Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Which of the following reactionis faster ?

Answer»



Both are EQUALLY FAST
REACTION (a)is not possible

Answer :A
Discussion
2.

When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A, B, C and D. Explain why the milkiness disappears in the last step.

Answer»

Solution :SolutionB turns milky ion passing`CO_(2)`, it is lime WATER `Ca(OH)_(2)`and comound C which gives milky appearance is `CaCO_(3)`. On passing excess of `CO_(2)`milkinessdisappears due to the FORMATION of compound D that is `Ca(HCO_(3))_(2)`. Compound A REACTS with water and gives B. It is CaO.
The reactions are :
(i) `underset((A)) (CaO) +H_(2)O tounderset((B))underset("Lime water")(Ca(OH)_(2))`
(ii)`underset((B))(Ca(OH)_(2))+underset((C ))underset(("Milky solution"))underset("Calcium carbonate")(CO_(2)+CaCO_(3)+H_(2)O)`
(iii)On passing high amount of `CO_(2)` through calcium carbonate solution, it forms calcium BICARBONATE, which remove milky colour of solution.
`underset((C ))underset("Milky colour")(CaCO_(3)) +CO_(2)+H_(2)O to underset((D))underset("(Soluble carbonate)")underset("Calcium carbonate")(Ca(HCO_(3))_(2))`
Discussion
3.

What is the maximum number of electrons that can be associated with the following set of quantum numbers ? n = 3, l = 1 and m = -1

Answer»

10
6
4
2

Solution :The GIVEN set of quantum NUMBER REPRESENTS one 3D orbital. An orbital can have maximum 2 electrons
Discussion
4.

Which are the cancer causing compounds ?

Answer»

Solution :Benzene and POLY nuclear HYDROCARBON with more than two benzene rings are CANCER causing substance. For e.g.,
(a) 1, 2-benzalthrecene
(b) 3-methylcolenthrene
(c) 1,3-benzapyrine
(d) 1,2,5,6-dibenzathrecene
(e) 9, 10-dimethyl-1,2 benzathrecene
Discussion
5.

What is the product of uncertainties in position and velocity of an electron ?

Answer»

Solution :`Deltax xx Delta V ge (H)/(4pim)=(6.625xx10^(-34)kgm^(2)s^(-1))/(4xx3.1416xx9.1xx10^(-31)KG)`
The product of uncertainties `=Deltax xx Deltav`
`=5.79xx10^(-5)m^(2)s^(-1)`
Discussion
6.

Which of the following dienes cannot undergo Diels-Alder reactions?

Answer»




Solution :A diene has to be ABLE to ASSUME an s-cis conformation in order to undergo Diels-Alder REACTIONS, and (d) has a CONSTRAINED s-trans structure.
Discussion
7.

What are gaseous solution ? Give its various types with example.

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SOLUTION :`{:("Solute" , "Gas", "EXAMPLE: AIR (A mixture of" N_(2) O _(2)` " and other GASES)"), ( "Solvent", "Gas",), ( "Solute" , "Liquid", "Example Humid OXYGEN"), ("Solute" , "Solid" , "Example Camphor in nitrogen gas"), ("Solvent", "Gas", ):}`
Discussion
8.

What are the concentration of Ag^(+), [Ag(NH_(3))]^(+) and [Ag(NH_(3))_(2)]^(+) in a solution prepared by adding 0.10 mole of AgNO_(3) to 1.0 litre of 3.0M NH_(3) ? Given : Ag_((aq.))^(+)+NH_(3(aq.))hArr [Ag(NH_(3))]_((aq.))^(+) , K_(1)=2.1xx10^(3)....(1) [Ag(NH_(3))]_((aq.))^(+)+NH_(3(aq.))hArr [Ag(NH_(3))_(2)]_((aq.))^(+) , K_(2)=8.1xx10^(3)....(2) Ag_((aq.))^(+)+2NH_(3(aq.))hArr[Ag(NH_(3))_(2)]_((aq.))^(+) , K_(3)=1.7xx10^(7).....(3)

Answer»


ANSWER :`[Ag^(+)]= 7.5xx10^(-10)M; [Ag(NH_(3))]^(+)= 4.4xx10^(-6)M;`
`[Ag(NH_(3))_(2)]^(+)=0.10M`;
Discussion
9.

