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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.
| 1. |
The density of liquid mercury is 13.6 g `cm^(-3)` . How many moles of mercury are there in 1 litre of the metal ? (Atomic mass of Hg = 200.) |
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Answer» Mass of mercury (Hg) in grams = density ( g `cm^(-3)) xx` volume `(cm^(3))` `= 13.6 g cm^(-3) xx 1000 cm^(3) = 13600 g .` `therefore` number of moles of mercury = `("mass of mercury in grams ")/("gram-atomic mass of mercury") = (13600 g)/(200 g) = 68.` |
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| 2. |
The potential difference between the cathode. And the target electrod in a coolidge tube is 24.75 kV. The minimum wavelength of the emitted X-rays isA. `0.1 Å`B. `0.5 Å`C. `1 Å`D. `5 Å` |
| Answer» Correct Answer - B | |
| 3. |
What does each of the following represent ?1. 2CO22. 2H2S3. 5H2SO44. 6NaNO3 |
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Answer» 1. 2CO2: 2 molecules of carbon dioxide. 2. 2H2S: 2 molecules of hydrogen sulphide. 3. 5H2SO4: 5 molecules of sulphuric acid. 4. 6NaNO3 : 6 molecules of sodium nitrate. |
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| 4. |
Elements X and Y have 1 and 7 electrons in their outermost shell respectively.1. Which element will lose electron ? 2. Which element will gain electron ? 3. Which will form positive ion ? 4. Which will form negative ion ?5. What will be the charge present of the molecule XY after chemical combination ? |
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Answer» 1. X 2. Y 3. X 4. Y 5. electrically neutral |
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| 5. |
Write the names of the elements present in the following compounds.1. Common salt 2. Ammonia3. Sulphuric acid4. Glucose5. Sodium hydroxide 6. Acetic acid |
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Answer» 1. Common salt: Sodium, chlorine. 2. Ammonia: Nitrogen, hydrogen. 3. Sulphuric acid: Hydrogen, sulphur, oxygen. 4. Glucose: Carbon, hydrogen, oxygen. 5. Sodium hydroxide: Sodium, oxygen, hydrogen. 6. Acetic acid: Carbon, hydrogen, oxygen. |
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| 6. |
An atom has 13 protons and 14 neutrons.1. What is the atomic number of the atom ?2. How many electrons would be present in this atom ?3. What is the atomic mass of the atom ? |
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Answer» 1. The atomic number of the atom 13 2. 13 electrons would be present in this atom 3. The atomic mass of the atom 27 |
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| 7. |
State Dalton’s atomic theory. |
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Answer» 1. Matter is made up of tiny particles called atoms 2. Atoms are indivisible. 3. Atom can be neither be created nor destroyed. 4. Atoms of same element are identical in mass and properties 5. Atoms of different elements differ in mass and properties |
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| 8. |
Write electronic configuration of oxygen, fluorine, sodium, silicon, argon. |
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| 9. |
What does the word ‘atom’ mean ? |
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Answer» “atom” means‘indivisible” |
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| 10. |
Who originated the word ‘atom’ ? |
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Answer» Democritus originated the word ‘atom |
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| 11. |
How did Neils Bohr visualize the atom ? |
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Answer» Niels Bohr explained that atom is like a solar system with an empty space with a minute central nucleus, like Sun and electrons life planets revolve around it. |
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| 12. |
Define :1. An Atom2. Atomic number3. Mass number4. Atomic weight |
