Bohr's Atomic Model

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   Bohr's three postulates are: i. In a hydrogen atom, the electron revolves round the nucleus in a fixed circular orbit with constant speed. Equation according to Bohr's first postulate, $$\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r_n^2}$$ ii. The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of $\dfrac{h}{2\pi}$, h being the Planck's constant. Equation according to Bohr's second postulate, $$m_er_nv_n=\dfrac{nh}{2\pi}$$ iii. An electron can make a transition from one of its orbits to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits. Equation according to Bohr's third postulate, $$E_m-E_n=h\nu$$ where, $E_m$ = Energy of electron in $m^{th}$ higher orbit, $E_n$ = Energy of electron in $n^{th}$ lower orbit

2. 2.↑ back to the question

   Expression for total energy of electron: i. Kinetic energy: Let, $m_e$ = mass of electron, $r_n$ = radius of $n^{th}$ orbit of Bohr's hydrogen atom, $v_n$ = velocity of electron, $-e$ = charge of electron, $+e$ = charge on the nucleus, Z = number of electrons in an atom. According to Bohr's first postulate, $$\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{Ze^2}{r_n^2}$$ where, $\varepsilon_0$ is permittivity of free space. $$\therefore m_ev_n^2=\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\qquad....(1)$$ The revolving electron in the circular orbit has linear speed and hence it possesses kinetic energy. It is given by, $K.E=\dfrac{1}{2}m_ev_n^2$ $$\therefore K.E=\dfrac{1}{2}\times\left(\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\right)\qquad....\text{[From equation (1)]}$$ $$\therefore K.E=\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(2)$$ ii. Potential energy: Potential energy of electron is given by, P.E $=V(-e)$ where, V = electric potential at any point due to charge on nucleus, $-e$ = charge on electron. In this case, $$\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{e}{r_n}\times(-Ze)$$ $$\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{-Ze^2}{r_n}$$ $$\therefore P.E=-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\qquad....(3)$$ iii. Total energy: The total energy of the electron in any orbit is its sum of P.E and K.E. $\therefore T.E=K.E+P.E$ $$=\left(\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\right)+\left(-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\right)\qquad....\text{[From equations (2) and (3)]}$$ $$\therefore T.E=-\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(4)$$ iv. But, $r_n=\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)\times n^2$ Substituting for $r_n$ in equation (4), $$\therefore T.E=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Ze^2}{\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2}=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Z^2e^2\pi m_ee^2}{\varepsilon_0h^2n^2}$$ $$\therefore T.E=-\left(\dfrac{m_eZ^2e^4}{8\varepsilon_0^2h^2}\right)\times\dfrac{1}{n^2}\qquad....(5)$$ This is required expression for energy of electron in $n^{th}$ orbit of Bohr's hydrogen atom. v. The negative sign in equation (5) shows that the electron is bound to the nucleus by an attractive force and hence energy must be supplied to the electron in order to make it free from the influence of the nucleus. vi. As m, e, $\varepsilon_0$, and h in equation (5) are constant, $$\therefore\dfrac{me^4}{8\varepsilon_0^2h^2}=\text{constant}$$ Using equation (5), $$T.E=\text{constant}\times\dfrac{1}{n^2}$$ $$\therefore T.E\propto\dfrac{1}{n^2}$$ Binding energy: Binding energy of an electron is the minimum energy required to make it free from the nucleus.

3. 3.↑ back to the question

   Expression for radius of Bohr orbit in atom: i. Let, $m_e$ = mass of electron, $-e$ = charge on electron, $r_n$ = radius of $n^{th}$ Bohr's orbit, $+e$ = charge on nucleus, $v_n$ = linear velocity of electron in $n^{th}$ orbit, Z = number of electrons in an atom, n = principal quantum number. ii. From Bohr's first postulate, Coulomb's force $F_e$ = Centripetal force $F_{cp}$ $$\therefore\dfrac{Ze^2}{4\pi\varepsilon_0r_n^2}=\dfrac{m_ev_n^2}{r_n}$$ $$\therefore v_n^2=\dfrac{Ze^2}{4\pi\varepsilon_0r_nm_e}\qquad....(1)$$ iii. According to Bohr's second postulate, $$m_er_nv_n=\dfrac{nh}{2\pi}$$ $$\therefore m_e^2v_n^2r_n^2=\dfrac{n^2h^2}{4\pi^2}$$ $$\therefore v_n^2=\dfrac{n^2h^2}{4\pi^2m_e^2r_n^2}\qquad....(2)$$ iv. From equations (1) and (2), $$\dfrac{n^2h^2}{4\pi^2m_e^2r_n^2}=\dfrac{Ze^2}{4\pi\varepsilon_0r_nm_e}$$ $$\therefore r_n=\dfrac{n^2h^2\varepsilon_0}{\pi m_eZe^2}$$ $$\therefore r_n=\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2=a_0n^2\qquad....(3)$$ This is the required expression for radius of $n^{th}$ orbit.

