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1. Bohr's theory is not applicable to which of the following?
a) H
b) H⁺
c) He⁺
d) Li²⁺
✅ Correct Answer: (b)
One of the limitations of Bohr's atomic model is that it does not explain the spectra of multi-electron atoms. Bohr's theory is applicable to hydrogen like atoms or hydrogenic ions (single electron system). All the given species like H, He⁺ and Li²⁺ are isoelectronic and have only one electron. Their electronic configurations are same and so their spectra is explained by Bohr's atomic model. But H⁺ has no electron at all and hence cannot form spectrum.
One of the limitations of Bohr's atomic model is that it does not explain the spectra of multi-electron atoms. Bohr's theory is applicable to hydrogen like atoms or hydrogenic ions (single electron system). All the given species like H, He⁺ and Li²⁺ are isoelectronic and have only one electron. Their electronic configurations are same and so their spectra is explained by Bohr's atomic model. But H⁺ has no electron at all and hence cannot form spectrum.
2. Nitrogen has the electronic configuration 1s² 2s² 2pₓ¹ 2pᵧ¹ 2pz¹ and 1s² 2s² 2pₓ² 2pᵧ¹. This is determined by
a) Aufbau principle
b) Pauli’s rule
c) Hund's rule
d) n+l rule
✅ Correct Answer: (c)
The filling of electrons in degenerate orbitals like p, d and f is governed by Hund’s rule of maximum electron multiplicity accordingly electrons in degenerate orbitals tend to remain singly with same spin until all of the degenerate orbitals become half-filled then pairing of electrons are allowed in them.
The filling of electrons in degenerate orbitals like p, d and f is governed by Hund’s rule of maximum electron multiplicity accordingly electrons in degenerate orbitals tend to remain singly with same spin until all of the degenerate orbitals become half-filled then pairing of electrons are allowed in them.
3. Quantum number values for 3s orbital are
a) n = 0, l = 1
b) n=1, l=0
c) n=3, l=1
d) n=3, l=0
✅ Correct Answer: (d)
In the notation 3s (nl), 3 stands for n (principal quantum number) while s represent l (azimuthal quantum number). For 3s orbital, n =3 and l=0.
In the notation 3s (nl), 3 stands for n (principal quantum number) while s represent l (azimuthal quantum number). For 3s orbital, n =3 and l=0.
4. The radius of first orbit of hydrogen atom is
a) 529Å
b) 52.9 Å
c) 5.29 Å
d) 0.529 Å
✅ Correct Answer: (d)
The radius of first orbit of hydrogen atom is 0.529 × 10⁻¹⁰m or 0.529 × 10⁻⁸ cm or 0.529 Å or 0.0529 nm or 52.9 pm.
The radius of first orbit of hydrogen atom is 0.529 × 10⁻¹⁰m or 0.529 × 10⁻⁸ cm or 0.529 Å or 0.0529 nm or 52.9 pm.
5. Line spectrum is used as a tool for the identification of
a) Colors
b) Electrons
c) Elements
d) Molecules
✅ Correct Answer: (c)
Samples of same element always produces same characteristic line spectrum. Each element emit of light of specific wavelength therefore the number of lines and the distance between them depends upon the nature of element, so line spectra is used as “Finger Print” for the identification of elements. For example, line spectrum of sodium contains two yellow coloured lines separated by a definite distance.
Samples of same element always produces same characteristic line spectrum. Each element emit of light of specific wavelength therefore the number of lines and the distance between them depends upon the nature of element, so line spectra is used as “Finger Print” for the identification of elements. For example, line spectrum of sodium contains two yellow coloured lines separated by a definite distance.
