Q. No.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 × 107
2. 6.153 × 107
3. 4.153 × 109
4. 6.153 × 103
1. 5.153 × 107
2. 6.153 × 107
3. 4.153 × 109
4. 6.153 × 103
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 |
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.
Symbols and can be written,
whereas symbols and are not acceptable because:
| 1. | The general convention for representing an element along with its atomic number (A) and atomic mass (Z) is |
| 2. | The general convention for representing an element along with its atomic mass (A) and atomic number (Z) is |
| 3. | The general convention for representing an element along with its wavelength (A) and frequency (Z) is |
| 4. | The general convention for representing an element along with its isotopes (A) and atomic number (Z) is |
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 |
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. The frequency of second transition and energy difference between two excited states are respectively -
1) ν = 9.5 × , E = 3.282 × J
2) ν = 1.0 × , E = 16.282 ×10-22 J
3) ν = 5.088 × 1014, E = 3.312 ×10-22 J
4) ν = 7.0 × 1014, E = 13.282 × J
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. | \(65.9 \times10^{14}~s^{-1}\) |
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) = 740 nm
2) = 540 nm
3) = 640 nm
4) = 840 nm
1. 32.22 × 10–16 J
2. 12.22 × 10–16 J
3. 22.27 × 10–16 J
4. 31.22 × 10–16 J
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
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