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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1851. |
Two electronic transitions were found to take place in a single electron species. One is deactivation of electron from the `5^(th)` shell to the `4^(th)` shell and the other is from the `2^(nd)`nd shell to the `l^(st)` shell. Do the energies emitted due to the above transitions have the same wavelength? justify |
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Answer» (i) calculation of energy emitted due to the above transition (ii) calculation of difference in energy between 5th and 4th and 2nd and 1st shells (iii) relation between the energy and the frequency of the electromagnetic radiation (iv) relation between frequency and the wave- length of the electromagnetic radiation |
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| 1852. |
Energy of an electron in a particular orbit of single electron species of beryllium is the same as the energy of an electron in the ground state of hydrogen atom . Identify the orbit of berylliumA. 1B. 2C. 3D. 4 |
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Answer» Energy of an electron in `Be^(+3)` ion = `(-13.6)/(n^2)xx4^2ev` Energy of the electron in ground state hydrogen atom = -13.6 ev `therefore(-13.6)/(n^2)xx4^2=-13.6` `rArrn^2=42` `n=42` |
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| 1853. |
One electron is made to revolve around a pro ton and it possesses the least possible energy and another electron is made to revolve around an `alpha`-particle with the same energy. Calculate the ratio of the distances of the electrons from the respective species |
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Answer» Revolution of an electron around a proton at the closest distance implies that it is a hydrogen atom in its ground state. Similarly, the other particle is `He^+`. The energy of the electron revolving around the nucleus of `He^+` should be -K. `therefore-k=(-kz^2)/(n^2)` `therefore n^2=4[thereforeZ=2]` `thereforen=2` The ratio of radii of the first orbit of hydrogen to the second orbit of `He^+=(kxx1^2)/((kxx2^2)/2)=1:2` |
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| 1854. |
What are isobars ? Give an example. |
| Answer» Different elements having same mass | |
| 1855. |
What is the lowest value of n that allows g orbitals to exist? |
| Answer» For g subshell `l = 4` The minimum value of n for which l can be `4` is `4 + 1 `or `5` (The value of l can be `4` for `n = 5,6,7, …..)` | |
| 1856. |
The ratio `e//m` i.e. specific , for a cathode rayA. How the smaller value when the discharge, tube is filled with `H_(2)`B. is constantC. Varies with the atomic number of gas in the discharge tubeD. Varies with the atomic number of an element forming the cathod |
| Answer» The ratio is constant | |
| 1857. |
The possible form of ` H_2` molecule on the basis fo three isotopes can be :A. `3`B. `6`C. `8`D. `19` |
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Answer» Correct Answer - B `H_(2), D_(2), T_(2)`, `HE,HT,DT`. |
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| 1858. |
What transition in `He^(o+)` ion shall have the same wave number as the first line in Balmer series of H atom ?A. `3rarr 2`B. `6 rarr4`C. `5rarr3`D. `7rarr5` |
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Answer» Correct Answer - B ` E_(3) - E_(2) ` for ` H= - E_(1H)/(9) + E_(1H)/(4) = (5 E_1H)/(36)` ` [because E_n = E_(1H)/(n^(2))]` ` E_6 - E_4 or He^+ = [ - ( E_(1H) xx 2^2 )/6^2 + (E_(1H) xx 2^(2))/4^2 ]` `:. E_(nHe^+) = (E_(1H)xx Z^2)/n^2 = + (5E_(1H))/( 36)`. |
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| 1859. |
Which of the following are true for cathode ray?A. It travels along a straght line.B. It emits X-rays when strikes a metal.C. It is an electromagnetic wave.D. It is not deflected by magnetic field. |
| Answer» Correct Answer - B | |
| 1860. |