What are the oxidation number of the underlined elements in each of the followingand how do you rationalize your results ? (a) KunderlineI_(3) (b) H_(2)underlineS_(4)O_(6) ( c) underlineFe_(3)O_(4) (d) underlineCH_(3)underlineCH_(2)OH (e) underlineCH_(3)underlineCOOH

Answer»

Solution :(a) `KunderlineI_(3)` = Oxidation number of K = +1
`THEREFORE` Oxidation number of `I=-1/3`, but oxidation number of I be in whole number.
`K^(+)[I-IlarrI]^(-)`
In `KI_(3),I_(2)andI^(-)` makes a covelant bond between them. Where oxidation number of `I_(2)=0`.
`therefore` Oxidation number of I is -1. In `KI_(3)` respectively, oxidation number of 3 IODINE are 0, 0 and -1.
(b) `H_(2)underlineS_(4)O_(6)` : Here, same oxidation number of four sulfur are not possible.
`H-O-underset(O)underset(||)overset(O)overset(||)(S)-S-S-underset(O)underset(||)overset(O)overset(||)(S)-O-H`
Here, 2, S - S bonding their oxidation number is zero. RATHER the two S left, their oxidation number is +1.
( c) `underlineFe_(3)O_(4)` : 3Fe + 4(O) = 0
3Fe + 4(-2) = 0
`x=8/3`
Now, from the stoichiometry :
`Fe_(3)O_(4)=FeO*Fe_(2)O_(3)`
`FeOtoFe+(-2)=0`
`thereforeFe=+2`
`Fe_(2)O_(3)to2Fe+3(-2)=0`
`therefore2Fe-6=0`
`therefore2Fe=+6`
`thereforeFe=+3`
(d) `underlineCH_(3)underlineCH_(2)OH=C_(2)H_(6)O`
`therefore2(C)+6(+1)+1(-2)=0`
`therefore2(C)+6-2=0`
`therefore2(C)+4=0`
`thereforeC=-2`
Structure : `H-underset(H)underset(|)overset(H)overset(|)(""^(2)C)-underset(H)underset(|)overset(H)overset(|)(C^(1))-OH`
Here, `C_(2)` attached with three hydrogen atom and ALSO with the group `CH_(2)OH`.
Oxidation number of `C_(2)=3(+1)+C_(2)+1(-1)=0`
`thereforeC_(2)=-2`
`C_(1)` attached with OH (Oxi. number -1) and - `CH_(3)` (Oxi. number +1).
Oxidation number of `C_(1)=+1+2(+1)+C_(1)+1(-1)=0`
`thereforeC_(1)=-2`
(E) `underlineCH_(3)underlineCOOH` : x + 1 - 2
`CH_(3)COOH:C_(2)H_(4)O_(2)`
`therefore2x+4(+1)+2(-2)=0`
`therefore2x+4-4=0`
`thereforex=0`
Structure : `H-underset(H)underset(|)overset(H)overset(|)(""^(2)C)-underset(H)underset(||)overset(1)(C)-OH`
Here, `C_(2)` is attached to three hydrogen atoms and with group -COOH.
Oxidation number of `C_(2)=3(+1)+C_(2)+1(-1)=0`
`thereforeC_(2)=-2`
`C_(1)` is attached with oxygen with double bond and other bond with OH and one `CH_(3)` group (Oxi. number = +1). Therefore,
Oxidation number of `C_(1)=+1+C_(1)+1(-2)+1(-1)=0`
`thereforeC_(1)=+2`
Discussion
10.

The solution of Schrodinger equation for hydrogen leads to three quantum numbers n, l , m_(1) (i) n can have any integral value (ii) When n=3 , l can have value of 0,1,2 (iii) When n=5, l=2, m_(1)can have values ranging between +2 to -2 through (iv) when n=2, l=1, three equivalent orbitals with a nodal plane containingnucleus of the H atom are possible. Out of these the correct one is /are

Answer»

(i), (ii) ,(III)
only (ii)
(ii)& (IV)
(i), (ii) , (iii) & (iv)

SOLUTION :STATEMENT (i) ,(ii) and (iii) are CORRECT.
Discussion
11.

