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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.
| 401. |
Determine the mass of the following : (i) `6.022 xx 10^(23)` number of `O_(2)` molecules (ii) `6.022 xx 10^(23)` number of O atoms . |
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Answer» (i) The number of moles = `("number of molecular")/("Avogadro constant") = (6.022 xx 10^(23))/(6.022 xx 10^(23)) = 1`. Now , mass = mole `xx` molecular mass = `1 xx 32 = 32` g (ii) The number of moles `= ("number of atoms")/("Avogadro constant") = (6.022 xx 10^(23))/(6.022 xx 10^(23)) = 1` . Now , mass = mole `xx` atomic mass = `1 xx 16 = 16` g |
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| 402. |
The volume of one mole of a gas at standard temperature and pressure isA. 11.2 litresB. 22.4 litresC. 100 litresD. none of these |
| Answer» Correct Answer - B | |
| 403. |
Determine the mass of the following : (i) 0.7 mole of `O_(2)` gas (ii) 0.7 mol of O atoms . |
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Answer» (i) Molecular mass of `O_(2) ` gas = 32 `therefore ` mass = mole `xx` molecular mass = `0.7 xx 32 = 22.4` g (ii) Atomic mass of O atom = 16 `therefore` mass = mole `xx` atomic mass = `0.7 xx 16 = 11.2` g . |
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| 404. |
Assertion: Hydrogen atom consists of anly one electron but its emission spectrum has may lines. Reason: Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.A. if both assertion and reason are true and reason is the correct explanation of assertion.B. if both assertion and reason are true but reason is not the correct explanation of assertion.C. if assertion is true but reason is false.D. if both assertion and reason are false. |
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Answer» Correct Answer - B Every atom has certain definite energy level. In the normal state, the electron in the hydrogen atom stays in lowest energy level. When the atom gets appropriate energy form outside,then this electron rises to some higher energy level i.e. atom is excited. Within nearly `10^(-8)`s, the electron leaves the higher energy level. Now, it can return either directly to the lowest energy level (or the ground state) or come to the ground after passing through other levels. Since there are a large number of atoms in a light source (hydrogen lamp),all possible transitions take place in the source and many lines are seen in the spectrum. The silt gives the shape of the spectrum and large number of lines are obtained because a large number of atoms are geeting excited and de -excited to different energy levels. |
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| 405. |
If `v_1` is the frequency of the series limit of lyman seies, `v_2` is the freqency of the first line of lyman series and `v_3` is the fequecny of the series limit of the balmer series, thenA. `v_1-v_2=v_3`B. `v_1=v_2-v_3`C. `(1)/(v_2)=(1)/(v_1)+(1)/(v_3)`D. `(1)/(v_1)=(1)/(v_2)+(1)/(v_3)` |
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Answer» Correct Answer - A For lyman series `upsilon=Rc[(1)/(1^2)-(1)/(n^2)]` where n=2,3,4,....... for the seires limit of lyman series , `n=oo` `therefore upsilon_1=Rc[(1)/(1^2)-(1)/(oo^2)]=Rc`.........(i) for the first line of lyman series, `n=2` `therefore upsilon_2=Rc[(1)/(1^2)-(1)/(2^2)]=(3)/(4)Rc`........(ii) for balmer series `upsilon=Rc[(1)/(2^2)-(1)/(n^2)]` where n= 3,4,5..... For the series limit of balmer serie , `n=oo` `thereforeupsilon_3=Rc[(1)/(2^2)-(1)/(oo^2)]=(Rc)/(4)`........(iii) from equations (i),(ii),(iii), we get `upsilon_1=upsilon_2+upsilon_3` or `upsilon_1-upsilon_2=upsilon_3` |
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| 406. |
Calculate the absolute mass of hydrogen atom. |
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Answer» The atomic mass of hydrogen is 1u We know 1u = 1.6605 x 10–24 g Therefore, Absolute mass of hydrogen = 1.6605 x 10–24 x 1 g = 1.6605 x 10–24 g |
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| 407. |
Find the absolute mass of Sodium. |
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Answer» The atomic mass of Sodium is 23u We know 1u = 1.6605 x 10–24 g Therefore, Absolute mass of Sodium = 1.6605 x 10–24 x 23 g = 38.191 x 10–24 = 3.8191 x 10–25 g |
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| 408. |
Calculate the absolute or actual mass of Nitrogen atom. |
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Answer» The atomic mass of Nitrogen is 14u We know 1u = 1.6605 x 10–24 g Therefore, Absolute mass of hydrogen = 1.6605 x 10–24 x 14 g = 23.247 x 10–24 g = 2.3247 x 10–25 g |
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| 409. |
Define Diatomic. |
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Answer» When molecule is formed by the diatomic: When molecule is formed by the combination of two atoms of it is called diatomic molecule. For example: Hydrogen (H2), Oxygen (O2) Nitrogen (N2), Chlorine (Cl2), etc. |
