IIT-JEE 2011 Paper 1 Offline | Atoms and Nuclei Question 37 | Physics

1

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)

+3

-1

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 $$\mathop A\limits^o $$

2

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

+3

-1

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

$${n^2}\left( {{{{h^2}} \over {8{\pi ^2}I}}} \right)$$

3

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

+3

-1

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 state 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

4

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

+3

-1

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