Atomic Structure: Rutherford, Planck, Bohr, and Heisenberg PDF

JELET CHEMISTRY Atomic Structure: Rutherford atomic Model, Planck’s Quantum theory, Bohr Atomic Model, de Broglie dual nature, Heisenberg’s Uncertainty Principle Th is...

JELET CHEMISTRY Atomic Structure: Rutherford atomic Model, Planck’s Quantum theory, Bohr Atomic Model, de Broglie dual nature, Heisenberg’s Uncertainty Principle Th is PD F be lo 1. The discovery of Electron, Proton and Neutron ng s to cz 1.1 A Brief History a6 vy (s The existence of atoms has been proposed since the time of early Indian and Greek philosophers (400 ub ha B.C.). m 37 20 The atomic theory of matter was first proposed on a firm scientific basis by John Dalton in 1808. His @ theory, called Dalton’s atomic theory, regarded the atom as the ultimate particle of matter. Dalton’s gm ai atomic theory was able to explain the law of conservation of mass, law of constant composition and l. co m law of multiple proportion very successfully. ,6 29 7 70 1.2 Discovery of Sub-Atomic Particles 40 09 ) 1.2.1 Discovery of Electron In 1830, Michael Faraday showed that if electricity is passed through a solution of an electrolyte, chemical reactions occurred at the electrodes. In mid 1850s many scientists mainly Faraday began to study electrical discharge in partially evacuated tubes, known as cathode ray discharge tubes (Fig.1). When sufficiently high voltage is applied across the electrodes, current starts flowing through a Fig.1 A cathode ray discharge tube stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were called cathode rays or cathode ray particles. In the presence of electrical or magnetic field, the behavior of cathode rays is similar to that expected from negatively charged particles, suggesting that the cathode rays consist of negatively charged particles, called electrons. The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube. Thus, we can conclude that electrons are basic constituent of all the atoms. 1.2.2 Charge to Mass Ratio of Electron In 1897, British physicist J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (me) by using cathode ray tube. P a g e 1 | 15 JELET CHEMISTRY Greater the magnitude of the charge on the particle, greater is the interaction with the electric or magnetic field and thus greater is the deflection. Lighter the mass of the particle, greater the deflection By carrying out accurate measurements on the amount of deflections observed by the electrons on the electric field strength or magnetic field strength, Thomson was able to determine the value of e/me as: 𝑒 = 1.758820 × 1011 C kg–1 𝑚𝑒 Th is PD 1.2.3 Charge and Mass on the Electron F be lo R.A. Millikan (1868-1953) devised a method known as oil drop experiment (1906-14), to determine ng s the charge on the electrons. to cz He found the charge on the electron to be –1.6 × 10–19 C. a6 vy The mass of the electron (me) was determined by combining these results with Thomson’s value of (s ub e/me ratio. ha m 𝑒 1.602176 X 10−19 C 37 𝑚𝑒 = 𝑒 = 1.758820 X 1011 C kg−1 = 𝟗. 𝟏𝟎𝟗𝟒 𝑿 𝟏𝟎−𝟑𝟏 𝒌𝒈 20 @ 𝑚𝑒 gm ai l. 1.2.4 Discovery of Protons and Neutrons co m ,6 Electrical discharge carried out in the modified cathode ray tube led to the discovery of canal rays 29 7 carrying positively charged particles. 70 40 The smallest and lightest positive ion was obtained from hydrogen and was called proton. This 09 ) positively charged particle was characterized in 1919. Unlike cathode rays, mass of positively charged particles depends upon the nature of gas present in the cathode ray tube. Neutrons were discovered by Chadwick (1932). 