6. The shape of orbital for which l = 0 is
a) Spherical
b) Dumbbell
c) Double dumbbell
d) Complicated
✅ Correct Answer: (a)
Its values show shape of orbitals. It has values l = 0 to (n – 1). e.g. l = 0, used for s-orbital, spherical in shape, l = 1, used for p-orbital, dumb-bell in shape l = 2, used for d-orbital, double dumb-bell in shape, l = 3, used for f-orbital, complicated shape
Its values show shape of orbitals. It has values l = 0 to (n – 1). e.g. l = 0, used for s-orbital, spherical in shape, l = 1, used for p-orbital, dumb-bell in shape l = 2, used for d-orbital, double dumb-bell in shape, l = 3, used for f-orbital, complicated shape
7. When 4d orbital is filled, the next electron enter into
a) 5s
b) 5p
c) 5d
d) 6s
✅ Correct Answer: (b)
According to n+ l rule, electrons are filled in various orbitals in the increasing order of (n+l) value. Those orbitals which have lower value of (n+l) value filled first. In case, if two orbitals which have same (n+l) value, then orbital having lower ‘n’ value will be filled first. The n+l value of 4d orbital is 4+2 =6. The next orbital to be filled must have n+l value either equal to 6 or greater than 6. The n+l value of 5d orbital is 5+2 =7, for 5s it is 5+0=5, for 5p it is 5+1=6 and for 6s it is 6+0=6. Since 5p and 6s orbitals have same n + l value, hence the electron first go to that orbital which has least n value i.e. 5p.
According to n+ l rule, electrons are filled in various orbitals in the increasing order of (n+l) value. Those orbitals which have lower value of (n+l) value filled first. In case, if two orbitals which have same (n+l) value, then orbital having lower ‘n’ value will be filled first. The n+l value of 4d orbital is 4+2 =6. The next orbital to be filled must have n+l value either equal to 6 or greater than 6. The n+l value of 5d orbital is 5+2 =7, for 5s it is 5+0=5, for 5p it is 5+1=6 and for 6s it is 6+0=6. Since 5p and 6s orbitals have same n + l value, hence the electron first go to that orbital which has least n value i.e. 5p.
8. Which of the following is not an iso electronic pair?
a) Na⁺ and Ne
b) Na⁺ and F⁻
c) Na and Ca
d) Na⁺ and Mg²⁺
✅ Correct Answer: (c)
The species having same number of total electrons and hence have same electronic configuration are called isoelectronic pair. Na⁺ and F⁻ are isoelectronic pair as both of them contains 10 electrons.
The species having same number of total electrons and hence have same electronic configuration are called isoelectronic pair. Na⁺ and F⁻ are isoelectronic pair as both of them contains 10 electrons.
9. Balmer series appears in the hydrogen spectrum if electron jumps from any appropriate higher energy orbit to
a) Second orbit
b) Third orbit
c) Fourth orbit
d) Fifth orbit
✅ Correct Answer: (a)
Balmer noted first that the Hydrogen spectrum consisted of some well-defined discrete lines in the visible region (i.e. having between 4000 to 7000Å) of spectrum. This series is obtained by the transition of electrons from any higher orbit (n₂ = 3, 4, 5, 6, 7 …. ∞) to 2nd orbit (n1 = 2).
Balmer noted first that the Hydrogen spectrum consisted of some well-defined discrete lines in the visible region (i.e. having between 4000 to 7000Å) of spectrum. This series is obtained by the transition of electrons from any higher orbit (n₂ = 3, 4, 5, 6, 7 …. ∞) to 2nd orbit (n1 = 2).
10. In 1935 A.D. James Chadwick was awarded Nobel Prize because
a) He discovered proton
b) He discovered neutron
c) He determined the radius of hydrogen atom
d) He gave the rules for electronic configuration
✅ Correct Answer: (b)
James Chadwick won the 1935 Nobel Prize in Physics for the discovery of the neutron.
James Chadwick won the 1935 Nobel Prize in Physics for the discovery of the neutron.
11. The e/m ratio of cathode rays (electrons) is:
a) 1.7588 × 10¹¹C/kg
b) 1.7588 × 10⁸ C/g
c) Both a and b
d) None of them
✅ Correct Answer: (c)
The e/m ratio for cathode rays is constant. It is equal to 1.76 ×10¹¹ C/kg 1.76 × 10⁸ C/g. It is independent of the nature of gas in discharge tube and nature of cathode or electrodes of tube. This shows that an electron is a universal fundamental particle. It is equal to the ratio of e/m for an electron as cathode rays consist of electrons. e/m ratio = Charge on electron/mass of electron = 1.6 × 10⁻¹⁹C/9.1 × 10⁻³¹kg = 1.758820 × 10¹¹ C/kg.
