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Q1. For the principle quantum no. n = 4; How many types of orbitals are there? How many electrons can be accommodated in the complete principle shell.

Solution

For n = 4, there are four possible values for l. They are; 0 .... s orbitals 1 .... p orbitals 2 .... d orbitals 3 .... f orbitals For each of these there are values for ml l=0, ml = 0 = 2 electrons l = 1, ml = -1, 0, +1 = 6 electrons l = 2, ml = -2, -1, 0, 1, 2 = 10 electros l = 3, ml = -3,-2, -1, 0, 1, 2,3 = 14 electrons Each orbital can accommodate 2 electrons hence total no. of electrons = 32

Q2. Write the ground state electron configuration of an atom of Tungsten. How many of the electrons in this atom have a principal quantum number of 4? How many electrons are in the valence shell of this atom?

Solution

Electronic configuration of Tungsten having at. no. 74 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁴ Total no. of possible electrons in 4^th Sub shell = 32 No. of electron in the valance shell = 2

Q3. A magnesium surface has a work function of 3.68 eV. Electromagnetic waves with a 215 nm wavelength strike the surface and eject photoelectrons. Calculate the energy of the photoelectrons in Joules and in electron volts.

Solution

E of 215 nm photon = h c / λ = (6.626 x 10^-34 J s) x (3.00 x 10⁸ m s^-1) / (215 x 10^-9 m) = 92.4 x 10^-19 J per photon which is equal to 5.77 eV Hence KE of photoelectrons = 2.09 eV = 3.35 x 10^-19 J

Q4. Which quantum number does not follow from the solutions of Schrodinger wave equation?

Solution

Spin quantum number does not follow from the solutions of Schrodinger wave equation.

Q5. Fill in the missing values or names of the following sets of quantum numbers.

n = 2, l = ?, ml = +1; name: 2p

n = ?, l = 0, ml = ?; name: 4s

n = 3, l = 1, ml = +1; name: ?

n = 3, l = ?, ml = 0; name: 3d

Solution

n = 2, l = 1, ml = +1; name: 2p

n = 4, l = 0, ml = 0; name: 4s

n = 3, l = 1, ml = +1; name: 3p

    d.  n = 3, l = 2, ml = 0; name: 3d

Q6. Calculate the wavelength of a bullet (mass 7.0 g) moving with a velocity of 1.1 x 10³ m s^-1.

Solution

The de Broglie wavelength associated with matter is found from λ = h / m v, where m is mass (kg) and v is velocity (m s-1).   For the bullet, m = 0.0070 kg and v = 1.1 x 10³ m s-1.   Using the de Broglie wavelength equation, λ = (6.626 x 10^-34 J s) / (0.0070 kg) x (1.1 x 10³ m s-1) = 8.6 x 10^-35 m

Q7. What is the effective mass in gms of one mole of photons that have a wavelength of 600 nm?

Solution

λ = 600 nm = 6.0 x 10^-7 m c = 3.0 X 10 ⁸ m/s   We know , from de Broglie relation ; λ = h/mv   Mass of photon = h/c λ   Substituting the values, we get , mass of photon = 3.6 x 10^-36 kg Hence , mass of one mole of photons = 2.2 x 10^-12 kg

Q8. How does Bohr’s theory account for stability of an atom?

Solution

According to Bohr, as long as an electron remains in a particular permitted circular orbit or stationary state, it neither emits nor absorbs energy. As a result, an electron can not spiral down towards the nucleus loosing energy continuously (as per Maxwell’s theory of electromagnetic radiation). This explains why atoms are stable and do not collapse due to electrostatic attraction between the nucleus and the electrons.

Q9. What is the maximum number of electrons in a given atom that can have n = 3, ml =1.

Solution

For n = 3, there are three possible values for l. They are; 0   s orbitals 1   p orbitals 2   d orbitals For each of these there are values for ml l = 0, ml = 0 l = 1, ml = -1, 0, +1 l = 2, ml = -2, -1, 0, 1, 2

Q10. Calculate the mass of a photon of electromagnetic radiation having wavelength 589 nm.

