The work function of the Cesium atom is \(1.9~\mathrm{eV}.\) What is the threshold frequency of radiation for this atom?

1. \(4.59 \times10^{14}~s^{-1}\) 2. \(8.59 \times10^{14}~s^{-1}\)
3. \(5.59 \times10^{-14}~s^{-1}\) 4. \(65.9 \times10^{14}~s^{-1}\)

Subtopic:  Photo Electric Effect |
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Following results are observed when sodium metal is irradiated with different wavelengths. The following value of threshold wavelength would be  -

λ (nm) 500 450 400
v × 10–5 (cm s–1) 2.55 4.35 5.35

1)  λ0 = 740 nm

2)  λ0 = 540 nm

3)  λ0 = 640 nm

4)  λ0 = 840 nm

Subtopic:  Photo Electric Effect |
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When a photon with a wavelength of 150 pm strikes an atom, one of its inner bound electrons is ejected at a velocity of 1.5 × 107 m s–1 The energy with which it is bound to the nucleus would be:

1. 32.22 × 10–16 J

2. 12.22 × 10–16 J

3. 22.27 × 10–16 J

4. 31.22 × 10–16 J
 

Subtopic:  Photo Electric Effect |
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The wavelength for the emission transition, the series to which this transition belongs and the region of the spectrum if it starts from the orbit having a radius of 1.3225 nm and ends at 211.6 pm would be respectively -

1) The transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. wavenumber for the transition is 800 nm

2) The transition is from the 4th orbit to the 2nd orbit. It belongs to the Balmer series. wavenumber for the transition is 634 nm

3) The transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. wavenumber for the transition is 434 nm

4) The transition is from the 6th orbit to the 2nd orbit. It belongs to the Balmer series. wavenumber for the transition is 534 nm

Subtopic:  Hydrogen Spectra |
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If the position of the electron were measured with an accuracy of +0.002 nm, the uncertainty in the momentum of the electron would be: 

1. 5.637 × 10–23 kg m s–1

2. 4.637 × 10–23 kg m s–1

3. 2.637 × 10–23 kg m s–1

4. 3.637 × 10–23 kg m s–1

Subtopic:  Heisenberg Uncertainty Principle |
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The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. 

1. n = 4, l = 2, ml = –2 , ms = –1/2

2. n = 3, l = 2, ml = 1 , ms = +1/2

3. n = 4, l = 1, ml = 0 , ms = +1/2

4. n = 3, l = 2, ml = –2 , ms = –1/2

5. n = 3, l = 1, ml = –1 , ms = +1/2

6. n = 4, l = 1, ml = 0 , ms = +1/2

1) 5(3p) < 2(3d) < 6(4p) < 1 (4d) = 4(3d) < 3(4p)

2) 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d)

3) 6(4p) < 1 (4d) < 5(3p) < 2(3d) < 4(3d) < 3(4p)

4) 5(3p) < 2(3d) < 4(3d) < 3(4p) < 6(4p) < 1 (4d)

Subtopic:  AUFBAU Principle |
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Which electron in a bromine (Br) atom experiences the lowest effective nuclear charge?
1. 2p and 3p
2. 4p
3. 2p
4. 3p
Subtopic:  Number of Electron, Proton & Neutron |
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Among the following pairs of orbitals the orbital that will experience the larger effective nuclear charge respectively is -

(i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.

1. 2s, 4d, 3p

2. 3s, 4f, 3d

3. 3s, 4f, 3p

4. 2s, 4d, 3d

Subtopic:  Pauli's Exclusion Principle & Hund's Rule |
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The unpaired electrons in Al and Si are present in the 3p orbital. The electrons that will experience a more effective nuclear charge from the nucleus are :

1. Electrons in the 3p orbital of silicon

2. Electrons in the 5d orbital of aluminium

3. Electrons in the 3p orbital of aluminium

4. Electrons in the 5p orbital of silicon

Subtopic:  Number of Electron, Proton & Neutron |
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Identify the element with the highest number of unpaired electrons in its ground state from the given options :

1. P

2. Fe

3. Kr

4. Cr
Subtopic:  Pauli's Exclusion Principle & Hund's Rule |
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