# 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 |
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Symbols ${}_{35}{}^{79}\mathrm{Br}$ and ${}^{79}Br$ can be written,
whereas symbols ${}_{79}{}^{35}\mathrm{Br}$ and ${}^{35}Br$ are not acceptable because
1. The general convention for representing an element along with its atomic number (A) and atomic mass (Z) is ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$
2. The general convention for representing an element along with its atomic mass (A) and atomic number (Z) is ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$

3. The general convention for representing an element along with its wavelength (A) and frequency (Z) is ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$
4. The general convention for representing an element along with its isotopes (A) and atomic number (Z) is ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$

Subtopic:  Introduction of Atomic Structure |
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The correct arrangement of the following electromagnetic spectrum in the increasing order of frequency is
1. Cosmic rays < Amber light < Radiation of FM radio < X-rays < Radiation from microwave ovens
2. Radiation from FM radio  < Radiation from microwave oven < Amber light < X- rays < Cosmic rays
3. Radiation from microwave ovens < Amber light < Radiation of FM radio < X-rays < Cosmic rays
4. Cosmic rays < X-rays < Radiation from microwave ovens < Amber light < Radiation of FM radio

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The work function for the Cesium atom is 1.9 eV. The threshold frequency of the radiation is

1. 4.59 × 1014 ${\mathrm{s}}^{-1}$

2. 8.59 × 1014 ${\mathrm{s}}^{-1}$

3. 5.59 × 10-14 ${\mathrm{s}}^{-1}$

4. 65.9 × 1014 ${\mathrm{s}}^{-1}$

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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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 electron of a Br atom that experiences the lowest effective nuclear charge is

1. 2p and 3p

2. 4p

3. 2p

4. 3p

Subtopic:  Number of Electron, Proton & Neutron |
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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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Among the given options, the element having the highest number of unpaired electrons in the ground state is

1. P

2. Fe

3. Kr

4. Cr

Subtopic:  Pauli's Exclusion Principle & Hund's Rule |
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The number of electrons that will be present in the subshells having ms value of –1/2 for n = 4 is:

1.  36

2.  4

3.  16

4.  2

Subtopic:  Shell & Subshell |
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