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| 410. |
Define Polyatomic. |
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Answer» When molecule is formed by the combination of more than two atoms, it is called polyatomic molecule. For example: Sulphur molecule (S8) |
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| 411. |
Define Existence of atoms. |
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Answer» Atoms of most of the elements exist in the form of molecule or ion, since they are most reactive. For example, hydrogen, oxygen, chlorine, etc. However, atoms of some elements, which are non-reactive, exist in free-state in nature. For example helium, neon, argon, etc. Usually atoms are exist in following two forms - • In the form of molecules • In the form of ions |
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| 412. |
Give the Characteristics of atoms. |
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Answer» Characteristics of atoms:
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| 413. |
Define Triatomic. |
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Answer» When moleucule is formed by the combination of three atoms it is called triatomic molecule. For example: molecule of ozone (O3) |
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| 414. |
Define Tetra-atomic. |
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Answer» When molecule is formed by the combination of four atoms it is called tetra-atomic molecule. For example: Phosphorous molecule (P4) |
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| 415. |
When was Dalton's atomic theory? |
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Answer» John Dalton, a British Chemists and scientists gave the Atomic Theory in 1808. This theory is popularly known as Dalton’s Atomic Theory in the honour of John Dalton. He gave the theory on the basis of Laws of Chemical Combination and explains them properly. In his theory he explains about atom. |
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| 416. |
Define the atomic mass unit. |
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Answer» Mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’. |
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| 417. |
How to calculate Molecular mass? Give Examples. |
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Answer» Generally we use relative atomic masses of atoms for calculating the molecular mass of 1 mole of any molecular or ionic substances. Example: Molecular mass of H2SO4 Atomic mass of Hydrogen = 1 Atomic mass of sulphur = 32 Atomic mass of oxygen = 16 Molecular mass of H2SO4 = 2(Atomic mass of Hydrogen) + 1 (Atomic mass of sulphur) + 4 (Atomic mass of oxygen) = 2×1 + 32 + 4× 16 = 98 u. Calculation of molecular mass of hydrogen chloride: Atomic mass of hydrogen + Atomic mass of chlorine = 1 + 35.5 = 36.5 u. |
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| 418. |
What is Molecular Mass of the Molecule? |
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Answer» The sum of the atomic masses of all the atoms in a molecule of a substance is called the molecular mass of the molecule. |
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| 419. |
Describe the term Molecule. |
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Answer» Molecule is the smallest particle of a compound. Atoms exist in free states in the form of molecule.
Most of the atoms exist in the form of molecule. Molecules are formed by the combination of two or more elements. Example: Molecule of hydrogen (H2, Molecule of oxygen (O2), Molecule of nitrogen (N2), etc. • Molecules of elements • Molecules of Compounds |
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| 420. |
Mention the three laws of chemical combination. |
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Answer» Before Dalton concept of atom was mere philosophical. Dalton explained about atom on the basis of Laws of Chemical Combinations. There are three laws of chemical combination. 1. Law of Conservation of Mass 2. Law of Constant Proportions 3. Law of Multiple Proportions |
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| 421. |
Define the term Molar mass.Give Example. |
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Answer» The molar mass of an element is equal to the numerical value of the atomic mass. However, in case of molar mass, the units change from ‘u’ to ‘g’. The molar mass of an atom is also known as gram atomic mass. For example, the atomic mass of carbon =12 atomic mass units. So, the gram atomic mass of carbon = 12 grams. |
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| 422. |
What is meant by the term chemical formula? |
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Answer» The chemical formula of a compound means the symbolic representation of the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For example, from the chemical formula CO2 of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide. |