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Answer» 1. An Atom: Smallest particle of an element that can exist and have properties of an element. 2. Atomic number : Number of protons present in the nucleus of an atom. 3. Mass number: Number of protons and neutrons present in the nucleus of an atom. 4. Atomic weight: It is the ratio that tells how many times an atom of an element is heavier than an atom of Hydrogen. |
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| 13. |
Define the following terms :1.Atom2.Molecule3.Radicals4.Valency 5.Periodic table |
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Answer» 1. Atom : An atom is the smallest indivisible unit of an OR Atom is the smallest unit of matter. 2. Molecule: Molecule is the smallest unit of a compound (or an element) which always has an independent existence. 3. Radicals: A radical is an atom of an element or a group of atoms of different elements that behaves as a single unit with a positive or negative charge on it. 4. Valency: It is the number of electrons donated or accepted by the valence shell of an atom during chemical combination. 5. Periodic table represents the tabular arrangement of elements in horizontal rows called periods and vertical columns called groups in order to classify the elements and their systematic study. |
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| 14. |
A hydrogen atom in ground state absorbs 12.09 eV of energy . The orbital angular momentum of the electronA. Is doublesB. Is halvedC. Remains sameD. Becomes three times |
| Answer» Correct Answer - D | |
| 15. |
The orbital angular momentum of electron in the `n_(1)th` shell of element of atomic number `Z_(1) is L_(1)` an the same in the `n_(2)th` shell of element of atomic number `Z_(2) is L_(2)` If `L_(2)gtL_(1)` thenA. `n_(2) gt n_(1)`B. `Z_(2) gtZ_(1)`C. `n_(2)Z_(2) gt n_(1)Z_(1)`D. Both (a) and (b) |
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Answer» Correct Answer - A `L=(nh)/(2pi)` it does not denpend on Z |
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| 16. |
What should be the maximum velocity of an electron orbiting around a hydrogen nucleus |
| Answer» Maximum velocity =`c/137` , where c=speed of light . | |
| 17. |
Total energy of electron in nth stationary orbit of hydrogen atom isA. `(-13.6)/(n)eV`B. `(-13.6)/(n^2)eV`C. `(-136)/(n)eV`D. `(-136)/(n^2)eV` |
| Answer» Correct Answer - B | |
| 18. |
Kinetic energy of electron in nth orbit is given byA. `(Rhc)/(2n^(2))`B. `(2Rhc)/(n)`C. `(Rhc)/(n)`D. `(Rhc)/(n^(2))` |
| Answer» Correct Answer - D | |
| 19. |
Potential energy `(PE_(n))` and kinetic energy `(KE_(n))` of electron in nth orbit are related asA. `PE_(n)=KE_(n)`B. `PE_(n)=-2KE_(n)`C. `PE_(n)=2KE_(n)`D. `PE_(n)=KE_(n)` |
| Answer» Correct Answer - B | |
| 20. |
The energy of the electron in the ground state of hydrogen atom is `-13.6 eV`. Find the kinetic energy and potential energy of electron in this state.A. 1.85 eVB. 13.6 eVC. 6.8 eVD. 3.4 eV |
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Answer» Correct Answer - B Kinetic energy of electron in ground state =-(Totoal energy in this state) `" " --(13.6 )eV=13.6 eV` |
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| 21. |
The energy of the electron in the ground state of hydrogen atom is `-13.6 eV`. Find the kinetic energy and potential energy of electron in this state. |
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Answer» As we khow, kinetic energy =-total energy `=-(-13.6) eV=13.6 eV` and potential energy =+(total energy ) `=2(-13.6) eV =-27.2 eV` |
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| 22. |
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon. |
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Answer» For ground state `n_(1)=1 and n_(2)=4` Energy of photon absorbed `E=E_(2)-E_(1)` `=(-13.6)/(n_(2)^(2))-((-13.6)/(n_(1)^(2)))eV=13.6 ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `=13.6 ((1)/(1^(2))-(1)/(4^(2)))eV` `=13.6xx(15)/(16) eV =(13.6xx15)/(16)xx1.6xx10^(-19) J` `E=2.04xx10^(-18)` Joule. From `E=(hc)/(lamda), lamda =(hc)/(E)=(6.6xx10^(-34)xx3xx10^(-8))/(2.04xx10^(-10))=9.7xx10^(-8) m` `lamda =97 nm` `v=(c)/(lamda)=(3xx10^(8))/(9.7xx10^(-8))=3.1xx10^(15) Hz.` |
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| 23. |
Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in `.^(1)H` and `.^(2)H`. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass `mu`, revolving around the nucleus at a distance equal to the electron -nucleus separation. Here `mu = m_(e) M//(m_(e)+M)`, where M is the nuclear mass and `m_(e)` is the electronic mass. Estimate the percentage difference in wavelength for the `1st` line of the Lyman series in `.^(1)H` and `.^(2)H`. (mass of `.^(1)H` nucleus is `1.6725 xx 10^(-27)` kg, mass of `.^(2)H` nucleus is `3.3374 xx 10^(-27)` kg, Mass of electron `= 9.109 xx 10^(-31) kg`.) |
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Answer» The total energy of the electron in the nth states of the hydrogen like atom of atomic number Z is given by `E_(n)^(.) = - (uZ^(2)e^(4))/(8epsi_(0)^(2)h^(2))((1)/(n^(2)))` where signs are as usual and the `mu` that occurs in the Bohr formula is the reduced mass of electron and proton. Let `mu_(H)` be the reduced mass of hydrogen and `mu_(D)` that of Deutrium. Then, the frequency of the `1st` Lyman line in hydrogen is `hv_(H) = (mu_(H)e^(4))/(8epsi_(0)^(2)h^(2)) (1-(1)/(4)) = (mu_(H)e^(4))/(8epsi_(0)^(2)h^(2)) xx 3/4` Thus, the wavelength of the transition is ` lambda_(H) = (3)/(4) (mu_(H)e^(4))/(4 8 epsi_(0)^(2)h^(3)c)`. The wavelength for the same line in Deutrium is `lambda_(D) = (3)/(4)(mu_(D)e^(4))/(8 epsi_(0)^(2)h^(3)c)` `:. Deltalambda = lambda_(D) - lambda_(H)` Hence, the percentage difference is `100 xx (Deltal)/(lambda_(H)) = (lambda_(D) - lambda_(H))/(lambda_(H)) xx 100 = (mu_(D) - mu_(H))/(mu_(H)) xx 100` `= ((m_(e)M_(D))/((m_(e)+M_(D)))- (m_(e)M_(H))/((m_(e)-M_(H))))/((m_(e)M_(H))/((m_(e)+M_(H))))xx 100` `=[((m_(e)+M_(H))/(m_(e)+M_(D)))(M_(D))/(M_(H))-1] xx 100` Since, `m_(e) lt lt M_(H) lt lt M _(D)` `(Deltalambda)/(lambda_(H)) xx 100=[(M_(H))/(M_(D))xx(M_(D))/(M_(H))((1+(m_(e))/(M_(H)))/(1+(m_(e))/(M_(D))))-1]xx100` `=[(1+(m_(e))/(M_(H)))(1+(m_(e))/(M_(D)))^(-1)-1]xx100=[1+(m_(e))/(M_(H))-(m_(e))/(M_(D))-1]xx100` [By binomial theorem, `(1+x)^(n) = 1+nx`is `|x| lt 1`] `~~m_(e)[(1)/(M_(H))-(1)/(M_(D))]xx100` `= 9.1 xx 10^(-31)[(1)/(1.6725 xx 10^(-27))- (1)/(3.3374 xx 10^(-27))] xx 100` `= 9.1 xx 10^(-4) [0.5979 - 0.2996] xx 100` `= 2.714 xx 10^(-2)%` |
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| 24. |
Assertion: Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate. Reason: According to classical physical all moving electrons radiate.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 bohr postulated that electron instead of revolving in any around the nucleus, revolve only in some specific orbits. These irbits are called the non-radiating orbits or the stationary orbits. The electrons revolving in these orbits do not radiate any energy. they radiate only when they go from one orbit to the next lower orbit. |
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| 25. |
The total energy of an electron in the first excited state of hydrogen atom is -3.4 eV. (a) What is kinetic energy of electron in this state? (ii) What is potential energy of electron in this state? (c) Which of the answers above would change if the choice of zero of potential energy is changed? |
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Answer» Total energy of the electron, E = - 3.4 eV Kinetic energy of the electron is equal to the negative of the total energy. implies K = - E `= - (-3.4) = + 3.4 eV` Hence, the kinetic energy of the electron in the given state is + 3.4 eV. (b) Potential energy (U) of the electron is equal tot he negative of twice of its kinetic energy. implies U = - 2 K `=-2 xx 3.4 = - 6.8 eV` Hence, the potential energy of a electron in the given state is = 6.8 eV. (c ) The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change. |
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| 26. |
An electron of a hydrogen like atom is in excited, state. If total energy of the electron is -4.6 eV, then evaluate (i) the kinetic energy and (ii) the de-Brogli wavelength of the electron . |
