The following questions are in continuation of the post dated August 26, 2006 (Questions on Bohr Atom Model):

You can work it out as n(n–1)/2 = 4(4–1)/2 =6.

The total energy is –13.6 Z

(5) If the binding energy of the electron in a hgydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li

(a) 122.4 eV (b) 30.6 eV (c) 13.6 eV (d) 3.4 eV

Questions of this type often appear in entrance test papers.

Since Li

E = –13.6 Z

The energy in the first excited state (second orbit) is – 13.6×9/4 eV = – 30.6 eV. The energy to be supplied to the electron to remove it from the first excited state is therefore + 30.6 eV [Option (b)].

E

In the elementary theory we take ‘m’ as the mass of the electron on the assumption that the nucleus has a very large mass compared to the mass of the electron and hence the electron is moving round with the nucleus at the centre. The real situation is that both the electron and the nucleus are moving along circular paths with the centre of mass as the common centre. Instead of the actual mss of the electron, the reduced mass of the electron and the nucleus is to be substituted in the expression for energy. The modified form of the expression is

The nucleus of deuterium contains a proton and a neutron and has very nearly twice the mass of the hydrogen nucleus (proton). So, the reduced mass and the energy levels of deuterium are slightly greater than those of hydrogen and this is the reason for the difference in wave length. [The wave lengths are slightly shorter].The correct option is (d).

************************************************************************

An important point you should remember in the light of the above discussion is the drastic change in the energy levels and the spectrum of

Now consider the following MCQ:

**(1) Suppose the energy required to remove all the three electrons from a lithium atom in the ground state is ‘E’ electron volt. What will be the energy required (in electron volt) to remove two electrons from the lithium atom in the ground state?**

(a) 2E/3 (b) E – 13.6 (c) E – 27.2 (d) E – 40.8 (e) E – 122.4

The energy of the electron in a hydrogen like atom in the(a) 2E/3 (b) E – 13.6 (c) E – 27.2 (d) E – 40.8 (e) E – 122.4

*ground state*is – 13.6Z^{2}electron volt. Therefore, after removing two electrons from the lithium (Z=3) atom, the third electron has energy equal to – 13.6×3^{2}eV = 122.4 eV. The energy needed to remove two electrons from the lithium atom in the ground state is therefore equal to (E –122.4) eV.**(2) How many revolutions does the electron in the hydrogen atom in the ground state make per second? (h = 6.63×10**

(a) 6.55×10

The angular momentum (Iω) of the electron is an integral multiple of h/2π. Therefore, Iω = nh/2π, from which, for the first orbit (n=1), ω = h/2πI = h/2πm^{-34}Js, mass of electron = 9.11×10^{-31}kg, Bohr radius = 0.53 A.U.)(a) 6.55×10

^{-15}(b) 3.28×10^{-15}(c) 3.28×10^{-16}(d) 1.64×10^{-15}(e) 9.11×10^{-15}_{e}r^{2}. The orbital frequency of the electron is given by f = ω/2π = h/4π^{2}m r^{2}= 6.55×10^{-15}per second, on substituting for h, m and r.**(3) The ionisation energy of the hydrogen atom is 13.6 eV. If hydrogen atoms in the ground state absorb quanta of energy 12.75 eV, how many discrete spectral lines will be emitted as per Bohr’s theory?**

(a) 1 (b) 2 (c) 4 (d) 6 (e) zero

On absorbing 12.75 eV, the energy of the electron in the hydrogen atom will become (–13.6 + 12.75) eV which is – 0.85 eV. This is an allowed state (with n=4) for the electron, since the energy in the 4th orbit is – 13.6/n(a) 1 (b) 2 (c) 4 (d) 6 (e) zero

^{2}= – 13.6/4^{2}= 0.85 eV. From the 4th orbit, the electron can undrego three transitions to the lower orbits (4→3, 4→2, 4→1). From the third orbit, the electron can undergo two transitions (3→2 and 2→1). The electron in the second orbit can undergo one transition (2→1). So, altogether 6 transitions are possible, giving rise to 6 discrete spectral lines [Option (d)].You can work it out as n(n–1)/2 = 4(4–1)/2 =6.

**(4) The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following siatements is true?**

(a) Its kinetic energy increases and its potential and total energies decrease.

(b)Its kinetic energy decreases, potential energy increases and its total energy remains the same.

(c) Its kinetic and total energies decrease and its potential energy increases.

(d) Its kinetic, potential and total energies decrease.

This question appeared in IIT 2000 entrance test paper. The correct option is (a). You should note that the kinetic energy is positive while the potential energy and total energy are negative. Further, the kinetic energy and total energy are numerically equal and the numerical value is equal to half the potential energy.(a) Its kinetic energy increases and its potential and total energies decrease.