Solution

λ = 589 nm = 5.89 x 10^-7 m                         c = 3.0 X 10 ⁸ m/s We know , from de Broglie relation ; λ = h/mv(c for photon)   Hence, Mass of photon =  h/c λ Substituting the values, we get , mass of photon = 3.7x10^-36 kg

Q11. A helium neon laser emits light of 632.8 nm wavelength. What is the energy of the photon in this beam?

Solution

λ = 632.8 nm = 6.328 x 10^-9 m = 6.32 x 10^-7 m We Know,      E = h c / λ    = (6.626 x 10^-34 J s) x (3.00 x 10⁸ m s^-1) / (6.32 x 10^-7 m)    = 3.14 x 10^-19 J    = 6.32 e V

Q12. Two isotopes of calcium (atomic number 20) are ⁴⁰Ca and ⁴⁴Ca. For both ⁴⁰Ca and ⁴⁴Ca²⁺ give the following information: atomic number & mass number.

Solution

The atomic number is unique for each element, and is the same for all isotopes of an element and it does not depend on the number of neutrons present. Therefore, the atomic number for both isotopes is 20. Atomic number depends only on the number of protons present, so the charge on the ⁴⁴Ca²⁺ ion (which arises from the loss of two electrons) does not alter this. The mass number is the number of (protons + neutrons) present For ⁴⁰Ca this is 40. Note that the charge present on the ⁴⁴Ca²⁺ ion does not affect the composition of the nucleus, and the mass number is 44.  

Q13. Calculate the frequency and wavelength of a body of mass 1 mg moving with a velocity of 10 ms^-1.

Solution

Mass = 10 mg = 10 x 10^-6 kg We know , from de Broglie relation ; λ = h/mv                                                       = 6.63x10^-34/10^-6x10                                                       = 6.63 x 10^-29 m We also know that, v = c/ λ Or , v = 3.0 X 10⁸ / 6.63 x 10^-29 = 4.52 x 10²⁰ s^-1

Q14. What do you mean by spectra? What are the basic types of spectra?

Solution

When light is passed through a prism or a diffraction grating it produces a spectrum which is basically a band of radiations of different colors in the visible zone and invisible zones arranged in order wavelength. There are three types of basic spectra depending upon what kind of object is producing light. They are 1) Continuous spectra, 2) Absorption Spectra and 3) Emission Spectra.

Q15. List the colors in the visible spectrum from lowest wavelength to highest wavelength.

Solution

Lowest wavelength: violet, indigo, blue, green, yellow, orange, red: highest wavelength

Q16. Write the detailed electronic configurations for the atoms of the following elements: Mn & Zn

Solution

  Electronic configurations of Mn: 1s² 2s² 2p⁶ 3s2 3p⁶ 3d⁵ 4s² Electronic configurations of Zn: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²  

Q17. What do you mean by Quantum Numbers?

Solution

Quantum Numbers specify the properties of atomic orbitals and their electrons. Quantum numbers can be used to write electron configurations which show how electrons are most likely distributed around the nucleus There are four quantum numbers: principal quantum number orbital quantum number magnetic quantum number spin quantum number The principal quantum number (n) specifies the main energy levels around the nucleus. As n increases, the distance from the nucleus increases. Currently the values for n are 1 to 7 Orbital Quantum Number (l) indicates the shape of the orbital where the electron can be found These orbitals are called sub shells or sublevels The four most common orbital quantum numbers are given letter abbreviations l" values range from 0 to (n-1) Orbital Quantum Numbers: l= 0, s orbital l = 1, p orbital l = 2, d orbital l = 3, f orbital Magnetic Quantum Number (m) indicates the orientation of an orbital about the nucleus it tells which axis that sublevel is located on (x, y, or z axis) ml ranges from - 1 to 1 Spin Quantum Number (ms) indicates the two possible states of an electron within an orbital their Values are + 1/2 or -1/2

Q18. Calculate the energy of photon when an electron in hydrogen atom jumps from an energy level with n=3 to an energy level with n=2.

Solution

We know from Bohr’s theory, E_n = -13.6/n² eV where E_n is the energy of nth orbit of H-atom     and ∆ E = E₃ - E₂                     E₂ = 1312/4 kjmol^-1                   E₃ = 1312/9 kjmol^-1                  Or, ∆ E = 182.2 kjmol^-1

Q19. What is the photoelectric effect? Give an example.

