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 × ${10}^8<mspace\; width="1em"></mspace>{\mathrm{s}}^{-1}$,  E = 3.282 × ${10}^{-10}$ J

2) ν = 1.0 × ${10}^8<mspace\; width="1em"></mspace>{\mathrm{s}}^{-1}$,  E = 16.282 ×10^-22 J

3) ν = 5.088 × 10¹⁴${\mathrm{s}}^{-1}$,  E = 3.312 ×10^-22 J

4) ν = 7.0 × 10¹⁴${\mathrm{s}}^{-1}$,  E = 13.282 × ${10}^{-10}$ J

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

Following results are observed when sodium metal is irradiated with different wavelengths. The following value of threshold wavelength would be  -

1)  ${\mathrm{\lambda }}_0$ = 740 nm

2)  ${\mathrm{\lambda }}_0$ = 540 nm

3)  ${\mathrm{\lambda }}_0$ = 640 nm

4)  ${\mathrm{\lambda }}_0$ = 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

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

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

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

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