Volume and pressure of balloon filled with hydrogen gas is 1 bar and 175 dm^(3). When balloon reached to height its pressure will be decreases to 0.8 bar than calculate volume of balloon.

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ANSWER :`218.75 DM^(3)`
Discussion
12.

Which are various forms of silica ?

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Solution :SILICA , OCCURS in several crystallographic FORMS LIKE Quartz, cristobalite and TRIDYMITE.
Discussion
13.

What is the % age of CO_2 in the pure dry air?

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SOLUTION :About 0.032%
Discussion
14.

The solubility in water of sulphates down the Be groupis : Be gt Mg gt Ca gt Sr gt Ba. This is due to ……

Answer»

Increase in MELTING point
High IONIZATION energy
Higher COORDINATION number
All of these

Answer :C
Discussion
15.

Which of the following given structure of Silicones is correct ?

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`-O[--UNDERSET(CH_3)underset|OVERSET(CH_3)overset|SI-]_n-O-underset(CH_3)underset|overset(CH_3)overset|(Si)-`
`-O[--O-underset(CH_3)underset|overset(CH_3)overset|(Si)-]_n-O-underset(CH_3)underset|overset(CH_3)overset|(Si)-O-`
`[--O-underset(CH_3)underset|overset(CH_3)overset|(Si)-]_n-O-underset(CH_3)underset|overset(CH_3)overset|(Si)-`
`-O[--underset(CH_3)underset|overset(CH_3)overset|(Si)-O--]_n underset(CH_3)underset|overset(CH_3)overset|(Si)-`

ANSWER :D
Discussion
16.

When a 25.00 mL volumetric flask weighing 20.340 g is filled partially with metal shot, the mas is 119.691g. The flask is then filled to the 25.00 mL mark with methanol (d=0.791 gcm^(-3)) and has a total mass of 130.410g. What is the density of the metal?

Answer»

`6.96 GCM^(-3)`
`8.68 gcm^(-3)`
`9.27 gcm^(-3)`
`11.7gcm^(-3)`

ANSWER :B
Discussion
17.

Which of the following salts exist(s) as dechydrated crystal ?

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WASHING sods
Glauber.s salt
Epsom salt
Gypsum

Solution :`underset("Washing SODA")(Na_(2)CO_(3).10H_(2)O)rArrunderset("Gluuber.s salt")(Na_(2)SO_(4).10H_(2)O)`
Discussion
18.

Which element of group 13 forms the most stable +1 oxidation state.

Answer»

SOLUTION :Inert pair effect is most PROMINENT in case of Tl and hence it forms the moststable +1 OXIDATIONSTATE.
Discussion
19.

Which one of the following ions has electronic configuration [Ar]3d^6 (Atomic number of Mn = 25, Fe = 26,C_@ = 27, N i = 28)

Answer»

`CO^(+3)`
`N i^(+2)`
`Mn^(+3)`
`Fe^(+3)`

Answer :A
Discussion
20.

What is sodium amalgam ?

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Solution :MIXTURE of SODIUM metal with mercury is CALLED sodium amalgam (Na-Hg).
Discussion
21.

Which one among the following sols is hydrophobic?

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Gum
Gelatin
Starch
Sulphur.

Answer :D
Discussion
22.

Which of the following reaction increases production of dihydrogen from synthesis gas?

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`CH_(4)(G) + H_(2)O(g) UNDERSET(Ni)OVERSET(1270K)to CO(g) + 3H_(2)(g)`
`C(s) + H_(2)O(g) overset(1270K)to CO(g) + H_(2)(g)`
`CO(g) +H_(2)O (g) underset("Catalyst")overset(673 K)to CO_(2)(g) + H_(2)(g)`
`C_(2)H_(6)+ 2H_(2)O underset(Ni)overset(1270K)to 2CO+5H_(2)`

Solution :To increase the production of `H_(2)` from SYNTHESIS gas, CO is oxidised to `CO_(2)` by passingit over steam at 673 K in presence of a catalyst.
`CO(g) + H_(2)O(g) underset("Catalyst")overset(637K) to CO_(2)(g) + H_(2)(g)`
Thus, option (c) is correct.
Discussion
23.

Write four characteristic properties of p-block elements.