4. 4.↑ back to the question

   An electron can make a transition from one of its orbits to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits.

5. 5.↑ back to the question

   i. Let, $E_m$ = Energy of electron in $m^{th}$ higher orbit, $E_n$ = Energy of electron in $n^{th}$ lower orbit. ii. According to Bohr's third postulate, $E_m-E_n=h\nu$ $$\therefore\nu=\dfrac{E_m-E_n}{h}\qquad....(1)$$ iii. But $E_m=-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2m^2}\qquad....(2)$ $$E_n=-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2n^2}\qquad....(3)$$ iv. From equations (1), (2) and (3), $$\nu=\dfrac{-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2m^2}-\left(-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2n^2}\right)}{h}$$ $$\therefore\nu=\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^3}\left[-\dfrac{1}{m^2}+\dfrac{1}{n^2}\right]$$ $$\therefore\dfrac{c}{\lambda}=\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^3}\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]\qquad\left[\because\nu=\dfrac{c}{\lambda}\right]$$ where, c = speed of electromagnetic radiation $$\therefore\dfrac{1}{\lambda}=\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^3c}\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]$$ v. But, $\dfrac{m_ee^4}{8\varepsilon_0^2h^3c}=R_H$ = Rydberg's constant $$=1.097\times10^7\text{ m}^{-1}$$ $$\therefore\dfrac{1}{\lambda}=R_HZ^2\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]\qquad....(4)$$ Equation (4) represents Rydberg's formula for atomic spectrum. vi. $\dfrac{1}{\lambda}$ is called wave number ($\bar\nu$) of the line. $$\therefore\bar\nu=\dfrac{1}{\lambda}=R_HZ^2\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]$$

6. 6.↑ back to the question

   Paschen series: i. The spectral lines which correspond to the transition of an electron from some higher energy state to 3rd orbit are termed as Paschen series. ii. For Paschen series, n = 3 and m = 4, 5,... The wave numbers and the wavelengths of the spectral lines constituting the Paschen series are given by, $$\bar\nu=\dfrac{1}{\lambda}=R_HZ^2\left(\dfrac{1}{3^2}-\dfrac{1}{m^2}\right)$$ iii. Paschen series lies in the infrared region of the spectrum which is invisible and contains infinite number of lines. iv. Wavelengths for m = 4 and 5 are 18750 Å and 12820 Å respectively. Brackett series: i. The spectral lines which correspond to the transition of an electron from a higher energy state to the $4^{th}$ orbit are termed as Brackett series. ii. For this series, n = 4 and m = 5, 6, 7,... The wave numbers and the wavelengths of the spectral lines constituting the Brackett series are given by, $$\bar\nu=\dfrac{1}{\lambda}=R_HZ^2\left(\dfrac{1}{4^2}-\dfrac{1}{m^2}\right)$$ iii. This series lies in the near infrared region of the spectrum and contains infinite number of lines. Wavelengths for m = 5 and 6, are 40518 Å and 26253 Å respectively.

7. 7.↑ back to the question

   [From 15.6, Q6]

   Paschen series: i. The spectral lines which correspond to the transition of an electron from some higher energy state to 3rd orbit are termed as Paschen series. ii. For Paschen series, n = 3 and m = 4, 5,... The wave numbers and the wavelengths of the spectral lines constituting the Paschen series are given by, $$\bar\nu=\dfrac{1}{\lambda}=R_HZ^2\left(\dfrac{1}{3^2}-\dfrac{1}{m^2}\right)$$ iii. Paschen series lies in the infrared region of the spectrum which is invisible and contains infinite number of lines. iv. Wavelengths for m = 4 and 5 are 18750 Å and 12820 Å respectively. Brackett series: i. The spectral lines which correspond to the transition of an electron from a higher energy state to the $4^{th}$ orbit are termed as Brackett series. ii. For this series, n = 4 and m = 5, 6, 7,... The wave numbers and the wavelengths of the spectral lines constituting the Brackett series are given by, $$\bar\nu=\dfrac{1}{\lambda}=R_HZ^2\left(\dfrac{1}{4^2}-\dfrac{1}{m^2}\right)$$ iii. This series lies in the near infrared region of the spectrum and contains infinite number of lines. Wavelengths for m = 5 and 6, are 40518 Å and 26253 Å respectively.for diagram and explanation of Brackett series. Balmer series: i. The spectral lines of this series correspond to the transition of an electron from some higher energy state to $2^{nd}$ orbit. ii. For Balmer series, n = 2 and m = 3, 4, 5, The wave numbers and the wavelengths of spectral lines constituting the Balmer series are given by, $$\bar\nu=\dfrac{1}{\lambda}=R_HZ^2\left(\dfrac{1}{2^2}-\dfrac{1}{m^2}\right)$$ iii. There are infinite number lines in this series out of which four lines are seen called $H_\alpha$, $H_\beta$, $H_\gamma$, $H_\delta$. iv. This series lies in the visible region. Wavelengths for m = 3 and 4 are 6563 Å and 4868 Å respectively.

