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 3^rd 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 3^rd 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 2^nd 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 2^nd 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 n_x 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 (E_n) of the electron in the orbit of quantum number n in a hydrogen-like atom is given by

E_n = –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.

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 = ∆Q – ∆W = 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 PV^γ = 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.

Electronics- Multiple Choice Questions involving Diodes

I am enough of an artist to draw freely upon my imagination. Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world.

– Albert Einstein

Here are three questions involving diodes:

(1) In the circuit shown in figure the inputs at A and B are respectively 7 V and 0.8 V. If the potential drops across the forward biased diodes are negligible and the breakdown voltage of the zener diode is 5V, the output voltage is

(a) 0.8 V

(b) 7 V

(c) 3.9 V

(d) 0 V

(e) 5 V

The diode connected to input A will be forward biased and the diode connected to input B will be reverse biased. The input of 7 V will be impressed at the junction between the diodes and the resistor. The zener diode will breakdown and the output voltage will be the breakdown voltage of the zener diode, which is equal to 5 V. (Two volts will be dropped across the resistor).

(2) If the logical 0 is 0 volt and logical 1 is + 5 volt, the circuit shown in the above question can function as

(a) AND gate

(b) NOT gate

(c) EXOR gate

(d) OR gate

(e) none of the above

If both inputs are 0 volt the output will be zero. If at least one input is 5V or all inputs are 5 V, the output will be 5 V and the circuit can function as an OR gate.

(3) The current through the 40 Ω resistor in the circuit shown in the adjoining figure is

(a) 50 mA

(b) 80 mA

(c) 40 mA

(d) 92.3 mA

(e) 44.4 mA

The diode D₂ is reverse biased and hence no current flows through its branch. The branches containing diodes D₁ and D₃ are in parallel and the diodes are forward biased. The effective resistance of these two branches in parallel is (20×20)/(20+20) = 10 Ω. Therefore, the total resistance in series with the 4 volts battery is 40 Ω +10 Ω = 50 Ω.

Therefore,the current through the 40 Ω resistor is 4 V/50 Ω = 0.08 A = 80 mA.

You will find some useful multiple choice questions (with solution) in electronics here as well as here.

Dates of Kerala Govt. Entrance Examinations (KEAM 2009) for the Engineering and Medical Streams Changed

The Commissioner for Entrance Examinations, Govt. of Kerala, has notified that the dates of the Entrance Examinations conducted by Govt. of Kerala for Admission to Medical/ Agriculture/ Veterinary/ Engineering Degree Courses 2009 (KEAM 2009), have been changed. This is the second time the dates get changed. Here is the renewed time table:

Dates of Exam:

Engineering Entrance Examination:

25.05.2009 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

26.05.2009 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

27.05.2009 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

28.05.2009 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

You will find complete details at the site http://www.cee-kerala.org/

The details regarding the National Aptitude Test in Architecture (NATA) required for those who apply for Architecture also can be seen at the site.

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions on Nuclear Physics

Here are two multiple choice questions which appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008:

(1) Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively. If initially they have the same number of nuclei then the ratio of the number of nuclei of X₁ to that of X₂ will be 1/e after a time

(1) λ/2

(2) 1/(4λ)

(3) e/λ

(4) λ

If the initial number of nuclei is N₀ we have

N₁ = N₀e^{–5λ t} and

N₂ = N₀e^{–λ t} where N₁ and N₂ are the number of nuclei of X₁ and X₁₂ at time t.

Therefore N₁/N₂ = e^–4λt

This will be equal to 1/e when t =1/(4λ).

(2) Two nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be

(1) 3:1

(2) (3)^1/3:1

(3) 1:1

(4) 1:3

The mass of a nucleus is directly proportional to the number (A) of the nucleons. The volume of the nucleus is (4/3)πR³ where R is the nuclear radius. But, R = R₀A^⅓ where R₀ is a constant (equal to 1.1 ×10^-15m). So, the volume of the nucleus also is directly proportional to the nucleon number A. Since the density is the ratio of mass to volume, it follows that the density of nuclear matter is independent of the nucleon number A so that the correct option is (c).

You will find some useful multiple choice questions (with solution) on nuclear physics at physicsplus