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Solution :The four important characteristic properties of p-block elements are the following,
(a) p-Block elements has both metals and non metals but the number of non-metals is much higher than that of metals. As metallic character increases from top to bottom WITHIN a group and non-metallic character increases from left to right along a period in this block.
(B) Ionisation ENTHALPY of p-block elements are higher as compared to s-block elements. (c) They mostly FORMS covalent compounds in nature.
(d) More than one oxidation states are seen in their compounds. OXIDISING character increases from left to right in a periodic table and reducing character is from top to bottom in a group.
Discussion
24.

The rotationof C-C single bond leads to different isomeric structure called as …………………….

Answer»

SOLUTION :CONFORMERS
Discussion
25.

Write the oxidation number of oxygen in (a) O_(3), (b) MgO, (c) H_(2)O_(2), (D) KO_(2) and (e) OF_(2)

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Solution :OXIDATION number of .O. in `O_(3)=0`
Oxidation number of .O. in `MgO=-2`
Oxidation number of .O. in `H_(2)O_(2)=-1`
Oxidation number of .O. in `KO_(2)=-1//2`
Oxidation number of .O. in `OF_(2)=+2`.
Discussion
26.

The secondionizationenergiesof Li ,Be , Band Care in theorder

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`Li gt C gt B gt Be`
`Li gt B gt C gt Be `
`B gt C gt Be gt Li`
`Be gt C gt B gt Li`

SOLUTION :`Li gt B gt C gt Be` i.e.,option ( b) is correctRefer toAns. To Q. 6,PAGE`3//94`
Discussion
27.

Which of the followin will show least dipole character?

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Water
Ethanol
Ethane
Ether

Answer :C
Discussion
28.

Which one of the following is applicable to the conformations of a hydrocarbon ?

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C- C DISTANCE changes
C-H distance changes
C-C-C and C-C-H BOND ANGLES change.
Only distance between non-bonded H-atoms changes

Answer :D
Discussion
29.

When primary alcohol is oxidised with Cl_(2), it gives

Answer»

`CH_(3)CHO`
`CH_(3)COCH_(3)`
`CH_(3)COCl`
`COCl_(2)`

Solution :`R-CH_(2)OHunderset("Oxidation")OVERSET(Cl_(2))rarr-CH_(2)CN`.
A primary alcohol on oxidation with `Cl_(2)` gives an ALDEHYDE `(CH_(3)CHO)`.
Discussion
30.

Which of the following gives on ozonolysis both aldehydes and ketones?

Answer»

`Me_2C=CHMe`
`Me_2C=CM e_2`
`MeCH_2-C(Me)=CM e_2`
`MECH(Me)-CH=CH Me`

Solution :`Me_2C=CHMe UNDERSET((ii)Zn// H_2O)overset((i)O_3//CH_2Cl_2)to underset"KETONE"(Me_2C=O)+underset"ALDEHYDE"(O=CHMe)`
Discussion
31.

What volume at S.T.P is occupied by (i) 3.50 g of nitrogen? (ii) 6.022 xx 10^(21)molecules of ammonia? (iii) 0.350 moles of oxygen? (iv) 39.9 g of argon?

Answer»

Solution :(i) The NUMBER of moles in 3.50 g of nitrogen
`=(3.50)/(28.02) = 0.125`
(since, the gram molecular mass of `N_(2)` is 28.02 g)
`therefore` One mole of a gas occupies 22.4 L at S.T.R
`therefore` Thevolumeoccupiedby0.125moles =` 0.125 xx 22.4 = 2.80 L`

Hence, 3.50 g of N2 occupy a VOLUME of 2.80 litres at S.T.P Ans.
(ii) One mole of a gas has 6.022 x 10 MOLECULES and occupies a volume of 22.4 L at S.T.P.
`therefore` The volume occupied by `6.022 xx 10^21` molecules.
(III) The volume occupied by 0.350 moles of `O_(2)` at S.T.P. = `0.350 xx 22.4 = 7.84` L (because the volume occupied by one mole at S.T.P. is 22.4 L)
Hence, 0.350 moles of `O_(2)` occupy 7.84 litres at S.T.P. Ans.
(iv) Argon is monoatomic in nature.
`therefore` 1 gram mole of Ar = 1 g atom of Ar = 39.95 g Hence, 39.95 g (1 mole) of argon will occupy a volume of 22.4 L at S.T.P.
Discussion
32.