8. 8.↑ back to the question

   [From 15.6, Q2] Expression for total energy of electron: i. Kinetic energy: Let, $m_e$ = mass of electron, $r_n$ = radius of $n^{th}$ orbit of Bohr's hydrogen atom, $v_n$ = velocity of electron, $-e$ = charge of electron, $+e$ = charge on the nucleus, Z = number of electrons in an atom. According to Bohr's first postulate, $$\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{Ze^2}{r_n^2}$$ where, $\varepsilon_0$ is permittivity of free space. $$\therefore m_ev_n^2=\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\qquad....(1)$$ The revolving electron in the circular orbit has linear speed and hence it possesses kinetic energy. It is given by, $K.E=\dfrac{1}{2}m_ev_n^2$ $$\therefore K.E=\dfrac{1}{2}\times\left(\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\right)\qquad....\text{[From equation (1)]}$$ $$\therefore K.E=\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(2)$$ ii. Potential energy: Potential energy of electron is given by, P.E $=V(-e)$ where, V = electric potential at any point due to charge on nucleus, $-e$ = charge on electron. In this case, $$\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{e}{r_n}\times(-Ze)$$ $$\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{-Ze^2}{r_n}$$ $$\therefore P.E=-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\qquad....(3)$$ iii. Total energy: The total energy of the electron in any orbit is its sum of P.E and K.E. $\therefore T.E=K.E+P.E$ $$=\left(\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\right)+\left(-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\right)\qquad....\text{[From equations (2) and (3)]}$$ $$\therefore T.E=-\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(4)$$ iv. But, $r_n=\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)\times n^2$ Substituting for $r_n$ in equation (4), $$\therefore T.E=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Ze^2}{\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2}=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Z^2e^2\pi m_ee^2}{\varepsilon_0h^2n^2}$$ $$\therefore T.E=-\left(\dfrac{m_eZ^2e^4}{8\varepsilon_0^2h^2}\right)\times\dfrac{1}{n^2}\qquad....(5)$$ This is required expression for energy of electron in $n^{th}$ orbit of Bohr's hydrogen atom. v. The negative sign in equation (5) shows that the electron is bound to the nucleus by an attractive force and hence energy must be supplied to the electron in order to make it free from the influence of the nucleus. vi. As m, e, $\varepsilon_0$, and h in equation (5) are constant, $$\therefore\dfrac{me^4}{8\varepsilon_0^2h^2}=\text{constant}$$ Using equation (5), $$T.E=\text{constant}\times\dfrac{1}{n^2}$$ $$\therefore T.E\propto\dfrac{1}{n^2}$$ Binding energy: Binding energy of an electron is the minimum energy required to make it free from the nucleus.for expression for energy of an electron in Bohr orbit. Expression for binding energy of electron in Bohr orbit: iii. When binding energy is supplied to an electron, the total energy of the system containing nucleus and electron becomes zero. $\therefore B.E+T.E=0$ $\therefore B.E=-T.E$ iv. But $T.E=-\dfrac{me^4}{8\varepsilon_0^2h^2n^2}$ $$\therefore B.E=-\left(-\dfrac{me^4}{8\varepsilon_0^2h^2n^2}\right)$$ $$\therefore B.E=\dfrac{me^4}{8\varepsilon_0^2h^2n^2}$$

9. 9.↑ back to the question

   Minimum angular momentum of electron in hydrogen atom is $\left(\dfrac{h}{2\pi}\right)$.

10. 10.↑ back to the question

    [From 15.6, Q1 (Statements only)] Bohr's three postulates are: i. In a hydrogen atom, the electron revolves round the nucleus in a fixed circular orbit with constant speed. Equation according to Bohr's first postulate, $$\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r_n^2}$$ ii. The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of $\dfrac{h}{2\pi}$, h being the Planck's constant. Equation according to Bohr's second postulate, $$m_er_nv_n=\dfrac{nh}{2\pi}$$ iii. An electron can make a transition from one of its orbits to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits. Equation according to Bohr's third postulate, $$E_m-E_n=h\nu$$ where, $E_m$ = Energy of electron in $m^{th}$ higher orbit, $E_n$ = Energy of electron in $n^{th}$ lower orbit [Any two]

11. 11.↑ back to the question

    Bohr's third postulate: $E_m-E_n=h\nu$ Where, $E_m$ = Energy of electron in $m^{th}$ higher orbit, $E_n$ = Energy of electron in $n^{th}$ lower orbit.

