CBSE-XI-Physics
43: Bohr's Model and Physics of the Atom
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- #Section : i
- Qstn #1How many wavelengths are emitted by atomic hydrogen in visible range (380 nm - 780 nm)? In the range 50 nm to 100 nm?Ans : Balmer series contains wavelengths ranging from 364 nm (for n2 = 3) to 655 nm (n2 = `` \infty ``).
So, the given range of wavelength (380-780 nm) lies in the Balmer series.
The wavelength in the Balmer series can be found by
`` \frac{1}{\lambda }=R\left(\frac{1}{{2}^{2}}-\frac{1}{{n}^{2}}\right)``
Here, R = Rydberg's constant = 1.097×107 m`` -``1
The wavelength for the transition from n = 3 to n = 2 is given by
`` \frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)``
`` {\lambda }_{1}=656.3\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 4 to n = 2 is given by
`` \frac{1}{{\lambda }_{2}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}}\right)``
`` {\lambda }_{2}=486.1\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 5 to n = 2 is given by
`` \frac{1}{{\lambda }_{3}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{5}^{2}}\right)``
`` {\lambda }_{3}=434.0\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 6 to n = 2 is given by
`` \frac{1}{{\lambda }_{4}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{6}^{2}}\right)``
`` {\lambda }_{4}=410.2\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 7 to n = 2 is given by
`` \frac{1}{{\lambda }_{5}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{7}^{2}}\right)``
`` {\lambda }_{5}=397.0\,\mathrm{\,nm\,}``
Thus, the wavelengths emitted by the atomic hydrogen in visible range (380-780 nm) are 5.
Lyman series contains wavelengths ranging from 91 nm (for n2 = 2) to 121 nm (n2 =`` \infty ``).
So, the wavelengths in the given range (50-100 nm) must lie in the Lyman series.
The wavelength in the Lyman series can be found by
`` \frac{1}{\lambda }=R\left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)``
The wavelength for the transition from n = 2 to n = 1 is given by
`` \frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)``
`` {\lambda }_{1}=122\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 3 to n = 1 is given by
`` \frac{1}{{\lambda }_{2}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)``
`` {\lambda }_{2}=103\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 4 to n = 1 is given by
`` \frac{1}{{\lambda }_{3}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{4}^{2}}\right)``
`` {\lambda }_{3}=97.3\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 5 to n = 1 is given by
`` \frac{1}{{\lambda }_{4}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{5}^{2}}\right)``
`` {\lambda }_{4}=95.0\,\mathrm{\,nm\,}``
The wavelength for the transition from n = 6 to n = 1 is given by
`` \frac{1}{{\lambda }_{5}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{6}^{2}}\right)``
`` {\lambda }_{5}=93.8\,\mathrm{\,nm\,}``
So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.
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- Qstn #2The first excited energy of a He+ ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?Ans : The energy of hydrogen ion is given by
`` {E}_{\,\mathrm{\,n\,}}=-\frac{(13.6\,\mathrm{\,eV\,}){Z}^{2}}{{n}^{2}}``
For the first excited state (n = 2), the energy of He+ ion (with Z = 2) will be `` -``13.6 eV. This is same as the ground state energy of a hydrogen atom.
Similarly, for all the hydrogen like ions, the energy of the (n `` -`` 1)th excited state will be same as the ground state energy of a hydrogen atom if Z = n.
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- Qstn #3Which wavelengths will be emitted by a sample of atomic hydrogen gas (in ground state) if electrons of energy 12.2 eV collide with the atoms of the gas?Ans : As the electron collides, it transfers all its energy to the hydrogen atom.
The excitation energy to raise the electron from the ground state to the nth state is given by
`` E=(13.6\,\mathrm{\,eV\,})\times \left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)``
Substituting n = 2, we get
E = 10.2 eV
Substituting n = 3, we get
E' = 12.08 eV
Thus, the atom will be raised to the second excited energy level.
So, when it comes to the ground state, there is transitions from n = 3 to n = 1.
Therefore, the wavelengths emitted will lie in the Lyman series (infrared region).
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- Qstn #4When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed in Lyman series only. Explain.Ans : White radiations are x-rays that have energy ranging between 5-10 eV.
When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed only in the Lyman series.
At room temperature, almost all the atoms are in ground state.
The minimum energy required for absorption is 10.2 eV (for a transition from n = 1 to n = 2).
The white radiation has photon radiations that have an energy of around 10.2 eV.
So, they are just sufficient to transmit an electron from n = 1 to n = 2 level.
Hence, the absorption lines are observed only in the Lyman series.
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- Qstn #5Balmer series was observed and analysed before the other series. Can you suggest a reason for such an order?Ans : The Balmer series lies in the visible range. Therefore, it was observed and analysed before the other series. The wavelength range of Balmer series is from 364 nm (for n2 =`` \infty ``) to 655 nm (for n2 =3).
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- Qstn #6What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write En = E1/n2, or rn = a0 n2?Ans : Energy of nth state of hydrogen is given by
`` {E}_{\,\mathrm{\,n\,}}=\frac{-13.6}{{n}^{2}}\,\mathrm{\,eV\,}``
`` ``
Energy of first excited state (n = 2) of hydrogen, E1 = `` \frac{-13.6}{4}\,\mathrm{\,eV\,}`` = -3.4 eV
This relation holds true when the refrence point energy is zero.Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton.In the given question, the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton so here our refrence point energy is 10 eV. Earlier The energy of first excited state was -3.4 eV when the refrence point had zero energy but now as the refrence point has shifted so The energy of the first excited state will also shift by the corresponding amount.Thus,
E1, = -3.4 eV-10 eV = -13.4 eV
We still write En = E1/n2, or rn = a0 n2 because these formulas are independent of the refrence point enegy.