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Answer» Given ,total energy of the electron `E=-4.6 EV` (i) Kinetic energy of electron K=-(Total energy E,) `rArr " " =-E=-(-4.6)=4.6 eV` (ii) de- Brogile wavelength `lambda_(d)=(h)/(sqrt(2 mK))=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx4.6xx1.6xx10^(-19)))` `0.57xx10^(-9)m=0.57 nm` |
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| 27. |
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. What is the kinetic energy of the electron in this state ? |
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Answer» We know kinetic energy of electron `=(Kze^(2))/(2r)` and P.E of electron `=(-Kze^(2))/(r)` P.E.=-2 (kinetic energy) In this calculation electric potential and hence potential energy is zero at infinity. Total energy = PE + KE = -2KE + KE = -KE In the first excited state total energy= -3.4 eV. `therefore K.E=-(-3.4 eV)=+3.4 eV.` |
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| 28. |
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. What is the potential energy of the electron in this state ? |
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Answer» We know kinetic energy of electron `=(Kze^(2))/(2r)` and P.E of electron `=(-Kze^(2))/(r)` P.E.=-2 (kinetic energy) In this calculation electric potential and hence potential energy is zero at infinity. Total energy = PE + KE = -2KE + KE = -KE P. E of electron in this first excited state `= -2KE = -2 xx 3.4-= 6.8 eV. ` |
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| 29. |
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. Which of the answers above would change if the choice of the zero of potential energy is changed? |
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Answer» We know kinetic energy of electron `=(Kze^(2))/(2r)` and P.E of electron `=(-Kze^(2))/(r)` P.E.=-2 (kinetic energy) In this calculation electric potential and hence potential energy is zero at infinity. Total energy = PE + KE = -2KE + KE = -KE If zero of potential energy is changed, KE does not change and continues to be + 3.4 eV. However, the P.E. and total energy of the state would change with the choice of zero of potential energy. |
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| 30. |
A neutron of kinetic energy 65 eV collides inelastically with a singly ionised helium atom at rest. It is scattered at an angle `90^@` with respect to original direction. If the energy of scattered neutron is 6.36 eV, find the frequency of emitted radiation from the helium atom after suffering collision. |
| Answer» Correct Answer - `9.85xx10^15` Hz | |
| 31. |
If the electron in a hydrogen atom is raised to one of its excited energy states, the electron’s (A) potential energy increases and kinetic energy decreases (B) potential energy decreases and kinetic energy increases (C) potential energy increases wi(D) potential energy decreases but kinetic energy remains constant.th no change in kinetic energy |
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Answer» (A) potential energy increases and kinetic energy decreases |
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| 32. |
`O_(2)`molecules consists of two oxygen atoms. In the molecules , nuclear force between the nuclei of the two atomsA. is not important because nuclear forces are short-rangedB. is as imortant as electrostatic force for binding the two atomsC. cancels the repulsive electrostatic force between the nucleiD. is not important because oxygen nucleus have equal number of neutrons protons |
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Answer» Correct Answer - A In the molecules , nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged. |
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| 33. |
What are stable nuclei? What decides nuclear stability? What are the properties of nuclear force? |