The e/m ratio for cathode rays is constant. It is equal to 1.76 ×10¹¹ C/kg 1.76 × 10⁸ C/g. It is independent of the nature of gas in discharge tube and nature of cathode or electrodes of tube. This shows that an electron is a universal fundamental particle. It is equal to the ratio of e/m for an electron as cathode rays consist of electrons. e/m ratio = Charge on electron/mass of electron = 1.6 × 10⁻¹⁹C/9.1 × 10⁻³¹kg = 1.758820 × 10¹¹ C/kg.
12. Which gas has a highest value of e/m ratio?
a) Hydrogen
b) Helium
c) Nitrogen
d) Oxygen
✅ Correct Answer: (a)
Since the e/m ratio depends upon the mass of particle, therefore, element with least mass would have highest e/m ratio. H is the lightest element, so e/m ratio of positive particle of H would be highest.
Since the e/m ratio depends upon the mass of particle, therefore, element with least mass would have highest e/m ratio. H is the lightest element, so e/m ratio of positive particle of H would be highest.
13. These are helium nuclei with double positive charge and atomic mass of 4 amu
a) Alpha rays
b) Beta-rays
c) Gamma rays
d) X-rays
✅ Correct Answer: (a)
Alpha rays or alpha particles are identical to helium nuclei. They are doubly ionized helium atom with double positive charge and mass number 4. Alpha rays are in fact stream of helium nuclei.
Alpha rays or alpha particles are identical to helium nuclei. They are doubly ionized helium atom with double positive charge and mass number 4. Alpha rays are in fact stream of helium nuclei.
14. The phenomenon of radioactivity was discovered by a French professor, Henry Becquerel in 1896 A.D. while working on uranium mineral called
a) Pitch-blende
b) Zinc sulphide
c) Fluorspar
d) Positron
✅ Correct Answer: (a)
Pitch-blende is a uranium mineral with composition U₃O₈.
Pitch-blende is a uranium mineral with composition U₃O₈.
15. Marie Currie and her husband, Pierre Currie isolated and separated two new radioactive elements named
a) Radium & radon
b) Radium & actinium
c) Radium & polonium
d) None of them
✅ Correct Answer: (c)
Radium and polonium are discovered by Marie Currie and her husband, Pierre Currie.
Radium and polonium are discovered by Marie Currie and her husband, Pierre Currie.
16. The ionization power of alpha rays is ………… times greater than beta rays.
a) 10,000
b) 1000
c) 100
d) None of them
✅ Correct Answer: (c)
The ionization power of alpha rays is Highest due to their greatest mass. Their ionization power is 100 times greater than beta rays and 10,000 times greater than gamma rays.
The ionization power of alpha rays is Highest due to their greatest mass. Their ionization power is 100 times greater than beta rays and 10,000 times greater than gamma rays.
17. Which isotope is used to measure the age of fossils and artefacts (archeology)?
a) Carbon-12
b) Iodine-131
c) Carbon-14
d) None of them
✅ Correct Answer: (c)
Carbon-14 is used to measure the age of fossils and artefacts (archeology).
Carbon-14 is used to measure the age of fossils and artefacts (archeology).
18. The wavelength of gamma rays is …………than most of the X-rays:
a) Shorter
b) Larger
c) Equal
d) None of them
✅ Correct Answer: (a)
The wavelength of gamma rays is shorter than most of the X-rays (wavelength of X-rays ranges between 10⁻⁸ to 10⁻¹⁰m). Gamma rays have a wavelength range below 100 pm (10⁻¹⁰to 10⁻¹⁴m). They are the most energetic form of electromagnetic radiation in a range greater than 100 keV.
The wavelength of gamma rays is shorter than most of the X-rays (wavelength of X-rays ranges between 10⁻⁸ to 10⁻¹⁰m). Gamma rays have a wavelength range below 100 pm (10⁻¹⁰to 10⁻¹⁴m). They are the most energetic form of electromagnetic radiation in a range greater than 100 keV.
19. Gamma rays are emitted as
a) Protons
b) Electrons
c) Neutrons
d) Photons
✅ Correct Answer: (d)
Gamma rays are photons emitted by an excited nuclei returning to its ground state.
Gamma rays are photons emitted by an excited nuclei returning to its ground state.
