the Sun.
These orbits are quantized, suggesting that only certain permissible orbits are allowed where electrons can stably exist without radiating energy. Bohr introduced the idea of quantized orbits to explain the hydrogen atom's spectral lines. - Each orbit corresponds to a specific energy level, governed by the principal quantum number \( n \). - An electron in a lower energy orbit is binded more strongly to the nucleus compared to one in a higher orbit. - Transition of electrons between these orbits involves absorption or emission of energy in the form of photons, a fundamental concept for understanding atomic spectra.In exercises analyzing ionization energies, recognizing which orbit an electron occupies is essential. The particular energy level determines the energy necessary to either move the electron between orbits or remove it entirely from the atom. This quantized view of electron orbits elegantly explains the discrete energy levels observed in atomic spectra and lays the foundation for modern quantum mechanics.

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Most popular questions from this chapter

The half-life of \({ }^{215}\) At is \(100 \mu \mathrm{s}\). The time taken for the cadioactivity of a sample of \({ }^{215}\) At to decay to \(\frac{1}{16}\) th of its initial value is (a) \(400 \mu \mathrm{s}\) (b) \(6.3 \mu \mathrm{s}\) (c) \(40 \mu \mathrm{s}\) (d) \(300 \mu \mathrm{s}\)

Two identical photo cathodes receive light of frequencies \(f_{1}\) and \(f_{2} .\) If the velocities of the photo electrons (of mass \(m\) ) coming out are respectively, \(v_{1}\) and \(v_{2}\), then (a) \(v_{1}-v_{2}=\left[\frac{2 h}{m}\left(f_{1}-f_{2}\right)\right]^{1 / 2}\) (b) \(v_{1}^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right)\) (c) \(v_{1}+v_{2}=\left[\frac{2 h}{m}\left(f_{1}+f_{2}\right)\right]^{1 / 2}\) (d) \(v_{1}^{2}+v_{2}^{2}=\frac{2 h}{m}\left(f_{1}+f_{2}\right)\)

\(M_{x}\) 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 \(Q_{2}=\left(M_{x}-M_{y}+2 m_{e}\right) c^{2}\)

Consider the following statements concerning electrons I. Electrons are universal constituents of matter. II. J J Thomson received the very first Nobel prize in Physics for discovering the electron. III. The mass of the electron is about \(1 / 2000\) of a neutron. IV. According to Bohr the linear momentum of the electron is quantised in the hydrogen atom. Which of the above statements are not correct? (a) 1 (b) 1 (c) 111 (d) IV

The binding energy of two nuclei \(p^{n}\) and \(Q^{2 n}\) are \(x\) joule and \(y\) joule respectively. If \(2 x>y\), then the energy released in the reaction \(p^{n}+p^{n}=Q^{2 n}\) will be (a) \(2 x+y\) (b) \(2 x-y\) (c) \(x y\) (d) \(x+y\)

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