Which of the following is a non-metal ?

Answer»

Gallium
Indium
BORON
Aluminium

Solution :Boron
Discussion
33.

What is C?

Answer»

`PbCI_(2)`
`BaSO_(4)`
`BaSO_(3)`
`PbSO_(4)`

Solution :(A) `CdSO_(4)` (B) `CDS` (C) `BaSO_(4)`
(D) `Cd(NO_(3))_(2)` (E) `K_(2)[Cd(CN)_(4)]` (F) `[Cd(NH_(3))_(4)]^(2+)`
Discussion
34.

When ethane-1, 2-diol is added to water, which of the following is observed ?

Answer»

The following of water increases
Water evaporature of water increases
Freezing POINT water is lowered
The viscosity of water decreases

Solution :Due to extensive intermolecular H-Bonding, ethlene GLYCOL is non-volatile. Adding a non-volatile SOLUTE (ethlyene glycol) to water LOWERS its freezing point.
Discussion
35.

Write expression for molar mass, M (in kg mol^(-1)) of a body-centred cubic crystal of an ionic compound if it has an edge length of 'a' metre and a density of 'd' kg m^(-3)

Answer»

SOLUTION :`d=(ZxxM)/(a^3xxN_A)` or `M=d xx a^3xxN_A` (as for ionic compound having BCC STRUCTURE, Z=1 )
Discussion
36.

White the produc t of following reactions. (a) CH_(3)-overset(H_(3)C)overset(|)(C)=overset(CH_(3))overset(|)(C)-CH_(3) overset(O_(3))underset(Zn//H_(2)O)to(b) CH_(3)-C-=C-CH_(3) overset(O_(3))underset(Zn//H_(2)O)to (c) CH_(3)-C-=C-CH_(3) overset(O_(3)//H_(2)O_(2))to(d) CH_(3)-overset(CH_(3))overset(|)(C)=CH-CH_(3)overset(O_(3)//H_(2)O_(2))to

Answer»


ANSWER :(a) `CH_(3)-overset(CH_(3))overset(|)(C)=O`(b) `CH_(3)-underset(O)underset(||)(C)-underset(O)underset(||)(C)-CH_(3)`
(c) `CH_(3)-CO OH`(d) `CH_(3)-overset(CH_(3))overset(|)(C)=O+CH_(3)-overset(O)overset(||)(C)-OH`
Discussion
37.

The structureof diborane (B_(2)H_(6))contains

Answer»

four `2C-2E` BONDS and two `3c-2e` bonds
two `2c-2e` bonds and four `3c-2e` bonds
two `2c-2e` bonds and four `3c-3e` bonds
four `2c - 2e` bonds and four `3c-2e` bonds

Solution :`B_(2)H_(6)`has four `2c-2e` and two `3c-2e` bonds.
Discussion
38.

Which oneis leaststablein fromfollowing

Answer»

`LI^(-)`
`B^(-)`
`Be^(-)`
`C^(-)`

ANSWER :C
Discussion
39.

Total number of antibonding electrons present in O_(2) will be-

Answer»

6
8
4
2

Solution :MO ELECTRONIC configuration of `O_(2):`
`(sigma_(1s))^(2)(sigma_(1s)^(**))^(2)(sigma_(2s))^(2)(sigma_(2s)^(**))^(2)(sigma_(2p_(Z)))^(2)(pi_(2p_(x)))^(2)(pi_(2p_(y)))^(2)`
`(pi_(2p_(x))^(**))^(1)(pi_(2p_(y))^(**))^(1)`
Thus, there are total 6 ELECTRONS in antibonding orbitals.
Discussion
40.

Write the IUPAC names of isomers of C_(2)H_(4)Cl_(2). Give one test to distinguish these.

Answer»

SOLUTION :1, 2-Dichloroethane, 1, 1-Dichloroethane HYDROLYSIS of 1, 2-Dichloroethane with aqueous NaOH GIVES ETHYLENE glycol while the hydrolysis of 1, 1-Dichloroethane gives ETHANAL.
Discussion
41.

Which of the following involves the redox reaction ?

Answer»

CORROSION of METALS.
Extraction of metals from their ores.
Respiration.
All of these

Answer :D
Discussion
42.