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- Qstn #7The difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series. Explain.Ans : The 'series limit' refers to the 'shortest wavelength' (corresponding to the maximum photon energy).
The frequency of the radiation emitted for transition from n1 to n2 is given by
`` f=k\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
Here, k is a constant.
For the series limit of Lyman series,
n1 = 1
n2 = `` \infty ``
Frequency, `` {f}_{1}=k\left(\frac{1}{{1}^{2}}-\frac{1}{\infty }\right)=k``
For the first line of Lyman series,
n1 = 1
n2 = 2
Frequency, `` {f}_{2}=k\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)=\frac{3k}{4}``
For series limit of Balmer series,
n1 = 2
n2 = `` \infty ``
`` {f}_{1}=k\left(\frac{1}{{2}^{2}}-\frac{1}{\infty }\right)=\frac{k}{4}``
f1 `` -`` f3 = f2
Thus, the difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series.
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- Qstn #8The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?Ans : The electron volt is the amount of energy given to an electron in order to move it through the electric potential difference of one volt.
1 eV = 1.6 × 10-19 J
The numerical value of ionisation energy in eV is equal to the ionisation potential in volts. The equality does not hold if these quantities are measured in some other units.
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- Qstn #9We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?Ans : When a photon of energy (E2 - E1 = hÏ…) is incident on an atom in the ground state, the atom in the ground state E1 may absorb the photon and jump to a higher energy state (E2). This process is called stimulated absorption or induced absorption.
Spontaneous absorption is the process by which an atom in its ground state spontaneously jumps to a higher energy state, resulting in the absorption of a photon.
We do not have any process such as spontaneous absorption. This is because for absorption, we need to incident a photon of sufficient energy on the atom to stimulate the atom for absorption.
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- Qstn #10An atom is in its excited state. Does the probability of its coming to ground state depend on whether the radiation is already present or not? If yes, does it also depend on the wavelength of the radiation present?Ans : When an atom transits from an excited state to ground state in the presence of an external radiation then it is called as stimulated transition.
When an atom transits from an excited state to ground state on its own then it is called as spontaneous transition.
Ratio of the coefficient for stimulated transition to spontaneous transition is given by
`` R=\frac{A}{B\rho \left(\nu \right)}={e}^{\frac{h\nu }{kT}}-1``
`` ``
For microwave region
`` \nu ={10}^{10}\left(\,\mathrm{\,say\,}\right)``
`` \,\mathrm{\,R\,}={\,\mathrm{\,e\,}}^{\frac{6.6\times {10}^{-34}\times {10}^{10}}{1.38\times {10}^{-23}\times 300}}-1``
`` ={\,\mathrm{\,e\,}}^{0.0016}-1``
`` =0.0016``
This implies that stimulated transition dominate in this region.
For visible region
`` \nu ={10}^{15}``
`` \Rightarrow R={e}^{160}-1``
`` \Rightarrow R>>1``
So here spontaneous transition dominate.
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- #Section : ii
- Qstn #1The minimum orbital angular momentum of the electron in a hydrogen atom is
(a) h
(b) h/2
(c) h/2Ï€
(d) h/λdigAnsr: cAns : (c) h/2π
According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplt of h/2Ï€.
∴ `` {L}_{\,\mathrm{\,n\,}}=\frac{nh}{2\pi }``
Here,
n = Principal quantum number
The minimum value of n is 1.
Thus, the minimum value of the orbital angular momentum of the electron in a hydrogen atom is given by
`` L=\frac{h}{2\pi }``
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- Qstn #2Three photons coming from excited atomic-hydrogen sample are picked up. Their energies are 12.1 eV, 10.2 eV and 1.9 eV. These photons must come from
(a) a single atom
(b) two atoms
(c) three atoms
(d) either two atoms or three atomsdigAnsr: dAns : (d) either two atoms or three atoms
The energies of the photons emitted can be expressed as follows:
`` 13.6\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\,\mathrm{\,eV\,}=10.2\,\mathrm{\,eV\,}``
`` 13.6\left(\frac{1}{{1}^{2}}-\frac{1}{{3}^{2}}\right)\,\mathrm{\,eV\,}=12.1\,\mathrm{\,eV\,}``
`` 13.6\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)\,\mathrm{\,eV\,}=1.9\,\mathrm{\,eV\,}``
The following table gives the transition corresponding to the energy of the photon:
Energy of photon Transition 12.1 eV n = 3 to n = 1 10.2 eV n = 2 to n = 1 1.9 eV n = 3 to n = 2
A hydrogen atom consists of only one electron. An electron can have transitions, like from n = 3 to n = 2 or from n = 2 to n = 1, at a time.
So, it can be concluded that the photons are emitted either from three atoms (when all the three transitions of electrons are in different atoms) or from two atoms (when an atom has n = 3 to n = 2 and then n = 2 to n = 1 electronic transition and the other has n = 3 to n = 1 electronic transition).
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