H has two natural isotopes fo `_1 H^_1` and (O) has two isotopes ` O^(16)` and ` O^(18)` Which of the following mol . Wt , o f ` H_(2) O` will not be possible ?A. `19`B. `20`C. `24`D. `22` |
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Answer» Correct Answer - C Possible mol wt may be ` 18, 20, 19, 20, 22, 21`, repectively for ` H^(1) H^(1) O^(16)` , `H^2 H^2 O^(16) ,H^(1) H^(2) O^(16) , H^(1) H^(1) O^(18) , H^(2) H^(2) o^(18) ,H^(1) H^(2) O^(18)`. |
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| 1861. |
Find the energy released (in erg) when `2.0 g` atom of hydrogen undergoes transition giving a spectral line of the lowest energy in the visible region of its atomic spectra |
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Answer» Correct Answer - A::B::C For H atom ,the wspectrum lines in visible region corresponding to Balmer lines of `n_(2) = 2` Now, for lower energy photon, the required transition will be from `3 rarr 2` Using the relation for `Delta E` `Delta E = 2.178 xx 10^(-18) (1)^(2)((1)/(2^(2)) - (1)/(3^(2))) (1)/("atom")` Now the `2.0 g `atom , the energy relation will be `E = (2 xx 6.023 xx 10^(23)) xx 2.18 xx 10^(-18) ((5)/(36)) J` `= 3.63 xx 10^(5) J = 3.63 xx 10^(12) erg` |
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| 1862. |
Three isotopes fo an elecment have mass numbers ` (m, (m + 1)` and `(M + 2)` . If the mean mass number is ( M + 0. 5) then which of the following ratios may be accepted for ` M`, `(M + 1)` and `( M+2)` in tha order ?A. `1: 1:1B. `4:1:1C. `3:2:1`D. `2:1:1 |
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Answer» Correct Answer - B Average mass ` = (M + 0. 5) = ( M xx 4+ (M + 1 ) xx 1 + (M + 2 ) xx 1)/6 = ( 6M +3)/6`. |
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| 1863. |
Three isotopes of an element have mass numbers ` (m), (m + 1)` and `(M + 2)` . If the mean mass number is ( M + 0. 5) then which of the following ratios may be accepted for ` M`, `(M + 1)` and `( M+2)` in the order ?A. `1:1:1`B. `4:1:1`C. `3:2:1`D. `2:1:1` |
| Answer» Correct Answer - B | |
| 1864. |
The wave number of the first line of Balmer series of hydrogen is `15200 cm ^(-1)` The wave number of the first Balmer line of `Li^(2+)` ion isA. `456200 cm^(-1)`B. `136800 cm^(-1)`C. `738720 cm^(-1)`D. `152000 cm^(-1)` |
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Answer» Correct Answer - C |
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| 1865. |
It travelling at same speeds, whichof the following mater waves have the shortest wavelength?A. ElectronB. Alpha particle `(He^(2+))`C. neutronD. Proton |
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Answer» Correct Answer - B `lambda = (h)/(mV)` shortest wave length for highest mass for `alpha`-particle |
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| 1866. |
A H-like species `Be^(3+)` is in a spherically symmetric state `S_(1)` with two radial nodes. Upon absorbing light the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the H-atom. Energy of state `S_(1)` in units of `He^(+)` ground state energy is :A. 0.75B. 1.33C. 0.44D. 2.25 |
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Answer» Correct Answer - c |
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| 1867. |
What will be the atomic number of an atom if its electronic configuration is `(n - 2)s^(2) (n - 1)s^(a) p^(b) ms^(2)p^(2)` where `n = 3, a= 2` and `b= 6` ? |
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Answer» The electronic configuraqtion of nitrogen atom is `(n-2)s^(2)(n-1)s^(a)p^(b)ms^(2)p^(2)`where` n= 3,a= 2` and `b= 6` ON substituting all the velues in the configuration we get `(3 - 2)s^(2) (3 - 1)s^(2)p^(6)3s^(2)p^(2) = 1s^(2)2s^(2)2p^(6)3s^(2)3p^(2)3p^(2) = 14` Thus the atomic number of the atom is `14` |
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| 1868. |
If the wavelength of series limit of Lyman series for `He^(+)` ions is `x overset(0)(A)`, then what will be the wavelength of series limit of Balmer series for `Li^(+2)` ion?A. `(9x)/(4) overset(0)(A)`B. `(16x)/(9) overset(0)(A)`C. `(5x)/(4)overset(0)(A)`D. `(4x)/(9)overset(0)(A)` |
