# A photon of wavelength 4 × 10–7 m strikes a metal surface, the work function of the metal being 2.13 eV.  The kinetic energy of emission would be: 1. 0.97 eV 2. 97 eV  3. 4.97 × ${10}^{-19}$ eV 4.  5.84 × 105 eV

Subtopic:  Photo Electric Effect |
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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

Subtopic:  Planck's Theory |
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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

Subtopic:  Shell & Subshell |
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The number of electrons in the species ${\mathrm{H}}_{2}^{+}$ $,$ ${\mathrm{H}}_{2}$ $,$ ${\mathrm{O}}_{2}^{+}$,  are respectively:

1. 15, 2, 1

2. 15, 1, 2

3. 1, 2, 15

4. 2, 1, 15

Subtopic:  Number of Electron, Proton & Neutron |
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Orbital that does not exist:

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

Subtopic:  Shell & Subshell |
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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 |
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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 |
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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 |
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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. ${\mathrm{n}}_{1}$ $=$ $3$ $\mathrm{to}$ ${\mathrm{n}}_{2}$ $=$ $4$

2. ${\mathrm{n}}_{2}$  = 3 to n1 = 2

3. ${\mathrm{n}}_{2}$  = 3 to n1 = 1

4. ${\mathrm{n}}_{2}$  = 2 to n1 = 1

Subtopic:  Hydrogen Spectra |
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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 |
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