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| 423. |
The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number of `Z` ) is the same as the shortest wavelength of the Balmer series of hydrogen atom. The value of `z` isA. 3B. 4C. 5D. 2 |
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Answer» Correct Answer - D According to the question, `" " lambda_(("shortest"))"Brackett"=(lambda_("shortest"))"Balmer"`" `rArr" "(1)/((lambda_("shortest"))"Brackett")=(1)/((lambda_("shortest"))"Balmer")` `rArrZ^(2)R((1)/(4^(2)))=R((1)/(4))rArrZ^(2)=4 rArrZ=2` |
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| 424. |
Let `v_(1)` be the frequency of series limit of Lyman series, `v_(2)` the frequency of the first line of Lyman series and `v_(3)` the frequency of series limit of Balmer series. Then which of the following is correct ?A. `v_(1)-v_(2)=v_(3)`B. `v_(1)=v_(2)-v_(3)`C. `(1)/(v_(2))=(1)/(v_(1))=(1)/(v_(3))`D. `(1)/(v_(1))=(1)/(v_(2))+(1)/(v_(3))` |
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Answer» Correct Answer - A We know that frequency, `v=Rc((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` This gives, `v_(1)=Rc(1-(1)/(oo))=RcrArrv_(2)=Rc (1-(1)/(4))=(3)/(4)Rc` `" " v_(3)=Rc((1)/(4)-(1)/(oo))=(Rc)/(4)rArrv_(1)-v_(2)=v_(3)` |
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| 425. |
How many grams of sodium will have the same number of atoms are 6 g of magnesium ? (Na = 23 , Mg = 24) |
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Answer» Number of grams- atoms of Mg = `("mass of Mg in atoms")/("gram-atomic mass") = (6g)/(24 g) = (1)/(4)` `therefore` gram - atoms of sodium should be `(1)/(4)` `because` 1 gram - atom of sodium = 23 g `therefore (1)/(4)` gram-atom of sodium = `23 xx 1/4 g = 5.75 g` . |
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| 426. |
How many grams of each of the following elements must be taken to get 1 mol of the element ? (a) sodium , (b) chlorine , (c) copper |
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Answer» The mass of 1 mol of an element is its atomic mass expressed in grams . Since the atomic masses of sodium , chlorine and copper are 23 g , 35.5 g and 63.5 g respectively , we get (a) 23 g (b) 35.5 g (c) 63.5 g |
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| 427. |
If `lambda_(Cu)` is the wavelength of `K_(alpha)` X-ray line of copper (atomic number `29`) and `lambda_(Mo)` is the wavelength of the `K_(alpha)`X-ray line of molybdenum (atomic number `42`),then the ratio `lambda_(Cu)//lambda_(Mo)` is close toA. 1.99B. 2.14C. 0.5D. 0.48 |
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Answer» Correct Answer - B `(lambda_(Cu))/(lambda_(Mo))=((Z_(Mo)-1)/(Z_(Cu)-1))^(2)=(41xx41)/(28xx28)=(1681)/(784)=2.144` |
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| 428. |
Name the following radicals :1. SO2-42. HCO-33. OH-4. Cr2O2-7 |
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Answer» 1. SO2-4: Sulphate 2. HCO-3: Bicarbonate 3. OH-: Hydroxide 4. Cr2O2-7: Dichromate |
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| 429. |
Name the following radicals and give their valencies :1. Sulphate SO42. Carbonate CO33. Hydroxide OH4. Dichromate Cr2O7 |
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| 430. |
Which of the following has valency 4 ?1. aluminium2. oxygen3. carbon4. phosphorus |
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Answer» phosphorus has valency 4. |
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| 431. |
Define Cations. |
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Answer» Ions having positive charge over them are called cations. For example: sodium ion (Na+), potassium ion (K+), etc. |
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| 432. |
What do you understand by the following terms valency, valence electrons and valence shell ? |
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Answer» 1. Valency— The capacity of the atom of an element to form chemical bonds. 2. Valence electrons— The electrons present in the valence shell are called valence electrons. 3. Valence shell— The outer most shell of an atom is known as its valence shell. |
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| 433. |
(i) Name three elements each whose ions have valency +1, +2 and +3 respectively.(ii) Name three non-metals/radicals which have valency -1, -2, and -3. |
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Answer» (i) +1 = Hydrogen, Sodium, Potassium. +2 = Calcium, Magnesium, Zinc. +3 = Aluminium, Chromium, Ferric [Fe (III)] (ii) -1 = Chlorine, Bromine, Iodine. -2 = Oxide, Sulphide, Sulphate. -3 = Nitrogen, Nitride Phosphate. |
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| 434. |
Define the following terms:1. Ions2. Chemical bond3. Atomic number4. Mass number5. Relative atomic mass |