12. 12.↑ back to the question

    [From 15.6, Q3] Expression for radius of Bohr orbit in atom: i. Let, $m_e$ = mass of electron, $-e$ = charge on electron, $r_n$ = radius of $n^{th}$ Bohr's orbit, $+e$ = charge on nucleus, $v_n$ = linear velocity of electron in $n^{th}$ orbit, Z = number of electrons in an atom, n = principal quantum number. ii. From Bohr's first postulate, Coulomb's force $F_e$ = Centripetal force $F_{cp}$ $$\therefore\dfrac{Ze^2}{4\pi\varepsilon_0r_n^2}=\dfrac{m_ev_n^2}{r_n}$$ $$\therefore v_n^2=\dfrac{Ze^2}{4\pi\varepsilon_0r_nm_e}\qquad....(1)$$ iii. According to Bohr's second postulate, $$m_er_nv_n=\dfrac{nh}{2\pi}$$ $$\therefore m_e^2v_n^2r_n^2=\dfrac{n^2h^2}{4\pi^2}$$ $$\therefore v_n^2=\dfrac{n^2h^2}{4\pi^2m_e^2r_n^2}\qquad....(2)$$ iv. From equations (1) and (2), $$\dfrac{n^2h^2}{4\pi^2m_e^2r_n^2}=\dfrac{Ze^2}{4\pi\varepsilon_0r_nm_e}$$ $$\therefore r_n=\dfrac{n^2h^2\varepsilon_0}{\pi m_eZe^2}$$ $$\therefore r_n=\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2=a_0n^2\qquad....(3)$$ This is the required expression for radius of $n^{th}$ orbit.

13. 13.↑ back to the question

    [From 15.6, Q1] Bohr's three postulates are: i. In a hydrogen atom, the electron revolves round the nucleus in a fixed circular orbit with constant speed. Equation according to Bohr's first postulate, $$\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r_n^2}$$ ii. The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of $\dfrac{h}{2\pi}$, h being the Planck's constant. Equation according to Bohr's second postulate, $$m_er_nv_n=\dfrac{nh}{2\pi}$$ iii. An electron can make a transition from one of its orbits to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits. Equation according to Bohr's third postulate, $$E_m-E_n=h\nu$$ where, $E_m$ = Energy of electron in $m^{th}$ higher orbit, $E_n$ = Energy of electron in $n^{th}$ lower orbit[(i) and (ii)] (Statements only)

14. 14.↑ back to the question

    [From 15.6, Q1 (Statements only)] Bohr's three postulates are: i. In a hydrogen atom, the electron revolves round the nucleus in a fixed circular orbit with constant speed. Equation according to Bohr's first postulate, $$\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r_n^2}$$ ii. The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of $\dfrac{h}{2\pi}$, h being the Planck's constant. Equation according to Bohr's second postulate, $$m_er_nv_n=\dfrac{nh}{2\pi}$$ iii. An electron can make a transition from one of its orbits to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits. Equation according to Bohr's third postulate, $$E_m-E_n=h\nu$$ where, $E_m$ = Energy of electron in $m^{th}$ higher orbit, $E_n$ = Energy of electron in $n^{th}$ lower orbit

15. 15.↑ back to the question

    [From 15.6, Q2 (vi)] Expression for total energy of electron: i. Kinetic energy: Let, $m_e$ = mass of electron, $r_n$ = radius of $n^{th}$ orbit of Bohr's hydrogen atom, $v_n$ = velocity of electron, $-e$ = charge of electron, $+e$ = charge on the nucleus, Z = number of electrons in an atom. According to Bohr's first postulate, $$\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{Ze^2}{r_n^2}$$ where, $\varepsilon_0$ is permittivity of free space. $$\therefore m_ev_n^2=\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\qquad....(1)$$ The revolving electron in the circular orbit has linear speed and hence it possesses kinetic energy. It is given by, $K.E=\dfrac{1}{2}m_ev_n^2$ $$\therefore K.E=\dfrac{1}{2}\times\left(\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\right)\qquad....\text{[From equation (1)]}$$ $$\therefore K.E=\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(2)$$ ii. Potential energy: Potential energy of electron is given by, P.E $=V(-e)$ where, V = electric potential at any point due to charge on nucleus, $-e$ = charge on electron. In this case, $$\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{e}{r_n}\times(-Ze)$$ $$\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{-Ze^2}{r_n}$$ $$\therefore P.E=-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\qquad....(3)$$ iii. Total energy: The total energy of the electron in any orbit is its sum of P.E and K.E. $\therefore T.E=K.E+P.E$ $$=\left(\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\right)+\left(-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\right)\qquad....\text{[From equations (2) and (3)]}$$ $$\therefore T.E=-\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(4)$$ iv. But, $r_n=\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)\times n^2$ Substituting for $r_n$ in equation (4), $$\therefore T.E=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Ze^2}{\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2}=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Z^2e^2\pi m_ee^2}{\varepsilon_0h^2n^2}$$ $$\therefore T.E=-\left(\dfrac{m_eZ^2e^4}{8\varepsilon_0^2h^2}\right)\times\dfrac{1}{n^2}\qquad....(5)$$ This is required expression for energy of electron in $n^{th}$ orbit of Bohr's hydrogen atom. v. The negative sign in equation (5) shows that the electron is bound to the nucleus by an attractive force and hence energy must be supplied to the electron in order to make it free from the influence of the nucleus. vi. As m, e, $\varepsilon_0$, and h in equation (5) are constant, $$\therefore\dfrac{me^4}{8\varepsilon_0^2h^2}=\text{constant}$$ Using equation (5), $$T.E=\text{constant}\times\dfrac{1}{n^2}$$ $$\therefore T.E\propto\dfrac{1}{n^2}$$ Binding energy: Binding energy of an electron is the minimum energy required to make it free from the nucleus. (Definition of Binding energy only)