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Answer» Correct Answer - B `bar(v) = (1)/(lambda) =[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]R_(H)z^(2)` |
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| 1869. |
In sample of hydrogen atoms , electron jump from `10^(th)` excited state to ground state .If x are the number of different ultraviolet radiations , y are number of different visible radiational and z are number of differnet imfrared radtions. The Value of `z-(x+y)`is. [Assume all the Baler lines lie with visble region ].A. 17B. 18C. 19D. 36 |
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Answer» Correct Answer - A `n_(1)`=11 higher state `n_(2)`=1 lowest state U.N..=11-1=10 visible =11-2=9 remaining =`(11xx10)/(2)-19=55-19` (z) I.R.=36 Hence `z-(x+y)=36-(9+10)=17` |
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| 1870. |
A H-like species `Be^(3+)` is in a spherically symmetric state `S_(1)` with two radial nodes. Upon absorbing light the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the H-atom. The orbital angular momentum of hte state `S_(2)` is :A. `sqrt(1.5h`B. `sqrt2h`C. 0D. h |
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Answer» Correct Answer - a |
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| 1871. |
Let `nu_(1)` be the frequency of the series limit of the lyman series `nu_(2)` be the frequency of the first line of th lyman series and `nu_(3)` be the frequency of the series limit of the Balmer series. ThenA. `v_(1)-v_(2)=v_(3)`B. `v_(2)-v_(1)=v_(3)`C. `v_(3)=1//2(v_(1)-v_(3))`D. `v_(1)+v_(2)=v_(3)` |
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Answer» Correct Answer - A `E_(oo)-E_(1)=hv_(1)," "impliesE_(1)implieshv_(1)` `E_(2)-E_(1)=hv_(2),` `E_(oo)-E_(2)=hv_(3)," "implies E_(2)implies hv_(3)` `-hv_(3)+hv_(1)=hv_(2)` `v_(2)=v_(1)-v_(3)` `v_(3)=v_(1)-v_(2)` |
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| 1872. |
The wavelength of series limit for lyman series for `He^(o+)` would beA. `911.7 Å`B. `227.9 Å`C. `1215.1 Å`D. `363.8 Å` |
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Answer» `Z= 2 (for HE^(Theta))` series limit for lyman series `n_(1) =1,n_(2) = oo` `bar v = (1)/(lambda) = RZ^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = R(2)^(2)((1)/(1^(2)) - (1)/(oo^(2))) = 4R` `lambda = (1)/(4R)= (1)/(4 xx 109677 cm^(-1)) ` `= 227.9 xx 10^(-10) m = 227.9Å` |
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| 1873. |
Let `nu_(1)` be the frequency of the series limit of the lyman series `nu_(2)` be the frequency of the first line of th lyman series and `nu_(3)` be the frequency of the series limit of the Balmer series. ThenA. `v_(1) - v_(2) = v_(3)`B. `v_(2) - v_(1) = v_(3)`C. `v_(3) = (1)/(2) (v_(2) - v_(3))`D. `v_(1) + v_(2) = v_(3)` |
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Answer» Correct Answer - A |
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| 1874. |
Amongst the following elements (whose electronic configuration an given below) the one having bighest ionization energy isA. `[Ne]3s^(2)3p^(1)`B. `[Ne]3s^(2)3p^(3)`C. `[Ne]3s^(2)3p^(2)`D. `[Ar]3d^(10)4s^(2)4p^(3)` |
| Answer» Correct Answer - B | |
| 1875. |
A nutral atom of an electron has `2K,8L` and `5M` electron .Find out the followingNumber of electron in valent shellA. 5B. 6C. 7D. 4 |
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Answer» Correct Answer - A The electronic configuration of an atom with `2K,8L` and `5M` electron is : `1s^(2)2s^(2)2p_(x)^(2)2p_(y)^(2)2p_(z)^(2)3s^(2)3p_(x)^(1)3p_(y)^(1)3p_(z)^(1)` Number of electron in valence shell`= 2+3 = 5` |
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| 1876. |
Let `v_(1)` be the frequency of the series limit of the lyman series `v_(2)` be the frequency of the first line of th elyman series and `v_(3)` be the frequency of the series limit of the Balmer seriesA. `nu_(1)` - `nu_(2)` = `nu_(3)`B. `nu_(2)` - `nu_(1)` = `nu_(3)`C. `nu_(2)` = `1/2( nu_(1) - nu_(3))`D. `nu_(1)` + `nu_(2)` = `nu_(3)` |