20. Which radioactive rays have greater penetrating power?
a) Alpha rays
b) Beta-rays
c) Gamma rays
d) None of them
✅ Correct Answer: (c)
The ability of radiation to pass through matter is expressed in terms of penetration power. In general, the greater mass present, the greater the ionizing power and the lower the penetration power. The penetrating power of radiation depends on the energy. Since gamma rays are most energetic, they have the highest penetration power. Order of penetrating power: γ > β > α
The ability of radiation to pass through matter is expressed in terms of penetration power. In general, the greater mass present, the greater the ionizing power and the lower the penetration power. The penetrating power of radiation depends on the energy. Since gamma rays are most energetic, they have the highest penetration power. Order of penetrating power: γ > β > α
21. Which of the following pairs have identical values of e/m?
a) A proton and a neutron
b) A proton and deuterium
c) Deuterium and an α-particle
d) An electron and γ-rays
✅ Correct Answer: (c)
e/m ratio depends on charge and mass. Deuterium and an alpha particle has an identical ratio of e/m ratio. The e/m ratio of the alpha particle is 4.82245 × 10⁷ C/kg which is identical with deuterium. e/m ratio of deuterium = +½. e/m ratio of α-particle = + ²/₄ = +½.
e/m ratio depends on charge and mass. Deuterium and an alpha particle has an identical ratio of e/m ratio. The e/m ratio of the alpha particle is 4.82245 × 10⁷ C/kg which is identical with deuterium. e/m ratio of deuterium = +½. e/m ratio of α-particle = + ²/₄ = +½.
22. The wavelength of X-rays ranges
a) 0.1 to 10Å
b) 0.1 to 100 Å
c) 0.2 to 10Å
d) None of them
✅ Correct Answer: (a)
Wavelength of X-rays ranging from 0.01 to 10 nanometers (10⁻⁸ to 10⁻¹²meter) or 0.1 to 10Å. Frequencies of X-rays ranging from 3 × 10¹⁹ Hz to 3 × 10¹⁶ Hz. Energies of X-rays ranging from in the range 100 eV to 100 keV.
Wavelength of X-rays ranging from 0.01 to 10 nanometers (10⁻⁸ to 10⁻¹²meter) or 0.1 to 10Å. Frequencies of X-rays ranging from 3 × 10¹⁹ Hz to 3 × 10¹⁶ Hz. Energies of X-rays ranging from in the range 100 eV to 100 keV.
23. Spectral line of short wavelength of X-rays is called
a) K-series
b) L-series
c) M-series
d) None of them
✅ Correct Answer: (a)
Spectral line of short wavelength of X-rays is called K-series. Spectral line of longer wavelength of X-rays is called L-series.
Spectral line of short wavelength of X-rays is called K-series. Spectral line of longer wavelength of X-rays is called L-series.
24. In 1913, Henry Moseley studied the different wavelengths (or frequencies) of X-rays produced from anodes of 38 different metals from
a) aluminium to gold
b) Calcium to gold
c) aluminium to tin
d) None of them
✅ Correct Answer: (a)
In 1913, Henry Moseley studied the different wavelengths (or frequencies) of X-rays produced from anodes of 38 different metals from aluminium to gold.
In 1913, Henry Moseley studied the different wavelengths (or frequencies) of X-rays produced from anodes of 38 different metals from aluminium to gold.
25. Spectrum given due to the transition of electron from M to L shell is
a) Absorption
b) Emission
c) Continuous
d) None of them
✅ Correct Answer: (b)
Electronic transition from high-energy orbit to low energy orbit in atom gives emission spectrum. Electronic transition from low energy orbit to high-energy orbit in atom gives absorption spectrum.
Electronic transition from high-energy orbit to low energy orbit in atom gives emission spectrum. Electronic transition from low energy orbit to high-energy orbit in atom gives absorption spectrum.