Solution

The photoelectric effect is where a photon (packet of light) can eject an electron from a metal. The photon must be at least at the threshold energy for ejecting a electron from metal surface. For example sodium metal.

Q20. What is the maximum number of electrons that may be present in all the atomic orbitals with principal quantum number 3 and azimuthal quantum number 2?

Solution

The maximum number of electrons that may be present in all the atomic orbitals with principal quantum number 3 and azimuthal quantum number 2 is Ten.

Q21. In the Quantum Mechanical Model, what are electrons considered to be?

Solution

In the Quantum Mechanical Model electrons are considered to be clouds of electromagnetic waves which occupy the space around the nucleus in different energy levels. The space around the nucleus where probability of finding an electron is maximum is called an orbital.

Q22. What do you mean by Shell or energy level according to Bohr’s model of atom?

Solution

Out of infinite number of circular orbits possible around the nucleus of atom, electrons revolve only in certain definite or fixed circular orbits called stationary states. The term stationary state implies that the motion of electron in this orbit does not cause any change in the energy of the electron and the effect is as if they are stationery. According to Bohr each orbit is associated with a certain definite amount of energy. Stationary states are also known as or Shell or Bohr’s energy levels. They are denoted by K, L, M, N and so on or by integers n =1, 2, 3, 4 and so on.

Q23. For radiation which consists of photons of energy 5.0 x 10^-16 J, find the wave length in nm

Solution

E = 5.0 x 10^-16 J We Know,    λ = h c / E = (6.626 x 10^-34 J s) x (3.00 x 10⁸ m s^-1) / (5.0 x 10^-16 J) = 4.0 x 10^-10 m = 0.40 nm.

Q24. What are the three types of the Electromagnetic Spectrum?

Solution

The Electromagnetic Spectrum is divided into three portions for convenience of study: Infra red zone consisting of waves greater than visible zone Visible zone consists of seven distinct colored light waves which are sensed by our eyes Ultra-Violet Zone consisting of low wavelengths but high frequency.

Q25. What is the energy of electron in Bohr’s n^th orbit of hydrogen like atom?

Solution

Energy of electron in any orbit for hydrogen like atom like He⁺, Li²⁺ etc is given by the following expression E_n = -13.6z²/n² eV where E_n is the energy of nth orbit of H-like atom and z is At. No. of the element.

Q26. Which of the following atoms/ions are diamagnetic Zn, Cu²⁺, Cd, Ti²⁺, Cu⁺

Solution

Zn, Cd and Cu⁺ are diamagnetic as they do not contain unpaired electrons.

Q27. How are emission, absorption and continuous spectra different?

Solution

Emission spectra are produced by thin gases in which the atoms do not experience many collisions (because of the low density). The emission lines correspond to photons of discrete energies that are emitted when excited atomic states in the gas make transitions back to lower-lying levels. An absorption spectrum occurs when light passes through a cold, dilute gas and atoms in the gas absorb at characteristic frequencies; since the re-emitted light is unlikely to be emitted in the same direction as the absorbed photon, this gives rise to dark lines (absence of light) in the spectrum. A continuum spectrum results when the gas pressures are higher. Generally, solids, liquids, or dense gases emit light at all wavelengths when heated.

Q28. Calculate the energy of one mole of photons of radiation having wavelength of 200 nm.

Solution

We know,   E (Energy of Photon) = hc/ λ                                = 6.63 x 10^-34 x 3.0 x 10 8/ 200 x 10^-9                                = 9.94 x 10^-19 joules Energy of one mole of Photons = 6.022 x 10²³ x 9.94 x 10^-19                                              = 5.8x 10⁵ joules

Q29. How atomic spectra of Hydrogen and Hydrogen like atom are explained by Bohr’s theory?

Solution

According to Bohr, when an atom is subjected to high temperature or electric discharge, an electron in an atom may jump from its normal energy levels i.e. from ground state to some higher energy level or excited state by absorbing photon of radiation of definite energy. The lifetime of such excited electrons is short and hence they return to some lower energy level or even o their stable ground state level in one or more steps. During each such jumping back energy is emitted I form of a photon of light or electromagnetic radiation of definite wavelength and frequency depending upon the energy difference between two energy levels. In Hydrogen spectra, corresponding to each wavelength of such photon emitted, there appears a bright line.