16. 16.↑ back to the question

    Given: $E_1=-13.6$ eV To find: Energy of electron in second Bohr orbit ($E_3$) Formula: $E\propto\dfrac{1}{n^2}$ Calculation: From formula, $$\dfrac{E_1}{E_2}=\dfrac{n_2^2}{n_1^2}$$ $$\therefore E_2=(-13.6)\times\left(\dfrac{1}{2}\right)^2$$ $$\therefore E_3=-3.4\text{ eV}$$ Ans: Energy of an electron in the third orbit is –3.4 eV.

17. 17.↑ back to the question

    Given: $h=6.63\times10^{-34}$ J s, $\lambda_\beta=4860$ Å $=4.86\times10^{-7}$ m, $c=3\times10^8$ m/s, 1 eV $=1.6\times10^{-19}$ J To find: i. Rydberg's constant ($R_H$) ii. Energy in eV ($E_3$) Formulae: i. $\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$ ii. $R_H=\dfrac{m_ee^4}{8\varepsilon_0^2ch^3}$ iii. $E_n=-\dfrac{m_ee^4}{8\varepsilon_0^2n^2h^2}$ Calculation: For the $H_\beta$ line of Balmer series, n = 2 and m = 4 Using formula (i) $$\dfrac{1}{\lambda_\beta}=R_H\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)=R_H\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3R}{16}$$ $$\therefore R_H=\dfrac{16}{3\lambda_\beta}=\dfrac{16}{3\times4.86\times10^{-7}}$$ $$\therefore R_H=1.097\times10^7\text{ m}^{-1}$$ Now, using formulae (ii) and (iii) we get, $$\dfrac{E_n}{R_H}=-\dfrac{ch}{n^2}$$ $$\therefore E_3=-\dfrac{R_Hch}{3^2}$$ $$=-\dfrac{1.097\times10^7\times3\times10^8\times6.63\times10^{-34}}{3^2}$$ $$=-\dfrac{21.819}{9}\times10^{-19}$$ $$=-2.424\times10^{-19}\text{ J}$$ $$=-\dfrac{2.424\times10^{-19}}{1.6\times10^{-19}}\text{ eV}$$ $$\therefore E_3=-1.515\text{ eV}$$ Ans: i. The value of Rydberg's constant is $1.097\times10^7$ m$^{-1}$. ii. Energy in eV of electron in the third orbit is –1.515 eV.

18. 18.↑ back to the question

    Given: $r=0.5$ Å $=0.5\times10^{-10}$ m $=5\times10^{-11}$ m To find: Period of revolution (T) Formula: $T=\dfrac{2\pi r}{v}$ Calculation: Using formula (i) we get, $$T=\dfrac{2\times3.14\times5\times10^{-11}}{2.24\times10^6}$$ $$=\dfrac{31.4}{2.24}\times10^{-17}$$ $$=14.02\times10^{-17}$$ $T=1.402\times10^{-16}$ s Ans: Period of revolution of electron in the first Bohr orbit is $1.402\times10^{-16}$ s.

19. 19.↑ back to the question

    Given: $R=1.097\times10^7$ m$^{-1}$, $h=6.63\times10^{-34}$ J-s, $c=3\times10^8$ m/s, $n=3$, $e=1.6\times10^{-19}$ C To find: Energy of electron in $3^{rd}$ orbit ($E_3$) Formula: $E=\dfrac{-hRc}{n^2e}$ (in eV) Calculation: From Formula, $$E_3=\dfrac{-6.63\times10^{-34}\times1.097\times10^7\times3\times10^8}{9\times1.6\times10^{-19}}=\dfrac{-6.63\times1.097\times3}{9\times1.6}$$ $$=-\{\text{antilog}[\log(6.63)+\log(1.097)+\log(3)-\log(9)-\log(1.6)]\}$$ $$=-\{\text{antilog}[0.8215+0.0402+0.4771-0.9542-0.2041]\}$$ $$=-\{\text{antilog}[0.1805]\}$$ $\therefore|E_3|=1.515$ eV Ans: The value of energy of electron in the third Bohr orbit of hydrogen atom is 1.515 eV.

