1
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

The $$\beta$$-decay process, discovered in around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e$$-$$) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, that is, n $$\to$$ p + e$$-$$ + $${\overline v _e}$$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($${\overline v _e}$$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 $$\times$$ 106 eV. The kinetic energy carried by the proton is only the recoil energy.

If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?

A
0 $$\le$$ K $$\le$$ 0.8 $$\times$$ 106 eV
B
3.0 eV $$\le$$ K $$\le$$ 0.8 $$\times$$ 106 eV
C
3.0 eV $$\le$$ K < 0.8 $$\times$$ 106 eV
D
0 $$\le$$ K < 0.8 $$\times$$ 106 eV
2
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 $$\mathop A\limits^o $$
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.

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

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
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