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Answer» Correct Answer - `nu_(1)` - `nu_(2)` = `nu_(3)` Lyman series limit `(n_(1) = 1,,n_(2)= oo)` `v_(1) = RZ^(2)((1)/(1^(2)) - (1)/(oo^(2))) = RZ^(2) = R(Z = 1)` First line of lyman `(n_(1) = 1,,n_(2)= 2)` `v_(2) = R((1)/(1^(2)) - (1)/(2^(2))) = (3)/(4)R = (3)/(4)(v_(1)) (v_(1) = R)` Balmer series limit `(n_(1) = 1,,n_(2)= oo)` `V_(3) = R((1)/(2^(2)) - (1)/(oo^(2))) = (R)/(4) = (v_(1))/(4)` `v_(3) = (v_(1) - v_(2)) = (v_(1) - (3)/(4) v_(1))= (v_(1))/(4)` `:. (v_(1) - v_(2)) = v_(3)` |
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| 1877. |
A nutral atom of an electron has `2K,8L` and `5M` electron .Find out the following Number of unpaired electronsA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - B The electronic configuration of an atom with `2K,8L` and `5M` electron is : `1s^(2)2s^(2)2p_(x)^(2)2p_(y)^(2)2p_(z)^(2)3s^(2)3p_(x)^(1)3p_(y)^(1)3p_(z)^(1)` Number of unpaired electron` = 3(3p_(x)^(1)3p_(y)^(1)3p_(z)^(1))` |
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| 1878. |
An atom has `2K,8L` and `5M` electron write its eklectronic configuration and indication the following in it :A. Number of sub-shellsB. Number of orbitalsC. Number of unpaired electronsD. Number of electron having `l= 1` |
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Answer» In the given atom then are `2 + 8 + 5 = 15` electron .The el,ectronic configuration of the atom is `1s^(2)2s^(2)2p^(6)3s^(2)3p_(x)^(1)3p_(y)^(1)3p_(z)^(1)` a.Number of sub-shell is `5` b.Number of orbital is `9` b..Number of unpaired electron is `3` b..Number of electron having `l= 1` is `9` |
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| 1879. |
The ratio of atomic numbers of two elements X and Y is 1 : 7. Radius of 5th orbit ofsingle electron species of X is 3.3 `Å`. Write the electronic con figuration of Y and four quantum numbers for the valence electrons |
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Answer» The atomic number of B is 73 because its electronic configuration is `[Xe]6s^24gamma^(14)5d^3` `therefore` Atomic number of A is x , that is 73-54=68 So , its electronic configuration is `[Xe]6s^2 5d^1 4f^(11)` |
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| 1880. |
An electron is revolving in an orbit with a veloc- ity of `5.46xx107 cm s^(-1)` . Calculate the wave length associated with the electron. |
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Answer» Radius `r_n=(0.529xxn^2)/ZrArr3.3` `(0.529xx25)/ZrArrZ=4` `therefore` Atomic number of x is 4 and atomic number of is 28. Electronic configuration of Y is `1s^22s^22p^63s^23p^64s^23d^8` and valenc electrons are `4s^2` electrons Electron : `1 rarr n= 4,1=0 m=0,s=-1/2` Electron `2 rarr n= 4,1=0,m=0,s=-1/2` |
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| 1881. |
Discovery of the nucleus of an atom was due to the experiment carried out by:A. BohrB. RutherrfordC. MoseleyD. Thomson |
| Answer» Correct Answer - B | |
| 1882. |
If the kinetic energy of a particle is doubled, de Broglie wavelength becomes:A. 2 timesB. 4 timesC. `sqrt(2)`timesD. `(1)/(sqrt(2))` times |
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Answer» Correct Answer - D `lamda=(h)/(sqrt(2Em))` where E=Kinetic energy of the particle `thereforelamda_(1)=(h)/(sqrt(2Em)),lamda_(2)=(h)/(sqrt(2xx2Em))` `(lamda_(1))/(lamda_(2))=sqrt(2)` i.e., `lamda_(2)=(lamda_(1))/(sqrt(2))` |
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| 1883. |
The mass number of two elements X and Z are 52 and 75 respectively. X contains 16.6% more neutrons compared to protons. Z contains 27.3% more neutrons compared to protons. X and Z are respectively:A. `.(24)Cr, ._(33)As`B. `._(24)Cr, ._(34)Zn`C. `._(19)K,._(33)As`D. `._(29)Cu,._(30)Zn` |
