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 would be:
A certain particle carries 2.5 × 10–16 C of static electric charge. The number of electrons present in it would be
1460
1350
1560
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 :
The same results will be observed
More deflection would be observed
There will not be enough deflection
None of the above
Symbols Br3579 and B79r can be written,whereas symbols Br7935 and B35r are not acceptable because
The correct arrangement of the following electromagnetic spectrum in the increasing order of frequency is
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 -
ν = 9.5 × 108 s-1, E = 3.282 × 10-10 J
ν = 1.0 × 108 s-1, E = 16.282 ×10-22 J
ν = 5.088 × 1014s-1, E = 3.312 ×10-22 J
ν = 7.0 × 1014s-1, E = 13.282 × 10-10 J
The work function for the Cesium atom is 1.9 eV. The threshold frequency of the radiation is
4.59 × 1014 s-1
8.59 × 1014 s-1
5.59 × 10-14 s-1
65.9 × 1014 s-1
Following results are observed when sodium metal is irradiated with different wavelengths. The following value of threshold wavelength would be -
1) λ0 = 740 nm
2) λ0 = 540 nm
3) λ0 = 640 nm
4) λ0 = 840 nm
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 -
22.27 × 10–16 J
32.22 × 10–16 J
12.22 × 10–16 J
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 -
The transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. wavenumber for the transition is 800 nm
The transition is from the 4th orbit to the 2nd orbit. It belongs to the Balmer series. wavenumber for the transition is 634 nm
The transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. wavenumber for the transition is 434 nm
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
5.637 × 10–23 kg m s–1
4.637 × 10–23 kg m s–1
2.637 × 10–23 kg m s–1
3.637 × 10–23 kg m s–1
O2 undergoes photochemical dissociation into one normal oxygen atom and one excited oxygen atom,1.976 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen requires 498 kJ mol-1. What is the maximum wavelength effective for photochemical dissociation of O2 ?
1440.92 Ao
2 . 1740.52 Ao
1900.72 Ao
1248.12 Ao
Iodine molecule dissociates into atoms after absorbing light of 4500 Å. If one quantum of radiation is absorbed by each molecule , calculate the kinetic energy of iodine atoms. (Bond energy of I2 = 240 kJ mol-1)
2.16 \(\times\) 10-21 J
4.16 \(\times\) 10-20 J
7.16 \(\times\) 10-23 J
2.16 \(\times\) 10-20 J
Calculate the shortest and longest wavelength in H spectrum of Lyman series (RH = 109678 cm-1) ?
911.7 Ao and 1215.7 Ao
511.7 Ao and 1215.7 Ao
911.7 Ao and 1415.7 Ao
1215.7 Ao and 911.7 Ao
Calculate the speed and de Broglie wavelength of an electron that has been accelerated by a potential difference of 500 V.
2.5 x 10-11 m
3.5 x 10-11 m
5.5 x 10-11 m
5.5 x 10-12 m
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
5(3p) < 2(3d) < 6(4p) < 1 (4d) = 4(3d) < 3(4p)
5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d)
6(4p) < 1 (4d) < 5(3p) < 2(3d) < 4(3d) < 3(4p)
5(3p) < 2(3d) < 4(3d) < 3(4p) < 6(4p) < 1 (4d)
The electron of a Br atom that experiences the lowest effective nuclear charge is
2p and 3p
4p
2p
3p
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.
2s, 4d, 3p
3s, 4f, 3d
3s, 4f, 3p
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
Electrons in the 3p orbital of silicon
Electrons in the 5d orbital of aluminium
Electrons in the 3p orbital of aluminium
Electrons in the 5p orbital of silicon
Among the given options, the element having the highest number of unpaired electrons in the ground state is
P
Fe
Kr
Cr
The number of electrons that will be present in the subshells having ms value of –1/2 for n = 4 is:
36
4
16
2
The atomic number of elements whose outermost electronic configuration are represented by 3s1 and 2p5 respectively, is:
9, 7
9, 11
11, 5
11, 9
The circumference of the Bohr orbit for the H atom is related to the de Broglie wavelength associated with the electron revolving around the orbit by which of the following relation?
1. 2πr=nλ2. πr = 2nλ3. mvr = nλ4. vr = 2nλ
Which among the following are isoelectronic species?
Na+, K+, Ca2+, Mg2+
Ca2+, Mg2+, S2-, K+
Ca2+, Ar, S2-, K+
The neon gas emits radiation of 616 nm.The number of quanta that are present in 2 J of energy is
1. 6.2 × 10-182. 5.6 × 10173. 6.2 × 10184. 32.2 × 10-20
The ultraviolet region in the above figure is indicated by -
D
B
A
C
The graph that represents the variation of ψ2 (probability of density of finding the electron) with distance (r) is -
The number of nodes as per the graph is-
zero
1
Cannot predict
A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light, c=3.0×108 ms-1]
2192 m
21.92 cm
219.3 m
219.2 m
When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 ×105 J mol–1. The minimum energy needed to remove an electron from sodium and the maximum wavelength that will cause a photoelectron to be emitted are respectively-
2.31 × 105 J mol–1, 517 nm
23.1 × 105 J mol–1, 517 nm
3.31 × 105 J mol–1, 417 nm
33.1 × 105 J mol–1, 417 nm
The frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom are respectively-
\(6.91\times10 ^{14} Hz\); 543 nm
\(7.91\times10 ^{14} Hz\); 434 nm
\(6.91\times10 ^{14} Hz\); 434
\(7.91\times10 ^{14} Hz\); 343 nm
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