26. The square root of the frequency is directly proportional to the atomic number of an element. This is the statement of
a) Henry’s law
b) Moseley law
c) Hund’s rule
d) None of them
✅ Correct Answer: (b)
According to Moseley law, the square root of the frequency is directly proportional to the atomic number of an element i.e. √v = a (Z-b) OR √v = aZ -ab
According to Moseley law, the square root of the frequency is directly proportional to the atomic number of an element i.e. √v = a (Z-b) OR √v = aZ -ab
27. Violet colour has a wavelength of
a) 4000-4200 Å
b) 4200-4600 Å
c) 4600-5000 Å
d) None of them
✅ Correct Answer: (a)
Violet colour has the shortest wavelength among all colours in visible region. The violet color in the visible spectrum typically corresponds to wavelengths in the range of approximately 380 nm to 450 nm, which is equivalent to 3800 Å to 4500 Å. In this case, the range of 4000-4200 Å (option a) falls within the violet range. Here’s a rough breakdown of visible light wavelength ranges: Violet: 380-450 nm (or 3800-4500 Å), Blue: 450-495 nm, Green: 495-570 nm, Yellow: 570-590 nm, Orange: 590-620 nm, Red: 620-750 nm
Violet colour has the shortest wavelength among all colours in visible region. The violet color in the visible spectrum typically corresponds to wavelengths in the range of approximately 380 nm to 450 nm, which is equivalent to 3800 Å to 4500 Å. In this case, the range of 4000-4200 Å (option a) falls within the violet range. Here’s a rough breakdown of visible light wavelength ranges: Violet: 380-450 nm (or 3800-4500 Å), Blue: 450-495 nm, Green: 495-570 nm, Yellow: 570-590 nm, Orange: 590-620 nm, Red: 620-750 nm
28. 1 Å is equal to:
a) 10⁻¹⁰m
b) 10⁻⁸ cm
c) 10⁻ nm or 10² pm
d) All of them
✅ Correct Answer: (d)
1 Ångström (1 Å) is a unit of length commonly used to express atomic and molecular dimensions. It is equivalent to: 1 Å = 10⁻¹⁰m = 10⁻⁸ cm = 10⁻¹ nm = 10² pm
1 Ångström (1 Å) is a unit of length commonly used to express atomic and molecular dimensions. It is equivalent to: 1 Å = 10⁻¹⁰m = 10⁻⁸ cm = 10⁻¹ nm = 10² pm
29. The value of Plank’s constant is:
a) 6.625 × 10⁻³²J-s
b) 6.625 × 10‒36 J-s
c) 6.625× 10⁻²⁸J-s
d) 6.625 × 10⁻³⁴J-s
✅ Correct Answer: (d)
The value of Plank’s constant is 6.625 × 10⁻³⁴J-s or 6.625 × 10⁻²⁷ergs-s
The value of Plank’s constant is 6.625 × 10⁻³⁴J-s or 6.625 × 10⁻²⁷ergs-s
30. The no of waves per unit distance is called:is:
a) Wave frequency
b) Wavelength
c) Amplitude
d) Wave number
✅ Correct Answer: (d)
Wave number is the no of waves per unit distance. It is the inverse of reciprocal of wavelength.
Wave number is the no of waves per unit distance. It is the inverse of reciprocal of wavelength.