Which of the following refers ton the original description of oxidation ?

Answer»

Addition ofoxygen
Addition foelectronegative ELEMENT
Removal fo hydrogen
Removal of ELECTROPOSITIVE element

Solution : Orginally, the term "oxidation" was used to DESCRIBE the addition of oxygen to an element or a COMPOUND :
`S(s) + O_2(g) RARR SO_2 (g)`
`CH_4 (g) + 2O_2 (g) rarr CO_2 (g) + 2H_2 O(1)`.
Discussion
43.

Which of the following paramagnetic ions would exhibit a magnetic moment (spin only) of the order of 5 BM? (At. No : Mn = 25, Cr = 24,V = 23, Ti = 22)

Answer»

`V^(2+)`
`TI^(2+)`
`Mn^(2+)`
`CR^(2+)`

Answer :C
Discussion
44.

Ethylene glycol (molar mass = 62 g mol ^(-1)) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4 of this substance in 100 g of water. Would it be advisable to keep this substance in the car radiator during summer? (K_(f) for water= 1.86 k kg//mol) (K_(b) for water = 0.512 K kg//mol.)

Answer»

SOLUTION :`Delta T_(b) = K _(b) xx (W_(B))/(M _(B)) xx ( 1000 )/( W _(A)) = 0.512 xx (12.4)/(62) xx (100)/(100) = 1. 024 K`
Since WATER boils at `100^(@)C,` so a solution CONTAINING ethylene glycol will boil at `101.024 ^(@)C,` so it is advisable to keep this substance in car RADIATOR during summer.
Discussion
45.

Which of the following are antacids ?

Answer»

Omeprazone
Lansoprazole
Sodium bicarbonate
Triprolidine

Answer :A::B::C
Discussion
46.

The value of Ke for the raction 2A hArr B +Cis 2 xx 10^(3) . At a given time . The composition of reaction mixture is [A]=[B]=[c]= 3 xx 10^(-4)M. In which direciton the reaction willproceed ?

Answer»

Solution :For the REACTION, the reaction QUOTIENT `Q_C` is given by `Q_C=[B][C]//[A]^2`
as `[A]=[B] = [C] = 3 xx 10^(-4)M`
`Q_(C) = (3XX 10^(-4))(3 xx 10^(4))//(3 xx 10^(-4))^2=1`
as `Q_(C) gt K_(C)` so, the reaction will proceed in the reverse direction.
Discussion
47.

Which of the following contains maximum number of nitrogen atoms ?

Answer»

`22.4 L` of `N_(2)` at STP
500 ML of `2.0 M NH_(3)`
`6.02 xx 10^(23)` MOLECULES of `NO_(2)`
1.00 MOL of`NH_(4)Cl`

SOLUTION :(A) 22.4 L of `N_(2)` at STP `~=` 1 mol of `N_(2) = 2` mol of N
(B) 500 mL of 0.2 M `NH_(3) = 1 mol " of " NH_(3)`
= 1 mol of N
(C) `6.02 xx 10^(23)` molecules of `NO_(2) =` 1 mol of N
(D) 1.00 mol of `NH_(4)Cl = 1` mol of N
Thus (A) contains maximum number of N atom
Discussion
48.

The stock notation for Mn_(2)O_(7) is

Answer»

MANGANESE (II) oxide
manganese (III) oxide
manganese (V) oxide
manganese (VII) oxide.

Answer :D
Discussion
49.

Which one of the following is an electron deficient hydride?

Answer»

`C_(2)H_(6)`
`B_(2)H_(6)`
`GeH_(4)`
`CH_(4)`

Answer :b
Discussion
50.

The wavelength of the radiation emitted when the electron jumps from 4th shell to 2nd shell is

Answer»

`4862 Å`
`2056Å`
`5241 Å`
`109700` CM

Solution :ACCORDING to Balmer EQUATION
Wave number `(barv)=109677(1/n_(1)^(2)-1/n_(2)^(2))cm^(-1)`
`barv=109677(1/((2)^(2))-1/((4)^(2)))cm^(-1)`
`=(109677xx3)/16cm^(-1)`
`lambda=1/v=16/(109677xx3)=4862xx10^(-8)`cm
`=4862xx10^(-10)m=4862 Å`
Discussion