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Answer» Correct Answer - A Element(X) `._(x)^(52)X` Number of protons=x Number of neutrons `=(52-x)` Number of neutrons `=(116.6)/(100)xx x` `therefore(52-x)=(116.6)/(100)xx x` `x=(52)/(2.166)=24` `therefore` Element x is `_(24)Cr`. Element (Z)` ._(x)^(75)Z` Number of protons=x Number of neutrons =(75-x) Number of neutrons`=(127.3)/(100)xx x` `(75-x)=1.273x` `x=(75)/(2.273)=33` `therefore` Element `Z=._(33)As` |
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| 1884. |
The sodium falme testhas a characteristic yellow colour due to the emission of a vavelength of `589 nm ` What is the mass equivalent of one photon of this wavelength of this wavelength |
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Answer» According of de Broglie equation `lambda = (h)/(mv)` Wavelength of photon `= 589 nm = 589 xx 10^(-9)m ` Velocity of photon `= 3 xx 10^(8) m s^(-1)` `m = (h)/(lambda v) = (6.626 xx 10^(-34) kg m^(2)s^(-1))/(589 xx 10^(-9) m xx 3 xx 10^(8) m s^(-1))` `3.75 xx 10^(-34) kg` |
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| 1885. |
A natural atom of an element has `2K,8L, 9M` and `2N` electrons .The atomic number of element is :A. 20B. 21C. 22D. 23 |
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Answer» Correct Answer - B `overset(2K,8L,9M,and,2N)underset(1s^(2),2s^(2)p^(6),3s^(2)3p^(6)3d^(1), 4s^(2))` `({:(K means n = 1, l means n = 2),(M means n = 3 , N means n = 4):})` `"Structure is" `3d^(1),4s^(2)` Atomic number 21 |
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| 1886. |
A natural atom of an element has `2K,8L, 9M` and `2N` electrons .The atomic number of element is : The valency of element isA. `+2`B. `+3`C. `Both +2 and +3`D. `-1` |
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Answer» Correct Answer - C `overset(2K,8L,9M,and,2N)underset(1s^(2),2s^(2)p^(6),3s^(2)3p^(6)3d^(1), 4s^(2))` `({:(K means n = 1, l means n = 2),(M means n = 3 , N means n = 4):})` `"Structure is" `3d^(1),4s^(2)` Atomic number 21 Structure is `3d^(4),4s^(2)` So `e^(bar)` can be excited from `4s` and `3d` both (since energy difference between `3d` and `4s` very very small ) So, veloncy is `+ 2` and `+3` both |
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| 1887. |
The sodium flame test has a characteristic yellow colour due to the emission of a wavelength of `589 nm`. What is the mass equivalent of one photon of this wavelength? |
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Answer» ` lambda = h/( m u)` or ` m = h/( lambda .u ) = ( 6.626 xx 10^(34))/( 589 xx 10^(-9) xx 3.0 xx 10^8)` ` 3. 74 xx 10^(-36) kg`. |
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| 1888. |
An iodine molecule dissociates into atom after absorbing light of wavelength `4500`Å. If quantum of radiation is absorbed by each molecule calculate the kinetic energy of iodine (Bond energy of `I_(2)` is `240`` kJ mol^(-1))`A. `8. 6 xx 10^(-36) J`B. `2.17 xx 10^(-20)J`C. 433 kJD. 4.3 J |
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Answer» Correct Answer - B |
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| 1889. |
Calculated the energy required to excite one litre of hydrogen gas at `1atm` and `298K` to the first excited state of atomic hydorgen. The enegry for the dissociation of `H-H` bond is `436 kJ mol^(-1)`. |
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Answer» Correct Answer - `98.44 kJ` |
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| 1890. |
Which diagram best represnts the apperance of the line spectrum of atomic hydorgen in the visible region? `overset(Increasing lambda)rarr`A. B. C. D. |
| Answer» Correct Answer - C | |
| 1891. |
A proton `(mass = 1.66 xx 10^(-27)kg)` is moving with kinetic energy `5 xx 10^(-27) J` calculate the de Broglie wavelength associated with it ? `(h = 6.6 xx 10^(-34)Js)` |
