The wavelength of the light emitted when the electron returns to the ground state in the H atom, from n = 5 to n = 1, would be: (The ground-state electron energy is –2.18 × 10-18 j)
1. Wavelength = 9.498 x 10-8 km
2. Wavelength = 12.498 x 10-8 m
3. Wavelength = 9.498 x 10-8 m
4. Wavelength = 9.498 x 10-8 cm
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The possible values of n, l, and m for the electron present in 3d would be respectively:
1. n = 3, l = 1, m = – 2, – 1, 3, 1, 2
2. n = 3, l = 3, m = – 2, – 1, 0, 1, 2
3. n = 3, l = 2, m = – 2, – 1, 0, 1, 2
4. n = 5, l = 2, m = – 2, – 1, 0, 1, 2
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The number of electrons in the species , are respectively:
1. 15, 2, 1
2. 15, 1, 2
3. 1, 2, 15
4. 2, 1, 15
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Orbital that does not exist:
1. | 6p | 2. | 2s |
3. | 3f | 4. | 2p |
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The set of quantum numbers which represent 3p is
1. n =1, l =0;
2, n = 3; l=1
3. n = 4; l =2;
4. n = 4; l = 3
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Which of the following sets of quantum numbers is possible-
1. n = 0, l = 0, ml = 0, ms = + ½
2. n = 1, l = 0, ml = 0, ms = – ½
3. n = 1, l = 1, ml = 0, ms = + ½
4. n = 3, l = 3, ml = –3, ms = + ½
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The total number of electrons in an atom with the following quantum numbers would be
(a) n = 4, ms = – ½ (b) n = 3, l = 0
1. 16, 2
2. 11, 8
3. 16, 8
4. 12, 7
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The transition in the hydrogen spectrum that would have the same wavelength as Balmer transition from n = 4 to n = 2 of He+ spectrum is -
1.
2. = 3 to n1 = 2
3. = 3 to n1 = 1
4. = 2 to n1 = 1
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2 ×108 atoms of carbon are arranged side by side. The radius of a carbon atom if the length of this arrangement is 2.4 cm would be
1. 7.0 x 10-11 m
2. 5.0 x 10-11 m
3. 8.0 x 10-11 m
4. 6.0 x 10-11 m
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The diameter of a zinc atom is 2.6 Å. If zinc atoms are arranged side by side lengthwise, number of atoms present in a length of 1.6 cm would be:
1. | 5.153 x 107 | 2. | 6.153 x 107 |
3. | 4.153 x 109 | 4. | 6.153 x 103 |
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