IIT-JEE 2008: Linked Comprehension Type Multiple Choice Questions on Bohr Model of Hydrogen-like Atoms

The IIT-JEE Physics question paper is no more a nightmare to students aspiring for a seat in one of the IIT’s, consequent on the changes effected from 2007. If you have good grasp of fundamentals, you should definitely take up the challenge for brightening your future. Students appearing for other entrance examinations such as Physics GRE, AP Physics, AIEEE, KEAM and the like also will find the IIT-JEE question paper very useful for their preparation.

Here are three multiple choice questions on Bohr atom model which appeared under Linked Comprehension Type in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

In a mixture of H – He⁺ gas (He⁺ is singly ionized He atom) H atoms and He⁺ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He⁺ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

(1) The quantum number n of the state finally populated in He⁺ ions is

(A) 2

(B) 3

(C) 4

(D) 5

The singly ionised helium atom is hydrogen like since it contains a single electron. In a hydrogen like atom the energy (E) of the electron in the n^th orbit is given by

E = – 13.6 Z²/n² electron volt where Z is the atomic number.

For hydrogen atom the energies are – 13.6 eV, – 3.4 eV, – 1.51 eV and – 0.85 eV for values of n equal to 1, 2, 3 and 4 respectively

For the helium ion the energies are four fold, equal to – 54.4 eV, – 13.6 eV, – 6.04 eV and – 3.4 eV for values of n equal to 1, 2, 3 and 4 respectively

The energy difference between the electrons in first excited states of hydrogen atom and helium ion is (– 3.4 eV) – (– 13.6 eV) = 10.2 eV.

On collision, the helium ion absorbs this energy and occupies the 3^rd excited state (n = 4)

of energy – 13.6 eV + 10.2 eV = –3.4 eV.

Therefore the correct option is (C).

(2) The wave length of light emitted in the visible region by He⁺ ions after collision with H atoms is

(A) 6.5×10^–7 m

(B) 5.6×10^–7 m

(C) 4.8×10^–7 m

(D) 4.0×10^–7 m

The visible photons have wave lengths in the range 4000 Ǻ to 7000 Ǻ and their energies lie in the range 3.1 eV to 1.771 eV.

[It will be useful to remember the range of energies in eV. You will definitely remember the wave length range of visible photons which you can convert into energy range using the equation, λ E = 12400  where λ is the wave length in Angstrom and E is the energy in electron volt].

The energy difference between the He⁺ ion levels for n = 3 and n = 4 correspond to visible photon energy: – 3.4 eV– (– 6.04 eV) =  2.64 eV

The wave length of the light emitted in the visible region is therefore 124000/2.64 = 4800 Ǻ, nearly. The answer is thus 4.8×10^–7 m [Option (C)].

(3) The ratio of the kinetic energy of the n = 2 electron for the H atom to that of the He⁺ ion is

(A) ¼

(B) ½

(C) 1

(D) 2

The kinetic energy of the electron in the n^th orbit is 13.6 Z²/n².

[Note that the kinetic energy and the total energy of the electron are equal in amount. But the kinetic energy is positive where as the total energy is negative].

The required ratio is Z_H²/Z_He² = ¼.

All posts on Bohr model of hydrogen atom on this site can be accessed by clicking on the label ‘Bohr model’ below this post.

You can find a few posts on hydrogen atom at apphysicsresources.blogspot.com also. One such post is here.

Kerala Engineering Entrance 2008 Questions on Heat & Thermodynamics

The following four questions on heat and thermodynamics appeared in KEAM (Engineering) 2008 question paper:

(1) Three rods made of the same material and having the same cross section have been joined as shown in the figure. Each rod is of same length. The temperature at the junction of the three rods is

(a) 45º C

(b) 90º C

(c) 30º C

(d) 20º C

(e) 60º C

The quantity of heat flowing through the two hotter rods has to flow through the colder rod so that we have

2×KA(90º – T)/L = KA(T– 0)/L where K, A and L are respectively the thermal conductivity, area of cross section and the length of each rod and T is the temperature of the junction in degree Celsius.

This gives 180º – 2T = T from which T = 60º

(2) The PV diagram of a gas undergoing a cyclic process (ABCDA) is shown in the graph where P is in units of Nm^–2 and V is in cm^–3. Identify the incorrect statement

(a) 0.4 J of work is done by the gas from A to B

(b) 0.2 J of work is done by the gas from C to D

(c) No work is done by the gas from B to C

(d) Net work is done by the gas in one cycle is 0.2 J

(e) Work is done by the gas in going from B to C and on the gas fromD to A

BC and DA represent isochoric processes (in which volume does not change) and hence no work is done. So, the incorrect statement is (e).

[Since Kerala entrance questions are single answer type there is no need of checking the other options when you are actually facing the examination. But it will be usful to know the correctness of the other options at the present moment.

Work (PdV) which is equal to 2×10⁵×2×10^–6 = 0.4 J is done by the gas from A to B since the gas expands.

Work (PdV) which is equal to 1×10⁵×2×10^–6 = 0.2 J is done on the gas from C to D since the gas gets compressed.

Since the cyclic process is clockwise, a net amount of work is done by the gas and its value is equal to the area of the closed curve which is 1×10⁵×2×10^–6 = 0.2 J].