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Answer» KE of proton = `(1)/(2) mv^(2) = 5 xx 10^(-27) J = 5 xx 10^(-27) kg m^(2)s^(-2)` `v^(2) = (2 xx 5 xx 10^(-27) kg m^(2)s^(-2))/(1.66 xx 10^(-27)) kg = 6.024 m^(2)s^(-2)` `v = sqrt(6.02m^(2)s^(-2)) = 2.454 m s^(-1)` `lambda = (h)/(mv) = (6.6 xx 10^(-34) Js)/(1.66 xx 10^(-27) kg xx 2.454 m s^(-1)) = 1.6202 xx 10^(-7)m` |
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| 1892. |
The mass of an electron is m, charge is e and it is accelerated from rest through a potential difference of V volts. The velocity acquired by electron will beA. `sqrt((2eV)/(m))`B. `sqrt((eV)/(m))`C. `sqrt((V)/(m))`D. `sqrt((eV)/(2m))` |
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Answer» Correct Answer - A `v alpha (1)/(n), :.vn=` constant |
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| 1893. |
A particle initially at rest having charge `q` coulomb, `&` mass `m kg` is accelerated by a potential difference of `V` volts. What would be its `K.E. & ` de broglie wavelength respectively after acceleration.A. `qV,(h)/(sqrt(2qVm))`B. `(h)/(sqrt(2qVm)),qV`C. `qV,(h)/(mV)`D. `(h)/(mV),qV` |
| Answer» Correct Answer - 1 | |
| 1894. |
What possibly can be the ratio of the de Broglie wavelength for two electrons each having zero initial energy and accelerated through 50 volts and 200 volts?A. `3:10`B. `10:3`C. `1:2`D. `2:1` |
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Answer» Correct Answer - D `lambda=(h)/(sqrt(2mqV))" " lambdaprop(1)/sqrt(V)` `(lambda_(1))/(lambda_(2))=sqrt((V_(2))/(V_(1)))=sqrt((200)/(500))=(2)/(1)` |
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| 1895. |
The mass of an electron is m, charge is e and it is accelerated form rest through a potential difference of V volts. The velocity acquired by electron will be :A. `sqrt((V)/(m))`B. `sqrt((eV)/(m))`C. `sqrt((2eV)/(m))`D. zero |
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Answer» Correct Answer - C `KE = (1)/(2) mv^(2) = eV_(0)` |
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| 1896. |
The wavelength associtated with a golf ball weight `200 g` and moving at a speed of `5 m h^(-1)` is of the orderA. ` 10^(10) m`B. ` 10^(-20) m`C. ` 10^(-30) m`D. ` 10^(-40) m` |
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Answer» Correct Answer - C `lambda = h/(mu) = (6.626 xx 10^(27))/( 200 xx 5)` ` = 6.626 xx 10 ^(30 ) m`. |
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| 1897. |
Number fo nodal plane in ` p_x` orbital is :A. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - A Nodal plane is p-orbital ` =l=1` |
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| 1898. |
The number of waves made by a Bohr electron in Hydrogen atom in one complete revolution in the `3^(rd)` orbit is. |
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Answer» Correct Answer - 3 no of waves in `n^(th)` orbit `=n` |
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| 1899. |
If velocity of an electron in `1^(st)` Bohr orbit of hydrogen atom us `x`, its velocity in `3^(rd)` orbit will be.A. `(x)/(3)`B. `3 x`C. `9 x`D. `(x)/(9)` |
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Answer» Correct Answer - A (a) `V_n = (2 pi Ze^2 K)/(nh)` `:. (V_3)/(V_1) = (1)/(3)` `:. V_3 = (V_1)/(3)` or `(x)/(3)`. |
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| 1900. |
The ratio of the wave number corresponding to the `1^(st)` line of Lyman series of `H-` atom and `3^(rd)` line of Pashcan series of a hydrogen like sample is `9:16`. Then find the third excitation potential in terms of volt for this `H-` like samples.A. 210B. 204C. 100D. 300 |
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Answer» Correct Answer - 2 Wave number `N=(1)/(lambda)=Rz^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `(N_(1))/(N_(2))=(z_(1)^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))))/(z_(2)^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))))=((1-(1)/(4)))/(z^(2)((1)/(9)-(1)/(36)))rArr" "(N_(1))/(N_(2))=(9)/(z^(2))=(9)/(16)" "rArr" "z=4` |
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