Q30. The mass to charge ratio for A⁺ ion is 1.97 x 10^-7 kg C^-1. Calculate the mass of A atom.

Solution

Given: m/e = 1.97 x 10^-7   Hence, m = 1.97 x 10^-7 x 1.6 x 10^-19 = 3.16 x 10^-26 kg

Q31. How were Cathode rays discovered and by whom?

Solution

Cathode rays were first discovered by Julius Plucker .The Ideal condition for producing cathode rays in discharge tube are 0.01 mm Hg and 10,000 V potential.

Q32. Give the electron configurations of potassium and calcium. Which of the following will have the greatest first ionization energy? Which of the following will have the greatest second ionization energy?

Solution

Electron configurations are as follows; K (alkali metal); [Ar] 4s¹ Ca (alkali earth metal); [Ar] 4s²   The effective nuclear charge experienced by the outermost valence electrons on Ca is greater than that on K. This explains why Ca is actually smaller than K and why the first ionization energy of Ca would be greater than that for K. The second ionization energy however is much greater for K than for Ca, since in removing the second electron one must remove one of the core electrons from K.

Q33. Explain de Broglie's wave length?

Solution

De Broglie hypothesized that all matter can exhibit wave like properties. He derived an equation (λ=h/ mv) to describe the wavelength of a particle where h is Planck's constant, m is the mass of the particle, and v is its velocity.

Q34. Account for the fact that manganese has a greater number of oxidation states than zinc even when they have similar electronic configurations.

Solution

Mn is a transition element with an incompletely filled 3d shell. The completely filled 3d shell in zinc leads to stability and removal of the 4s electrons with only a relatively low ionization energy to form the 2+ ion is the only common oxidation state of zinc (+II). The filled 3d orbital cannot participate in bonding. For manganese, both the 3d and 4s sub shells are available for occupation by bonding electrons, so a wide range of oxidation states is observed.

Q35. For one atom of the isotope ³⁴X (atomic number 16) give the number of protons the number of electrons.

Solution

The number of protons equals atomic number, which is 16. We assume here that the atom in question is neutral, since no charges have been explicitly mentioned. In a neutral atom the number of electrons equals the number of protons, which is 16.

Q36. Explain where the different emission lines in the hydrogen spectrum come from?

Solution

The electron in the hydrogen atom can be excited to various energy levels by absorbing energy. As it relaxes back to a lower state, it releases energy. By relaxing from different levels, different amounts of energy will be released, which produce a different color of light.

Q37. Determine the maximum number of electrons that can be found in each of the following sub shells: 3s,3d and 4p

Solution

3s Each orbital can accommodate 2 electrons.  The number of orbitals in a sub shell is given by 2ℓ + 1.   An s sub shell (1 orbital) can hold 2 electrons. 3d A d sub shell (ℓ = 2) can hold 10 electrons. 4p A p sub shell (ℓ = 1) can hold 6 electrons.

Q38. Calculate the total no. spectral lines and total spectral lines in Balmer series if electron undergoes a transition from 7^th energy level to ground state.

Solution

Total no. of spectral lines = n(n-1)/2           Or, 7(7-1)/2 = 21   Total no. of spectral lines in Balmer series = 7-2 = 5 (this is because for Balmer series n₁= 2 and n₂ = 3, 4, 5…)

Q39. What is the difference between a 2p_x and a 2p_y orbital?

Solution

The orbitals have the same size (energy) and shape, but are oriented differently in space (different magnetic quantum numbers).  The 2p_x orbital lies on the x-axis while the 2p_y orbital lies along the y-axis.

Q40. What are isotope, isobars and isotones?

Solution

Isotopes are atoms of the same element having same atomic number but different mass numbers. They have similar chemical properties but different physical properties. Isobars are atoms of different elements having same mass numbers (i.e. the sum of their Protons and Neutrons are same). Isotones are atoms of different elements having same number of Neutrons.