31. The photons of which one of the following colours are more energetic than the blue colour?
a) Red
b) Indigo
c) Orange
d) Yellow
✅ Correct Answer: (b)
In the visible spectrum, which is represented by the acronym VIBGYOR (Violet → Indigo → Blue → Green → Yellow → Orange → Red), the energy of light increases as you move from Red to Violet (i.e. from right to left) because photon energy is inversely proportional to wavelength (E=h/cλ):Violet light has the highest energy, while Red has the lowest energy. Indigo lies before blue (shorter wavelength, higher frequency), so its photons are more energetic than blue. This is because, as we move from the red end (lower energy) to the violet end (higher energy) of the spectrum, the energy of light increases. Red, orange, and yellow lie to the right of blue, meaning they have lower energy photons. ✨ This makes Indigo the correct choice. The order of energy from high to low in the visible spectrum is: Violet > Indigo > Blue > Green > Yellow > Orange > Red
In the visible spectrum, which is represented by the acronym VIBGYOR (Violet → Indigo → Blue → Green → Yellow → Orange → Red), the energy of light increases as you move from Red to Violet (i.e. from right to left) because photon energy is inversely proportional to wavelength (E=h/cλ):Violet light has the highest energy, while Red has the lowest energy. Indigo lies before blue (shorter wavelength, higher frequency), so its photons are more energetic than blue. This is because, as we move from the red end (lower energy) to the violet end (higher energy) of the spectrum, the energy of light increases. Red, orange, and yellow lie to the right of blue, meaning they have lower energy photons. ✨ This makes Indigo the correct choice. The order of energy from high to low in the visible spectrum is: Violet > Indigo > Blue > Green > Yellow > Orange > Red
32. Centimeter inverse is the unit of:
a) Boltzman constant
b) Gas constant
c) Rydberg’s constant
d) Plank’s constant
✅ Correct Answer: (c)
The Rydberg constant (Rң) is used in atomic spectroscopy to calculate the wavelengths of spectral lines. Its unit is reciprocal length i.e. Centimeter inverse or meter inverse: Rң ≈ 1.097×10⁷ m⁻¹ or 1.097×10⁵ cm⁻¹
The Rydberg constant (Rң) is used in atomic spectroscopy to calculate the wavelengths of spectral lines. Its unit is reciprocal length i.e. Centimeter inverse or meter inverse: Rң ≈ 1.097×10⁷ m⁻¹ or 1.097×10⁵ cm⁻¹
33. The spectrum of molecular form of the substance is called
a) Band spectrum
b) Line spectrum
c) Absorption spectrum
d) All of them
✅ Correct Answer: (a)
Band spectra is the name given to groups of lines so closely spaced that each group appears to be a band. Band spectra, or molecular spectra, are produced by molecules radiating their rotational or vibrational energies
Band spectra is the name given to groups of lines so closely spaced that each group appears to be a band. Band spectra, or molecular spectra, are produced by molecules radiating their rotational or vibrational energies
34. The spectrums have dark spaces is called:
a) Absorption spectrum
b) Continuous spectrum
c) Emission spectrum
d) Atomic spectrum
✅ Correct Answer: (a)
Absorption spectrum have dark spaces. An absorption spectrum is produced when white light passes through a cold gas or solution. Certain wavelengths are absorbed by the atoms or molecules, leaving dark lines or bands in the spectrum.
Absorption spectrum have dark spaces. An absorption spectrum is produced when white light passes through a cold gas or solution. Certain wavelengths are absorbed by the atoms or molecules, leaving dark lines or bands in the spectrum.
35. Line spectrum is produced by:
a) O₂
b) H
c) H₂
d) N₂
✅ Correct Answer: (b)
Elements form line spectrum, therefore, the line spectrum is associated with atomic hydrogen. A line spectrum is produced when an excited atom emits radiation at specific discrete wavelengths. Hydrogen atom (H), being a single electron system, produces a characteristic line spectrum (Balmer, Lyman, Paschen series, etc.).Molecules like O₂, H₂, and N₂ produce band spectra instead, due to vibrational and rotational transitions.
Elements form line spectrum, therefore, the line spectrum is associated with atomic hydrogen. A line spectrum is produced when an excited atom emits radiation at specific discrete wavelengths. Hydrogen atom (H), being a single electron system, produces a characteristic line spectrum (Balmer, Lyman, Paschen series, etc.).Molecules like O₂, H₂, and N₂ produce band spectra instead, due to vibrational and rotational transitions.
36. Electromagnetic radiations produced from nuclear reactions are
a) X-rays
b) Gamma rays
c) Beta rays
d) Alpha rays
✅ Correct Answer: (b)
Both X-rays and gamma rays are electromagnetic radiations. However, gamma rays are produced from disintegration of atomic nuclei through nuclear reactions while X-rays are produced due to electronic transitions.
Both X-rays and gamma rays are electromagnetic radiations. However, gamma rays are produced from disintegration of atomic nuclei through nuclear reactions while X-rays are produced due to electronic transitions.
37. The velocity of photon is
a) Dependent on its source
b) Equal to square of its amplitude
c) Dependent on its wavelength
d) Independent of its wavelength
✅ Correct Answer: (d)
The velocity of photon is independent of its wavelength. A photon is a quantum of electromagnetic radiation. All photons, regardless of their wavelength or frequency, travel at the speed of light in vacuum (c = 3×10⁸ m/s). The energy of a photon depends on its frequency (E=hν) or wavelength (E=h/cλ), but its velocity does not. Therefore, photon velocity is independent of wavelength, frequency, amplitude, or source.