(b)Its kinetic energy decreases, potential energy increases and its total energy remains the same.

(c) Its kinetic and total energies decrease and its potential energy increases.

(d) Its kinetic, potential and total energies decrease.

The total energy is –13.6 Z

^{2}/n^{2}. In lower orbits (with smaller n), the potential energy is smaller since it has a larger negative value. The total energy also is therefore smaller. But, the kinetic energy is greater since it has a larger positive value.(5) If the binding energy of the electron in a hgydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li

^{++}is(a) 122.4 eV (b) 30.6 eV (c) 13.6 eV (d) 3.4 eV

Questions of this type often appear in entrance test papers.

Since Li

^{++}is a hydrogen like system with a single electron revolving round a nucleus of proton number Z = 3, the energy of the electron in orbit of quantum number n isE = –13.6 Z

^{2}/n^{2}eV.The energy in the first excited state (second orbit) is – 13.6×9/4 eV = – 30.6 eV. The energy to be supplied to the electron to remove it from the first excited state is therefore + 30.6 eV [Option (b)].

**(6) The wave lengths involved in the spectrum of deuterium (**

(a) the attraction between the electron and the nucleus is different in the two cases

(b) the size of the two nuclei are different

(c) the nuclear forces are different in the two cases

(d) the masses of the two nuclei are different.

This MCQ appeared in AIEEE 2003 questionn paper. The answer to this will not be easy if you stick on to the elementary theory of the Bohr model in which the energy (E_{1}D^{2}) are slightly different from that of hydrogen spectrum, because(a) the attraction between the electron and the nucleus is different in the two cases

(b) the size of the two nuclei are different

(c) the nuclear forces are different in the two cases

(d) the masses of the two nuclei are different.

_{n}) of the electron of quantum number n (n^{th}orbit) is given byE

_{n}= – me^{4}/8ε_{0}n^{2}h^{2}where m is the mass of the electron, e is its charge, ε_{0}is the permittivity of free space and h is Planck’s constant.In the elementary theory we take ‘m’ as the mass of the electron on the assumption that the nucleus has a very large mass compared to the mass of the electron and hence the electron is moving round with the nucleus at the centre. The real situation is that both the electron and the nucleus are moving along circular paths with the centre of mass as the common centre. Instead of the actual mss of the electron, the reduced mass of the electron and the nucleus is to be substituted in the expression for energy. The modified form of the expression is

**E**where μ is the reduced mass of electron and the nucleus, given by_{n}= – μ e^{4}/8ε_{0}n^{2}h^{2}**μ = Mm**, M and m_{e}/(M+m_{e})_{e}being the masses of the nucleus and the electron respectively. [Generally, for a hydrogen like system with proton number Z, the expression for energy is**E**_{n}=**– μ Z**].^{2}e^{4}/8ε_{0}n^{2}h^{2}The nucleus of deuterium contains a proton and a neutron and has very nearly twice the mass of the hydrogen nucleus (proton). So, the reduced mass and the energy levels of deuterium are slightly greater than those of hydrogen and this is the reason for the difference in wave length. [The wave lengths are slightly shorter].The correct option is (d).

************************************************************************

An important point you should remember in the light of the above discussion is the drastic change in the energy levels and the spectrum of

*compared to hydrogen. Positronium is a highly unstable neutral atom with an electron revolving round a positron. [ You can as well say, a positron revolving round an electron!]. The concept of reduced mass is absolutely necessary in this case since the positron has the same mass as that of the electron so that the reduced mass of positronium is mm/(m+m) = m/2 where ‘m’ is the mass of positron as well as the electron.***positronium**Now consider the following MCQ:

**A positronium atom undergoes a transition from the state n = 4 to n = 2. The energy of the photon emitted in this process is**

(a) 1.275 eV (b) 2.55 eV (c) 3.4 eV (d) 5.1 eV (e) 13.6 eV

The expression for energy of positronium is E(a) 1.275 eV (b) 2.55 eV (c) 3.4 eV (d) 5.1 eV (e) 13.6 eV

_{n}= – μ e^{4}/8ε_{0}n^{2}h^{2}where μ is the reduced mass of positron and electron, given by μ = mm/(m+m) = m/2. Therefore, the mass of the electron (m) used in the expression for the energy of a hydrogen atom (Bohr’s theory) is to be replaced by m/2. All energy levels are therefore reduced to half of the hydrogen levels. Since the energies for states n=4 and n=2 for hydrogen are –13.6/16 eV(= –0.85 eV) and –13.6/4 eV (= –3.4 eV) respectively, the energy of the photon emitted in the case of hydrogen is [(–0.85) – (–3.4)] eV = 2.55 eV. In the case of positronium, the energy will be half of this. So, the answer is 2.55/2 eV = 1.275 eV.