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1

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $$\mathop A\limits^o $$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

A
1215 $$\mathop A\limits^o $$
B
1640 $$\mathop A\limits^o $$
C
2430 $$\mathop A\limits^o $$
D
4687 $$\mathop A\limits^o $$

Explanation

$${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$$

For singly-ionized helium atom, Z = 2. For hydrogen atom Z = 1.

For Balmer series n1 = 2.

For first spectral line of hydrogen :

$${1 \over {6561}} = {R_H} \times {(1)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] = {{5R} \over {36}}$$

$$ \Rightarrow {R_H} = {{36} \over {5 \times 6561}}$$

For second spectral line of helium,

$${1 \over \lambda } = {R_H} \times {(2)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right] = {{3{R_H}} \over 4}$$

$$ = {3 \over 4} \times {{36} \over {5 \times 6561}}$$

$$ \Rightarrow \lambda = 1215\,\mathop A\limits^o $$

2

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.

In a CO molecule, the distance between C (mass = 12 amu) and O (mass = 16 amu), where 1 amu $$ = {5 \over 3} \times {10^{ - 27}}$$ kg, is close to

A
2.4 $$\times$$ 10$$-$$10 m
B
1.9 $$\times$$ 10$$-$$10 m
C
1.3 $$\times$$ 10$$-$$10 m
D
4.4 $$\times$$ 10$$-$$11 m

Explanation

The moment of inertia of CO molecule is

I = $$\mu$$r2 ....... (i)

where, $$\mu$$ = reduced mass of the CO molecule, r = distance between C and O or bond length

The reduced mass $$\mu$$ of the CO molecule is

$$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = \left[ {{{(12)(16)} \over {12 + 16}}} \right] \times {5 \over 3} \times {10^{ - 27}}$$ kg

But $$I = 1.87 \times {10^{ - 46}}$$ kg m2 (From the above question)

From equation (i), we get

$${r^2} = {I \over \mu }$$

Substituting the values of I and $$\mu$$ in above equation, we get

$${r^2} = {{1.87 \times {{10}^{ - 46}}} \over {\left[ {{{12 \times 16} \over {28}} \times {5 \over 3} \times {{10}^{ - 27}}} \right]}}$$

or $${r^2} = {{1.87 \times {{10}^{ - 46}} \times 28 \times 3} \over {12 \times 16 \times 5 \times {{10}^{ - 27}}}}$$ or $$r = 1.3 \times {10^{ - 10}}$$ m

3

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.

It is found that the excitation frequency from ground to the first excited stat of rotation for the CO molecule is close to $${4 \over \pi } \times {10^{11}}$$ Hz. Then, the moment of inertia of CO molecule about its centre of mass is close to (Take h = 2$$\pi$$ $$\times$$ 10$$-$$34 J-s)

A
2.76 $$\times$$ 10$$-$$46 kg m2
B
1.87 $$\times$$ 10$$-$$46 kg m2
C
4.67 $$\times$$ 10$$-$$47 kg m2
D
1.17 $$\times$$ 10$$-$$47 kg m2

Explanation

The energy of photon is equal to the energy difference between the ground level and first excited level.

$$hv = {E_2} - {E_1}$$

$$hv = {{(4 - 1){h^2}} \over {8{\pi ^2}I}} \Rightarrow I = {{3h} \over {8{\pi ^2}v}}$$

$$I = {{3 \times 2\pi \times {{10}^{ - 34}}} \over {[(8{\pi ^2}4)/\pi ] \times {{10}^{11}}}} = {3 \over {16}} \times {10^{ - 45}}$$ kg-m2 = 1.87 $$\times$$ 10$$-$$46 kg-m2

4

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.

A diatomic molecule has moment of inertia I. By Bohr's quantization condition, its rotational energy in the nth level (n = 0 is not allowed) is

A
$${1 \over {{n^2}}}\left( {{{{h^2}} \over {8{\pi ^2}I}}} \right)$$
B
$${1 \over n}\left( {{{{h^2}} \over {8{\pi ^2}I}}} \right)$$
C
$$n\left( {{{{h^2}} \over {8{\pi ^2}I}}} \right)$$
D
$${n^2}\left( {{{{h^2}} \over {8{\pi ^2}I}}} \right)$$

Explanation

Bohr's quantization rule states that angular momentum L is an integral multiple of $$h/(2\pi )$$ i.e.,

$$L = nh/(2\pi )$$.

The angular momentum of a system rotating with angular velocity $$\omega$$ and having moment of inertia I is

$$L = I\omega $$.

and its rotational kinetic energy is

$${E_n} = {1 \over 2}I{\omega ^2} = {1 \over 2}I{\left( {{L \over I}} \right)^2} = {{{n^2}{h^2}} \over {8{\pi ^2}I}}$$.

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