20. 20.↑ back to the question

    Given: $m_e=9.1\times10^{-31}$ kg, $e=1.6\times10^{-19}$ C, $h=6.63\times10^{-34}$ J-s, $\varepsilon_0=8.85\times10^{-12}$ C²/Nm², $n=2$ To find: Radius of $2^{nd}$ Bohr orbit ($r_2$) Formula: $r_n=\dfrac{\varepsilon_0n^2h^2}{\pi me^2}$ Calculation: From formula, $$r_2=\dfrac{8.85\times10^{-12}\times2^2\times(6.63\times10^{-34})^2}{3.14\times9.1\times10^{-31}\times(1.6\times10^{-19})^2}=\dfrac{8.85\times4\times(6.63)^2\times10^{-12}\times10^{-68}}{3.14\times9.1\times(1.6)^2\times10^{-31}\times10^{-38}}=\dfrac{8.85\times4\times(6.63)^2}{3.14\times9.1\times(1.6)^2}\times10^{-11}$$ $$=\{\text{antilog}[\log(8.85)+\log(4)+2\log(6.63)-\log(3.14)-\log(9.1)-2\log(1.6)]\}\times10^{-11}$$ $$=\{\text{antilog}[0.9469+0.6021+2\times0.8215-0.4969-0.9590-2\times0.2041]\}\times10^{-11}$$ $$=\{\text{antilog}[1.3279]\}\times10^{-11}$$ $$=21.27\times10^{-11}=2.127\times10^{-10}\text{ m}$$ $\therefore r_2=2.127$ Å Ans: Radius of second Bohr orbit is 2.127 Å.

21. 21.↑ back to the question

    Given: $E_1=-13.6$ eV, $n_2=5$, $n_1=1$ To find: Ionisation energy Formulae: i. $\dfrac{E_5}{E_1}=\dfrac{n_1^2}{n_2^2}$ ii. Ionisation energy $=E_\infty-E_n$ Calculation: Using formula (i), $$\dfrac{E_5}{-13.6}=\dfrac{1^2}{5^2}$$ $$\therefore E_5=\dfrac{-13.6}{25}=-0.544\text{ eV}$$ Using formula (ii), Ionisation energy $=E_\infty-E_5=0-(-0.544)=0.544$ eV Ans: Ionisation energy for given atom is 0.544 eV.

22. 22.↑ back to the question

    Given: For longest wavelength m = 2, for shortest wavelength m = $\infty$ To find: i. Longest wavelength ($\lambda_L$) ii. Shortest wavelength ($\lambda_S$) Formula: $\dfrac{1}{\lambda}=R_H\left[\dfrac{1}{1^2}-\dfrac{1}{m^2}\right]$ Calculation: The ultraviolet region for hydrogen atom lies in the Lyman series. For the longest wavelength ($\lambda_L$), from formula, $$\dfrac{1}{\lambda_L}=R_H\left[\dfrac{1}{1^2}-\dfrac{1}{2^2}\right]$$ $$\dfrac{1}{\lambda_L}=\dfrac{3R_H}{4}$$ $$\therefore\lambda_L=\dfrac{4}{3R_H}\qquad....(1)$$ For Lyman series shortest wavelength ($\lambda_S$), from formula, $$\dfrac{1}{\lambda_S}=R_H\left[\dfrac{1}{1^2}-\dfrac{1}{\infty^2}\right]$$ $$\dfrac{1}{\lambda_S}=R_H$$ $$\therefore\lambda_S=\dfrac{1}{R_H}\qquad....(2)$$ Ans: The expression for longest and shortest wavelength of spectral lines in ultraviolet region for hydrogen atom is $\dfrac{4}{3R_H}$ and $\dfrac{1}{R_H}$ respectively.

23. 23.↑ back to the question

    Given: $r_2=2.14\times10^{-10}$ m, n = 2, $v_2=1.09\times10^6$ m/s To find: Frequency of revolution ($\nu_2$) Formula: $v=r\omega=r(2\pi\nu)$ $$\therefore\nu=\dfrac{v}{2\pi r}$$ Calculation: From formula, $$\nu_2=\dfrac{v_2}{2\pi r_2}=\dfrac{1.09\times10^6}{2\times3.142\times2.14\times10^{-10}}$$ $$=\{\text{antilog}[\log(1.09)-\log(3.142)-\log(4.28)]\}\times10^{16}$$ $$=\{\text{antilog}[0.0374-0.4972-0.6314]\}\times10^{16}$$ $$=\{\text{antilog}[\overline2.9088]\}\times10^{16}$$ $$=8.106\times10^{-2}\times10^{16}$$ $\therefore\nu_2\approx8.11\times10^{14}$ Hz Ans: The frequency of revolution of electron in $2^{nd}$ Bohr orbit is $8.11\times10^{14}$ Hz.