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Answer» 1. Ions— An atom which becomes charged by losing or gaining electrons is called an ion. 2. Chemical bond— A chemical bond is the binding force between two or more atoms of a molecule. 3. Atomic number— Atomic number refers to the number of protons present in an atom. It is denoted by Z. 4. Mass number— Mass number is the sum of the number of protons and neutrons present in the nucleus of an atom. It is denoted by A. 5. Relative atomic mass— Relative atomic mass is the mass of an atom of an element as a multiple of the standard atomic mass unit. |
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| 435. |
Define Monoatomic ions. |
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Answer» Ions formed by one atom are called monoatomic ions. For example: sodium ion (Na+), potassium ion (K+), Chloride ion (Cl–), Fluoride ion (F–), etc. |
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| 436. |
Define Monoatomic. |
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Answer» When molecule is formed by single atom only, it is called monoatomic molecule. Generally noble gas forms monoatomic molecules. For example: Helium (He), Neon (Ne), Argon (Ar), Kr (Krypton), Xenon (Xe), Randon (Rn). |
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| 437. |
The half-life of \(^{226}_{88}Ra\) is 1620 y. Find its decay constant in SI unit. |
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Answer» Data: T = 1620 y = 1620 × 365 × 8.64 × 104 S = 5.109 × 1010 s The decay constant of \(^{226}_{88}Ra\) is λ = \(\cfrac{0.693}T\) = \(\cfrac{0.693}{5.109\times10^{10}}\) = 1.356 × 10-11 s-1 |
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| 438. |
An X-ray tube is operated at 30 eV. If a particular electron loses 10% of its kinetic energy to emit an X-ray photon during the collision. Find the wavelength and maximum frequency assoicated with this photon. |
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Answer» Kinetic energy of electron =30 eV Acccording to the question , 10% of this energy is converted to photon which is given by `(10)/(100)xx30=3 kV` `:. " " lambda_("min")("in"Å)=(12375)/(V("in volt"))=(12375)/(3xx10^3)=4.12Å` Also, `" " V("max")=(c)/(lambda_("min"))=(3xx10^(8))/((4.12xx10^(-10)))=7.2xx10^(17)Hz` |
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| 439. |
(A) according to electromagnetic theory an accelerated particle continuously emits radiation. (R) according to classical theory, the proposed path of an electron in rutherford atom model will be parabolic.A. if both assertion and reason are true and reason is the correct explanation of assertion.B. if both assertion and reason are true but reason is not the correct explanation of assertion.C. if assertion is true but reason is false.D. if both assertion and reason are false. |
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Answer» Correct Answer - C According to classical electronmagnetic theory, an accelerated charge continuously emits radiation. As electrons revolving in circular paths are constantly experiencing centripetal acceleration,hence thay will be losing their energy continuously and the orbital radius will go on decreasing and spial and finally the electron will fall on the nuclus. |
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| 440. |
According to classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom. |
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Answer» From Example 3 we kriow that velocity of electron moving around a proton in hydrogen atom in an orbit of radius `5.3 xx 10^(-11) m" is "2.2 xx 10^(-6)m//s`. Thus, the frequency of the electron moving around the proton is `v=(v)/(2pi r)=(2.2xx10^(6) m s^(-1))/(2pi (5.3xx10^(-11) m)` `=6.6xx10^(15) Hz`. According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is `6.6 xx 10^(15)` Hz. |
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| 441. |
Using Balmer empirical formula, obtain the wavelengths of Hα, Hβ, Hγ, Hδ......... H∞ |
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Answer» Balmer empirical formula is given by 1/λ = R(1/22 - 1/n2) For Hα line, put n = 3, 1/λα = R (5/36) where R = 1.097 x 107 m-1 this gives λα = 656.3 nm. For Hβ line, put n = 4, 1/λβ = R(3/16), this gives line = 486.1 nm For Hγ line, put n = 5, then λγ = 434.1 nm For Hδ line put n = 6, then λγ = 434.1 nm For H∞ line is called series limit, put n = ∞; λ∞ = 364.6 nm |
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| 442. |
Why do we use gold in Rutherford’s α-particle scattering experiment? |
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Answer» The reasons are : (i) gold is malleable (ii) Gold nucleus is heavy and produce large deflections of α – particles. |
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| 443. |
if an electron is revolving in its bohr orbit having bohr radius of `0.529Å`, then the radius of third orbit isA. `4234Å`B. `4496Å`C. `4.761Å`D. `5125nm` |