(3) The plots of intensity of radiation versus wave length of three black bodies at temperatures T₁, T₂ and T₃ are shown. Then

(a) T₃> T₂> T₁

(b) T₁> T₂> T₃

(c) T₂> T₃> T₁

(d) T₁> T₃> T₂

(e) T₃> T₁> T₂

According to Wien’s law the product λ_mT is a constant (equal to 0.29 cmK) in the case of any black body (where λ_m and T are respectively the wave length of emitted radiation of maximum intensity and T is the temperature of the black body). Obviously, option (d) is correct.

(4) A bubble of 8 moles of helium is submeged at a certain depth in water. The temperature of water increases by 30º C. How much heat is added approximately to helium during expansion?

(a) 4000 J

(b) 3000 J

(c) 3500 J

(d) 4500 J

(e) 5000 J

The heat δQ added to helium is given by

δQ = mC_PδT wher m is the mass of helium, C_P is its specific heat at constant pressure and δT is its rise in temperature.

[You have to use C_P (and not the specific heat at constant volume C_V) since the bubble absorbs the heat at constant pressure. Further, since we are given the amount of the gas in moles, the molar specific heat is to be used in the above relation].

Therefore, δQ = 8×(5/2)R×30, noting that helium is mono atomic so that its C_P = (5/2)R where R is universal gas constant which is approximately equal to 8.3 J mol^–1K^–1

Thus δQ = 8×(5/2) ×8.3×30 = 5000 J, nearly.

You will find a useful post on the equations to be remembered in thermodynamics here

Kerala Engineering Entrance 2008 Questions on Rotational Motion

Here are the two questions on rotational motion which appeared in KEAM (Engineering) 2008 question paper:

(1) A thin circular ring of mass M and radius R rotates about an axis through the centre and perpendicular to its plane, with a constant angular velocity ω. Four small spheres each of mass m (negligible radius) are kept gently at the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

(a) 4ω

(b) Mω/4m

(c) [(M+4m)/M]ω

(d) [M/(M–4m)]ω

(e) [M/(M+4m)]ω

The angular miomentum of the system is conserved since there is no external torque. Therefore we have I₁ω₁ = I₂ω₂ where I₁ and I₂ are the initial and final moments of inertia and ω₁ and ω₂ are the initial and final angular velocities of the system.

We have I₁= MR², ω₁ = ω and I₂ = MR²+4mR².

Therefore, MR²ω = (M+4m)R²ω₂ from which ω₂ = [M/(M+4m)]ω

[Note that the four masses can be placed anywhere on the ring and you will get the same answer as above].

(2) The angular velocity of a wheel increases from 100 rps to 300 rps in 10 seconds. The number of revolutions made during that time is

(a) 600

(b) 1500

(c) 1000

(d) 3000

(e) 2000

The angular acceleration (α) of the wheel is given by

α = Change of angular velocity/Time = (300–100)/10 = 20 revolutions per second².

[We have retained the angular velocity in rps itself since the number of revolutions is to be found out]

Substituting the known values in the equation, ω² = ω₀² + 2αθ (which is similar to v² = v₀² +2as in linear motion), we have

300² = 100² + 2×20×θ

[The angular displacement θ will be in revolutions for obvious reasons]

From the above equation, θ = 2000 revolutions.

You can find all posts on rotational motion on this site by clicking on the label ‘rotation’ below this post.

You will find many useful multiple choice questions (with soluton) involving rotational motion at apphysicsresources here as well as here

Kerala Medical Entrance 2008 Questions on Rotational Motion

Here are the two questions on angular motion which appeared in the Kerala Medical Entrance 2007 question paper:

(1) When a ceiling fan is switched off, its angular velocity reduces to half its initial value after it completes 36 rotations. The number of rotations it will make further before coming to rest is (Assume angular retardation to be uniform)

(a) 10

(b) 20

(c) 18

(d) 12

(e) 16

You have to use the equation, ω² = ω₀² + 2αθ for finding the angular acceleration α and hence the number of further rotations. Note that this equation is the rotational analogue of the equation v² = v₀² + 2as (or, v² = u² + 2as) in linear motion.

Since the angular velocity has reduce to half of the initial value ω₀ after 36 rotations, we have

(ω₀ /2) ² = ω₀² + 2α×36 from which α = – ω₀²/96

[We have expressed the angular displacement θ in rotations itself for convenience]

If the additional number of rotations is x, we have

0 = (ω₀ /2) ² + 2αx = (ω₀ /2) ² + 2×(– ω₀²/96)x

This gives

x = 12

(2) Two particles starting from a point on a circle of radius 4 m in horizontal plane move along the circle with constant speeds of 4 ms^–1 and 6 ms^–1 respectively in opposite directions. The particles will collide with each other after a time of

(a) 3 s

(b) 2.5 s

(c) 2.0 s

(d) 1.5 s

(e) 3.5 s

This is a very simple question. The sum of the distances to be traveled by the two particles is 2πR = 2π×4 = 8π.

The relative speed is 10 ms^–1 so that the time required for the particles to collide is 8π/10 = 2.5 s.

You can find all posts on rotational motion on this site by clicking on the label ‘rotation’ below this post.