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Answer» Those nuclei which for certain combinations of neutrons and protons do not spontaneously disintegrate are called stable nuclei. There are two aspects that decide the stability of a nucleus. Firstly, the existence of nuclear energy levels implies certain configurations to achieve potential energy minimum, and secondly, the balance of forces. Just like energy levels in atoms, nuclear energy levels are filled in sequence obeying the exclusion principle. Thus, there is a tendency for N to equal Z, or to have both even Z and even N. Properties of the nuclear force : (1) The nucleons in a nucleus are held .together by the attractive strong nuclear force. This force is much stronger than gravitational force and electromagnetic force. (2) Nucleons interact strongly only with their nearest neigh bours because the nuclear force has an extremely short range. Gravitational force and electromagnetic force are long range forces. They tend to zero only when the separation between two particles tends to infinity. (3) Inside a nucleus, this force appears to be the same between two protons, a proton and a neutron, and two neutrons. However, between two protons there is also Coulomb repulsion which has a much longer range and, therefore, has appreciable magnitude throughout the entire nucleus. In nuclei having 2 ≤ Z ≤ 83, with neutrons present, the nuclear force is strong enough to overcome the Coulomb repulsion. For light nuclei (A < 20), N ≥ Z, but is never smaller (except in \(^1_1H\) and \(^3_2H\)) However, with more than about 10 protons, an excess of neutrons is required to form a stable nucleus; for high atomic numbers, N/Z = 1.6. For Z > 83, even an excess of neutrons cannot prevent spontaneous disintegration and there are no stable nuclei. [Note : The strength of the nuclear force is evident from the nuclear binding energy.] |
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| 34. |
`O_(2)`molecules consists of two oxygen atoms. In the molecules , nuclear force between the nuclei of the two atomsA. is not important because nuelear forces are short-rangedB. is not important as electrostaic force for binding the two atomsC. cancels the replusive electrosatatic force between the nucleiD. is not important because oxygen nucleus have euqal number of netrons and protons |
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Answer» Correct Answer - A In the molecules, nuclear force between the nuclei of the two atom is not important because nuclear forces are short-ranged. |
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| 35. |
`O_(2)`molecules consists of two oxygen atoms. In the molecules , nuclear force between the nuclei of the two atomsA. is not important because nuclear forces are short-ranged.B. is as important as electrostatic force for binding the two atomsC. cancels the repulsive electrostatic force between the nucleiD. is not important because oxygen nucleus has equal number of neutrons and protons. |
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Answer» Correct Answer - A In the given oxygen molecule, nuclear force between the nuclei of two atoms is not important because nuclear forces being short ranged are confined only within one particular. The distance between the nuclei to two atoms is large. So nuclear forces between two nuclei is not effective. |
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| 36. |
A uranium nucleus (atomic number `92`, mass number `231`) emits an `alpha`-particle and the resultant nucleus emits a `beta`-particles. What are the atomic and mass numbers of the final nucleus? . |
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Answer» Correct Answer - `Z = 91, A = 234` |
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| 37. |
Monochromatic radiation of wavelength `lambda` are incident on a hydrogen sample in ground state. Hydrogen atoms absorb the light and subsequently emit radiations of 10 different wavelength . The value of `lambda` is nearly : |
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Answer» From nth energy state a total of `(n(-1))/(2)` lines are emitted. `:. " " (n(n-1))/(2)=10or n=5` Further , `" " E_(n)=(E_(1))/(n^(2))` `:." " E_(5)=(-13.6)/((5^(2)))=-0.544 eV` The energy needed to take a hydrogen atom form its ground state to n=5 is `DeltaE=E_(5)-E_(1)` `=-0.54-(-13.6)` =13.06 eV The corresponding wavelength of light, `lambda=(12375)/(DeltaE)Å=(12375)/(13.06)=947.5Å` |
