If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Tuesday, April 21, 2009

Multiple Choice Questions on Bohr Atom Model

Questions on Bohr model of hydrogen-like atoms usually appear in medical and engineering and other degree entrance examinations. You can access all posts on Bohr model on this site by clicking on the label ‘Bohr model’ below this post. Here are a few more questions on Bohr model of hydrogen atom:

(1) The electron in a hydrogen atom in the ground state absorbs energy equal to 12.1 eV and gets elevated to the highest possible excited state. What will be change in the angular momentum of the electron? (h = Planck’s constant)

(a) h/π

(b) 2h/π

(c) 3h/π

(d) 4h/π

(e) h/2π

In the case of the hydrogen atom the energies of the electron are – 13.6 eV, – 3.4 eV, – 1.51 eV, – 0.85 eV etc. for values of n equal to 1, 2, 3, 4 etc. respectively. After absorbing 12.1 eV of energy, the electron in the innermost orbit having energy – 13.6 eV gets elevated to the 3rd orbit where its energy is – 1.51 eV.

The angular momentum of the electron in the innermost orbit (of quantum no. n = 1) is h/2π. In the 3rd orbit its angular momentum is 3h/2π. Therefore, the change in the angular momentum is (3h/2π) – (h/2π) = h/π

(2) If the ionisation potential of hydrogen atom is 13.6 volt, the energy required to remove an electron from the second orbit of hydrogen atom is

(a) 0.54 eV

(b) 0.85 eV

(c) 1.51 eV

(d) 3.4 eV

(e) 13.6 eV

By stating that the ionisation potential of hydrogen atom is 13.6 volt, you are informed that 13.6 eV of energy is required to remove an electron from the innermost orbit of the hydrogen atom. This is because of the energy –13.6 eV possessed by the electron in the innermost orbit. Since the electron in the 2nd orbit possesses energy equal to –13.6/22 = –3.4 eV, the energy required to remove an electron from the second orbit of hydrogen atom is 3.4 eV.

(3) In the hydrogen atom the transition that gives radiation in the visible region is

(a) from n > 1 to n = 1

(b) from n > 1 to n = 1

(c) from n > 2 to n = 1

(d) from n > 3 to n = 1

(e) from n > 2 to n = 2

The hydrogen atom gives visible spectral lines in the balmer series because of the transitions from outer orbits to the 2nd orbit. So the correct option is (e).

(4) In the hydrogen spectrum the frequency of a line resulting from the transition of the electron from the orbit of quantum number nx to quantum number n1 is f. In a hydrogen- like atom the same transition gives rise to a spectral line of frequency 9f. The hydrogen- like atom has atomic number

(a) 1

(b) 2

(c) 3

(d) 6

(e) 9

The energy (En) of the electron in the orbit of quantum number n in a hydrogen-like atom is given by

En = –13.6 z2/n2 where z is the atomic number.

The energy difference between two states of the hydrogen-like atom is therefore z2 times the energy difference in the case of the hydrogen atom. The frequency of the resulting spectral line also is z2 times the frequency in the case of the hydrogen atom. Since the frequency is 9 times, the atomic number of the hydrogen like atom is 3.

[The hydrogen-like atom in this question is doubly ionised lithium (Li++)].

You will find many useful questions in this section here as well as here.

Saturday, April 11, 2009

All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 (AIPMT 2009) Questions from Thermodynamics

The following questions were included from thermodynamics in the All India Pre-Medical/Pre-Dental 2009 Entrance Examination (Preliminary):

1. The internal energy change in a system that has absorbed 2 Kcals of heat and done 500 J of work is

(1) 6400 J

(2) 5400 J

(3) 7900 J

(4) 8900 J

We have Q = U + W where Q is the heat supplied to the system, U is the increase in internal energy of the system and W is the work done by the system.

Therefore, the increase in internal energy of the system is given by

U = QW = 2000×4.2 J – 500 J = (8400 – 500) J = 7900 J

[Note that 1 calorie = 4.2 joule, nearly]

2. In thermodynamic process which of the following statements is not true?

(1) In an isochoric process pressure remains constant.

(2) In an isothermal process temperature remains constant.

(3) In an adiabatic process PVγ = constant.

(4) In an adiabatic process the system is insulated from the surroundings.

Statement (1) is not true since an isochoric process is one in which volume remains constant. So option (1) is the answer.

You will find many useful questions on thermodynamics at physicsplus here and at other locations.

You may click on the label ‘thermodynamics’ below relevant posts to access all posts related to thermodynamics. Or, you may try a search using the ‘search blog’ box at the top of the site.