24. 24.↑ back to the question

    Let, $\lambda_S$ = shortest wavelength, $\lambda_L$ = longest wavelength $$\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$$ Longest wavelength in Paschen series is obtained when n = 3, m = 4 For longest wavelength, $$\dfrac{1}{\lambda_L}=R\left[\dfrac{1}{3^2}-\dfrac{1}{4^2}\right]$$ $$\dfrac{1}{\lambda_L}=R\left[\dfrac{1}{9}-\dfrac{1}{16}\right]$$ $$\dfrac{1}{\lambda_L}=R\left[\dfrac{16-9}{9\times16}\right]$$ $$\dfrac{1}{\lambda_L}=\dfrac{7R}{9\times16}$$ $$\therefore\lambda_L=\dfrac{9\times16}{7R}=\dfrac{144}{7R}$$ Shortest wavelength in Balmer series is obtained when n = 2, m = $\infty$ For shortest wavelength, $$\dfrac{1}{\lambda_S}=R\left[\dfrac{1}{2^2}-\dfrac{1}{\infty^2}\right]$$ $$\therefore\lambda_S=\dfrac{4}{R}$$ $$\therefore\dfrac{\lambda_L}{\lambda_S}=\dfrac{144}{7R}\times\dfrac{R}{4}=\dfrac{36}{7}$$ $$\therefore\dfrac{\lambda_L}{\lambda_S}=5.131$$ Ans: The ratio of longest wavelength in Paschen series to shortest wavelength in Balmer series is 5.131.

25. 25.↑ back to the question

    Given: $v=2.3\times10^6$ m/s, $r=0.53$ Å $=0.53\times10^{-10}$ m To find: Period of revolution (T) Formula: $T=\dfrac{2\pi r}{v}$ Calculation: Using formula, $$T=\dfrac{2\times3.142\times0.53\times10^{-10}}{2.3\times10^6}$$ $$=\{\text{antilog}[\log(2)+\log(3.142)+\log(0.53)-\log(2.3)]\}\times10^{-16}$$ $$=\{\text{antilog}[0.3010+0.4972+\overline1.7243-0.3617]\}\times10^{-16}$$ $$=\{\text{antilog}[0.1608]\}\times10^{-16}$$ $\therefore T=1.448\times10^{-16}$ s Ans: Period of revolution of electron is $1.448\times10^{-16}$ s.

26. 26.↑ back to the question

    Given: $h=6.63\times10^{-34}$ J.s, $E_1=-13.6$ eV, $E_n=-0.85$ eV To find: Angular momentum of the electron (L) Formulae: i. $E_n=\dfrac{E_1}{n^2}$ ii. $L=\dfrac{nh}{2\pi}$ Calculation: Using formula (i) $$-0.85=\dfrac{-13.6}{n^2}=16$$ $$n^2=\dfrac{-13.6}{-0.85}$$ $\therefore n=4$ Using formula (ii) $$L=\dfrac{4\times6.63\times10^{-34}}{2\times3.142}=\dfrac{2\times6.63}{3.142}\times10^{-34}$$ $$=\{\text{antilog}[\log(2)+\log(6.63)-\log(3.142)]\}\times10^{-34}$$ $$=\{\text{antilog}[0.3010+0.8215-0.4972]\}\times10^{-34}$$ $$=\{\text{antilog}[0.6253]\}\times10^{-34}$$ $\therefore L=4.22\times10^{-34}$ kg m²/s Ans: The angular momentum of the electron in the fourth orbit in a hydrogen atom is $4.22\times10^{-34}$ kg m²/s.

27. 27.↑ back to the question

    In hydrogen spectrum, $H_\gamma$ line corresponds to m = 5 line of Balmer series (n = 2). Wavelength of $H_\gamma$ line is given by formula, $$\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$$ $$\dfrac{1}{\lambda}=1.093\times10^7\times\left(\dfrac{1}{2^2}-\dfrac{1}{5^2}\right)=1.093\times10^7\times0.21=2.295\times10^6$$ $$\therefore\lambda=\dfrac{1}{2.295\times10^6}=4.3567\times10^{-7}\text{ m}\approx4357\text{ \text{Å}}$$ For series limit of Brackett series, n = 5 and m = $\infty$, $$\dfrac{1}{\lambda_{Br}}=R\left(\dfrac{1}{5^2}-\dfrac{1}{\infty^2}\right)=\dfrac{1.093\times10^7}{25}=437200$$ $$\therefore\lambda_{Br}=\dfrac{1}{437200}=2.2873\times10^{-6}=22873\text{ \text{Å}}$$ Ans: The wavelengths corresponding to $H_\gamma$ line and series limit of Brackett series are 4357 Å and 22873 Å respectively.

28. 28.↑ back to the question

    Given: $E_2=-3.4$ eV To find: Energy of electron in third orbit ($E_3$) Formula: $E\propto\dfrac{1}{n^2}$ Calculation: From formula, $$\dfrac{E_3}{E_2}=\dfrac{n_2^2}{n_3^2}$$ $$\therefore E_3=\left(\dfrac{2}{3}\right)^2\times(-3.4)$$ $\therefore E_3=-1.51$ eV Ans: Energy of an electron in the third orbit is –1.51 eV.