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| 38. |
When an electron revolving in the ground state of a hydrogen atom jumps to its `2^"nd"` excited state.A. It frequency of revolution becomes `(3)^(2//3)` timesB. It frequency of revolution becomes `(3)^(-3)` timesC. Its speed becomes `(1/3)` timesD. Its speed becomes `(1/9)` times |
| Answer» Correct Answer - B::C | |
| 39. |
A hydrogen atom in its ground state is excited to the state of energy E3 by an electron colliding with it. The minimum energy that the colliding electron must have is (A) 10.2 eV (B) 12.09 eV (C) 12.5 eV (D) 13.6 eV. |
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Answer» (B) 12.09 eV |
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| 40. |
Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model ? |
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Answer» According to Bohr model electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as `L = (nh)/(2pi)` or `L prop n` |
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| 41. |
The speed of an electron in the `4^"th"` orbit of hydrogen atom isA. cB. `c/137`C. `c/2192`D. `c/548` |
| Answer» Correct Answer - D | |
| 42. |
Ionisation potential (IP) and ionisation (IE) are related aA. `IP=(IE)e`B. `IP=(IE)/(e)`C. `IE=(IP)/(e^(2))`D. `IP=(IE)/(e^(2))` |
| Answer» Correct Answer - B | |
| 43. |
Which one of the series of hydrogen spectrum is in the visible region ?A. Lyman seriesB. Balmer seriesC. Paschen seriesD. Brackett series |
| Answer» Correct Answer - B | |
| 44. |
The ratio of energies of hydrogen atom in its first excited state to third excited state isA. `1/4`B. `4/1`C. `3/4`D. `4/3` |
| Answer» Correct Answer - B | |
| 45. |
How many spectral lines are emitted by atomic hydrogen excited to the `n-th` energy level?A. nB. 2nC. `(n^2-n)/2`D. `(n^2+n)/2` |
| Answer» Correct Answer - C | |
| 46. |
Atomic hydrogen is excited to the `n^(th)` energy level . The maximum number of spectral lines which it can emit while returning to ground state, is:A. `(n(n+1))/(2)`B. `(n(n-1))/(2)`C. `(n(n-1)^(2))/(2)`D. `(n(n+1)^(2))/(2)` |
| Answer» Correct Answer - B | |
| 47. |
What is the maximum number of spectral line emitted by a hydrogen atom when it is in the third excited state ? |
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Answer» If n is the quantum number of highest energy level, then the total number of possible spectral line emitted is N = n(n - 1) / 2 Here third excited state means fourth energy level, i.e.n = 4 N = 4(4 - 1) / 2 = 6 |
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| 48. |
What is the maximum number of spectral line emitted by a hydrogen atom when it is in the fourth excited state ? |
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Answer» n = 5 N = n(n -1) / 2 = 5 x 4 / 2 N = 10 |
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| 49. |
What are hydrogenic atoms ? |
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Answer» Hydrogenic (Hydrogen like ) atoms are the atoms consisting of a nucleus with positive charge ‘+Ze’ and a single electron. Here ‘Z’ is atomic mass number and ‘e’ is the quantised unit of charge. E.g., singly ionised helium, doubly ionised lithium. |
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| 50. |
If elements with principal quantum number `ngt 4` were not allowed in nature, the number of possible elements would be:A. 32B. 60C. 18D. 4 |
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Answer» Correct Answer - B If all the elements having `ngt4` are removed the number of elements that will be present in the periodic table are calculated as n=1 represents K shell and the number of elements having K shell =2 [ in accoredance with `2n^(2)`] n=2 , repreents L shell and the number of elements having L shell =8 n=3 represent M shell and the number of elements having M shell =18 n=4 represents N shell and the number of elements having N shell =32 So, the number of elelments of elements having `nlt5` are `" " 2+8+18+32=60` |
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