Orbital that does not exist:

1. 6p 2. 2s
3. 3f 4. 2p

Subtopic:  Shell & Subshell |
 89%
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The set of quantum numbers which represent 3p is
1. n =1, =0;
2, n
= 3; l=1
3. n
= 4; l =2;
4. n 
= 4; = 3 

Subtopic:  Shell & Subshell |
 94%
From NCERT
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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

Subtopic:  Quantum Numbers & Schrodinger Wave Equation |
 84%
From NCERT
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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

Subtopic:  Quantum Numbers & Schrodinger Wave Equation |
 81%
From NCERT
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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. n1 = 3 to n2 = 4

2. n2  = 3 to n1 = 2

3. n2  = 3 to n1 = 1

4. n2  = 2 to n1 = 1

Subtopic:  Hydrogen Spectra |
 64%
From NCERT
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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

Subtopic:  Introduction of Atomic Structure |
 71%
From NCERT
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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
Subtopic:  Introduction of Atomic Structure |
 77%
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A certain particle carries 2.5 × 10–16 C of static electric charge. The number of electrons present in it would be:

1. 1460  2. 1350
3. 1560 4. 1660

Subtopic:  Introduction of Atomic Structure |
 75%
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In Rutherford's experiment, generally, the thin foil of heavy atoms like gold, platinum, etc. have been used to be bombarded by the α-particles.

 If the thin foil of light atoms like aluminum etc. is used in Rutherford’s experiment, the difference that would be observed from the above results is :

1. The same results will be observed.

2. More deflection would be observed.

3. There will not be enough deflection.

4. None of the above.

Subtopic:  Introduction of Atomic Structure |
 61%
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Symbols Br3579 and B79r can be written,
whereas symbols Br7935 and B35r are not acceptable because:

1. The general convention for representing an element along with its atomic number (A) and atomic mass (Z) is XZA
2. The general convention for representing an element along with its atomic mass (A) and atomic number (Z) is XZA
3. The general convention for representing an element along with its wavelength (A) and frequency (Z) is XZA
4. The general convention for representing an element along with its isotopes (A) and atomic number (Z) is XZA
Subtopic:  Introduction of Atomic Structure |
 81%
From NCERT
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