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Answer» Correct Answer - C As`r_n=n^2a_0`, Here, `a_0=0.529Å` and `n=3` `therefore r_(n=3)=(3)^2a_0=9xx0.529Å=4.761Å` |
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| 444. |
Find the expression of radius of an orbit of electron in terms of nucleus charge `Q_2`, tangential velocity v and mass of electron `m_e` ? |
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Answer» `F_e=F_c` `(KQ_1Q_2)/r^2=(m_ev^2)/r` `r=(KQ_1Q_2)/(m_eV^2)`, where `K=1/(4piepsilon_0)` |
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| 445. |
A slow neutron strikes a nucleus of `._(92)U^(235)` splitting it into lighter nuclei of barium and krypton and releasing three neutrons. Write the corresponding nuclear reaction. Also calculate the energy released on this reaction Given `m(._(92)U^(235))=235.043933 "amu" m(._(0)n^(1))=1.008665` amu `m(._(56)Ba^(141))=140.917700 "amu" m(._(36)Kr^(92))=91.895400` amu |
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Answer» Correct Answer - `198.77` MeV |
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| 446. |
The wavelength interval between `k_alpha` line and `lambda_"min"` of continuous X-ray spectrum of metal becomes 2 times when operating voltage is charged from 10 kV to x kV. Find x. What will happen to wavelength interval between `k_alpha` line and `k_beta` line in this case ? (Z=28) |
| Answer» Correct Answer - 15.3 | |
| 447. |
For he ground state , the electron int eh H-atom has an angular momentum `=h`, accord-ing to the simple Bohr model. Angular momentum is a vector ans hence there will be infi-nitely many orbits with the vector pointing in alll possible direction . In actuality , this is not true,A. because bohr model gives incorrect values of angular momentumB. because only one of these would have a minimum energyC. angular momentum must be in the direction of spin of electronD. because electrons go around only in horizontal orbits |
| Answer» Correct Answer - A | |
| 448. |
An isolated hydrogen atom emits a photon of `10.2 eV`. (i) Determine the momentum of photon emitted (ii) Calculated the recoil momenium of the atom (ii) Find the kinetic energy of the recoil atom [Mass of proton `= m_(0) = 1.67 xx 10^(-27) kg`] |
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Answer» (i)Momentum of the photon is `P_1=L^2/C=(10.2xx1.6xx10^(-19))/(3xx10^8)=5.44xx10^(-27)` kg m/s (ii)Applying the momentum conservation `P_2=P_1=5.44xx10^(-27)` kg m/s (iii)`k=1/2mV^2` (V=recoil speed of atom , m=mass of hydrogen atom ) `k=1/2m(P/m)^2=P^2/"2m"` Substituting the value of the momentum of atom we get `k=((5.44xx10^(-27))^2)/(2xx1.67xx10^(-27))=8.86xx10^(-27)` J |
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| 449. |
(a)Use the Bohr model to determine the ionization energy of the `He^+` ion, which has a single electron. (b)Also calculate the maximum wavelength a photon can have to cause ionization. |
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Answer» Strategy:We want to determine the minimum energy required to lift the electron from its ground state and to barely reach the free stable at E=0. The ground state energy of `He^+` is given by `E=-(me^4Z^2)/(8epsilon_0^2h^2n^2)` with n=1 and Z=2 . (a)Since all the symbols in above equation are the same as for the calculation for hydrogen , except that Z is 2 instead of 1 , we see that `E_1` will be `Z^2=2^2=4` times the `E_1` for hydrogen. That is `E_1` =4(-13.6 eV)=-54.4 eV Thus, to ionize the `He^+` ion should require 54.4 eV, and this value agrees with experiment. (b)The maximum wavelength photon that can cause ionization will have energy hf=54.4 eV and wavelength `lambda=c/f="hc"/"hf"` `=((6.63xx10^(-34) J.s)(3.00xx10^8 m//s))/((54.4 eV)(1.60xx10^(-19) J//eV))` =22.8 nm If `lambda gt 22.8` nm , ionization cannot occur. |
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| 450. |
A hydrogen like atom with atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition ot quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is – 13. 6 eV. |
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Answer» The energy released during de-excitation in hydrogen like atoms is given by : `E_(n_2)-E_(n_1)=(me^4)/(8epsilon _0^2h^2)[1/n_1^2 -1/n_2^2]Z^2` Energy released in de-excitation will be maximum if transition takes place from nth energy level to ground state. `i.e., E_(2n)-E_1 =13.6 [1/1^2-1/(2n)^2]Z^2=204` eV....(i) and also `E_(2n)-E_n =13.6 [1/n^2-1/(2n)^2]Z^2=40.8` eV....(ii) Taking ratio of (i) to (ii) , we will get `(4n^2-1)/3=5 rArr n^2=4` `rArr` n=2 Putting n=2 in equation (i) we get `Z^2=(204xx16)/(13.6xx15)` `rArr` Z=4 `because E_n=-13.6 Z^2/n^2` `rArr E_1=-13.6xx4^2/1^2=-217.6` eV = ground state energy `DeltaE ` is minimum if transition will be form 2n to 2n-1. i.e., between last two adjacent energy levels. `therefore DeltaE_"min"=E_"2n"-E_"2n-1"=13.6 [1/3^2-1/4^2] 4^2` =10.57 eV is the minimum amount of energy released during de-excitation. |
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