The velocity of photon is independent of its wavelength. A photon is a quantum of electromagnetic radiation. All photons, regardless of their wavelength or frequency, travel at the speed of light in vacuum (c = 3×10⁸ m/s). The energy of a photon depends on its frequency (E=hν) or wavelength (E=h/cλ), but its velocity does not. Therefore, photon velocity is independent of wavelength, frequency, amplitude, or source.
38. Which statement incorrectly describes the relation between photon energy and its wavelength, frequency, or wave number?
a) Greater the wavelength → greater is its energy
b) Greater the wavelength → smaller is its energy
c) Greater the wave number → greater is its energy
d) Greater the frequency → greater is its energy
✅ Correct Answer: (a)
Photon energy is given by:E=hν=h/cλ=hc⋅ν~. Wavelength: Longer wavelength → smaller energy.Wave number: Larger wave number → greater energy. Frequency: Higher frequency → greater energy. Thus, options (b), (c), and (d) are correct, while (a) is wrong.
Photon energy is given by:E=hν=h/cλ=hc⋅ν~. Wavelength: Longer wavelength → smaller energy.Wave number: Larger wave number → greater energy. Frequency: Higher frequency → greater energy. Thus, options (b), (c), and (d) are correct, while (a) is wrong.
39. Which nature of electron was explained by Bohr’s theory?
a) Particle
b) Wave
c) Dual
d) None of them
✅ Correct Answer: (a)
Bohr’s theory explained the particle nature of electron. The wave nature of electron was completely ignored.
Bohr’s theory explained the particle nature of electron. The wave nature of electron was completely ignored.
40. According to Bohr’s atomic model, the angular momentum of an electron in the nth orbit is quantized. Which of the following correctly expresses this quantization condition?
a) L=nh/2π
b) L=2π/h
c) L=2h/π
d) L=π/2h
✅ Correct Answer: (a) L=nh/2π
Bohr’s postulate states that the angular momentum of an electron in orbit is quantized: L=mvr=nh/2π,n=1,2,3,…This condition ensures that only certain discrete orbits are allowed, leading to quantized energy levels. Options (b), (c), and (d) are incorrect because they do not represent multiples of h/2π. Thus, the angular momentum is always an integral multiple of h/2π.
Bohr’s postulate states that the angular momentum of an electron in orbit is quantized: L=mvr=nh/2π,n=1,2,3,…This condition ensures that only certain discrete orbits are allowed, leading to quantized energy levels. Options (b), (c), and (d) are incorrect because they do not represent multiples of h/2π. Thus, the angular momentum is always an integral multiple of h/2π.
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🌙 صبح کے بعد تو کہاں شام کے بعد ہم کہاں
🌙 صبح کے بعد تو کہاں شام کے بعد ہم کہاں
🎶 یہ جو نوا گری کے ساتھ چیخ نہیں رہا ہوں میں
💭 سینہ نے نواز میں آخر اب اتنا دم کہاں
💭 سینہ نے نواز میں آخر اب اتنا دم کہاں
👀 اب یہ جبین و چشم و لب تجھ کو نظر نہ آئیں گے
🕊️ غور سے دیکھ لو ہمیں آج کے بعد ہم کہاں
🕊️ غور سے دیکھ لو ہمیں آج کے بعد ہم کہاں
🌟 اب تری جلوہ گاہ میں خود نگروں کا ذکر کیا
🌺 اب تری سیر گاہ میں عشوہ گیانِ رم کہاں
🌺 اب تری سیر گاہ میں عشوہ گیانِ رم کہاں
🔥 شوقِ فزوں کو آج شام، صاف جواب مل گیا
💫 اب وہ معاشِ خدمتِ گیسوئے خم بہ خم کہاں
💫 اب وہ معاشِ خدمتِ گیسوئے خم بہ خم کہاں
Tags
Aptitude test (MDCAT/ECAT)
Atomic Structure Practice Test
First year
FSC Chapter-wise MCQs
MDCAT ECAT FSC Atomic Structure
MDCAT ECAT FSC Chemistry Atomic Structure Quiz