29. 29.↑ back to the question

    Given: Series limit of Lyman Series, ($\lambda_L$) = 912 Å To find: Shortest wavelength of i. Balmer series ii. Paschen series Formula: $\dfrac{1}{\lambda}=R\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]$ Calculation: From formula, for shortest wavelength, m = $\infty$ $$\therefore\lambda=\dfrac{n^2}{R}$$ For Lyman series, n = 1 $$\lambda_L=\dfrac{n^2}{R}=\dfrac{1^2}{R}=\dfrac{1}{R}=912\text{ \text{Å}}$$ i. For Balmer series, n = 2 $$\lambda_B=\dfrac{2^2}{R}=\dfrac{4}{R}=4\times912=3648\text{ \text{Å}}$$ ii. For Paschen series, n = 3 $$\lambda_P=\dfrac{3^2}{R}=\dfrac{9}{R}=9\times912=8208\text{ \text{Å}}$$ Ans: The shortest wavelength of i. Balmer series is 3648 Å. ii. Paschen series is 8208 Å.

30. 30.↑ back to the question

    For hydrogen: $\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$ For longest wavelength of Lyman series, n = 1 and m = 2 $$\therefore\dfrac{1}{\lambda_L}=R_H\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)\implies\lambda_L=\dfrac{3}{4}R_H$$ For longest wavelength of Balmer series, n = 2 and m = 3 $$\therefore\dfrac{1}{\lambda_B}=R_H\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)\implies\lambda_B=\dfrac{5}{36}R_H$$ $$\therefore\dfrac{\lambda_L}{\lambda_B}=\dfrac{4}{3R_H}\times\dfrac{5R_H}{36}=\dfrac{5}{27}$$

31. 31.↑ back to the question

    Radius of Bohr orbit, $r_n=a_0n^2$ $$\therefore\dfrac{r_2}{r_3}=\dfrac{(2)^2}{(3)^2}=\dfrac{4}{9}$$ $$\therefore r_2=r_3\times\dfrac{4}{9}=\dfrac{0.477\times4}{9}=0.212\text{ nm}$$

32. 32.↑ back to the question

    For hydrogen: $\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$ The shortest wavelength is obtained when m = $\infty$. For shortest wavelength of Paschen series, n = 3 and m = $\infty$ $$\therefore\dfrac{1}{\lambda_L}=R_H\left(\dfrac{1}{3^2}-\dfrac{1}{\infty^2}\right)\implies\lambda_L=\dfrac{9}{R_H}=820\text{ nm}$$ For longest wavelength of Balmer series, n = 2 and m = 3 $$\therefore\dfrac{1}{\lambda_B}=R_H\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)\implies\lambda_B=\dfrac{36}{5R_H}=656.3\text{ nm}$$

33. 33.↑ back to the question

    Angular momentum of electron in first orbit, $L_1=mv_1r_1=\dfrac{n_1h}{2\pi}$ Angular momentum of electron in third orbit, $L_3=mv_3r_3=\dfrac{n_3h}{2\pi}$ $\therefore$ Change in angular momentum, $L_3-L_1=mv_3r_3-mv_1r_1$ $$=\dfrac{n_3h}{2\pi}-\dfrac{n_1h}{2\pi}=\dfrac{h}{2\pi}(n_3-n_1)$$ $$=\dfrac{h}{2\pi}(3-1)=\dfrac{h}{\pi}=\dfrac{6.63\times10^{-34}}{3.142}$$ $\therefore L_3-L_1=2.11\times10^{-34}$ kg m²/s

34. 34.↑ back to the question

    Given: $R_H=1.097\times10^7$ m$^{-1}$, We know that, for Balmer series, n = 2 To find: i. Wavelength of first line of Balmer series ii. Wavelength of second line of Balmer series Formula: For Balmer series, $\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{2^2}-\dfrac{1}{m^2}\right)$ Calculation: i. For first line in Balmer series, m = 3 From formula, $$\dfrac{1}{\lambda}=1.097\times10^7\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)=1.097\times10^7\times\left(\dfrac{5}{4\times9}\right)=1.524\times10^6\text{ m}^{-1}$$ Using reciprocal table, $\lambda=6.563\times10^{-7}$ m $=$ 6563 Å ii. For second line in Balmer series, m = 4 From formula, $$\dfrac{1}{\lambda}=1.097\times10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)=1.097\times10^7\times\left(\dfrac{12}{4\times16}\right)=0.2057\times10^7\text{ m}^{-1}$$ Using reciprocal table, $\lambda=4.862\times10^{-7}$ m $=$ 4862 Å Ans: i. The wavelength of the first line of the Balmer series is 6563 Å. ii. The wavelength of the second line of the Balmer series is 4862 Å.