2. Atomic Models The major problems before the scientists after the discovery of sub-atomic particles were: i. To account for the stability of atom ii. To compare the behavior of elements in terms of both physical and chemical properties iii. To explain the formation of different kinds of molecules by the combination of different atoms iv. To understand the origin and nature of the characteristics of electromagnetic radiation absorbed or emitted by atoms 2.1 Thomson Model of Atom J. J. Thomson, in 1898, proposed that an atom possesses a spherical shape (radius approximately 10–10 m) in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement (Fig.2) An important feature of this model is that the mass of the atom is assumed to be uniformly distributed over the atom Many different names are given to this model, for example, plum pudding, raisin pudding or watermelon. P a g e 2 | 15 JELET CHEMISTRY Th is PD F be lo ng s to cz a6 vy (s ub ha m 37 20 Fig.2 Thomson model of atom @ gm ai l. co m ,6 29 7 70 2.2 Rutherford’s Nuclear Model of Atom 40 09 ) 2.2.1 Experiment and Observation Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with –particles. Rutherford’s famous 𝛼–particle scattering experiment is represented in Fig.3. A. Rutherford’s scattering experiment B. Schematic molecular view of the gold foil Fig.3 Schematic view of Rutherford’s scattering experiment. When a beam of alpha (𝛼) particles is “shot” at a thin gold foil, most of them pass through without much effect. Some, however, are deflected. It was observed that: (i) Most of the 𝛼–particles passed through the gold foil undeflected. (ii) A small fraction of the 𝛼–particles was deflected by small angles. (iii) A very few 𝛼–particles (∼1 in 20,000) bounced back, that is, were deflected by nearly 180°. P a g e 3 | 15 JELET CHEMISTRY On the basis of the observations, Rutherford drew the following conclusions regarding the structure of atom: (i) Most of the space in the atom is empty as most of the 𝛼–particles passed through the foil undeflected. (ii) A few positively charged 𝛼–particles were deflected. The deflection must be due to enormous Th is PD repulsive force showing that the positive charge of the atom is not spread throughout the F atom but concentrated in a very small volume that repelled and deflected the positively be lo charged 𝛼–particles. ng s to (iii) Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly cz small as compared to the total volume of the atom. The radius of the atom is about 10–10 m, a6 vy while that of nucleus is 10–15 m. (s ub ha m On the basis of above observations and conclusions, Rutherford proposed the nuclear model of atom: 37 20 (i) The positive charge and most of the mass of the atom was densely concentrated in extremely @ gm small region. This very small portion of the atom was called nucleus by Rutherford. ai l. co (ii) The nucleus is surrounded by electrons that move around the nucleus with a very high speed m ,6 in circular paths called orbits. 29 7 (iii) Electrons and the nucleus are held together by electrostatic forces of attraction. 70 40 09 ) Fig.4 Rutherford Atomic Model 2.2.2 Drawback i. According to the electromagnetic theory of Maxwell, charged particles when accelerated should emit electromagnetic radiation. Therefore, an electron in an orbit will emit radiation continuously. The orbit will thus continue to shrink. Calculations show that it should take an electron only 10–8 s to spiral P a g e 4 | 15 JELET CHEMISTRY into the nucleus. But this does not happen. Thus, the Rutherford model cannot explain the stability of an atom. ii. Another serious drawback of the Rutherford model is that it says nothing about distribution of the electrons around the nucleus and the energies of these electrons. iii. Line Spectra of atom can’t be explained by this model. Th is 2.3 Bohr’s Model of Atom PD F be 2.3.1 Development leading to the Bohr’s Model lo ng s Two developments played a major role in the formulation of Bohr’s model of atom. These were: to cz (i) Dual character of the electromagnetic radiation which means that radiations possess both wave like and a6 vy particle like properties (s ub (ii) Experimental results regarding atomic spectra. ha m 37 20 2.3.2 Wave Nature of Electromagnetic Radiation @ gm In earlier days (Newton) light was supposed to be made of particles (corpuscules). ai l. co James Maxwell (1870) suggested that when electrically charged particle moves under acceleration, m ,6 alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in 29 7 70 the forms of waves called electromagnetic waves or electromagnetic radiation. 40 09 Maxwell was again the first to reveal that light waves are associated with oscillating electric and ) magnetic character (Fig.5) Fig.5 The electric and magnetic field components of an electromagnetic wave. These components have the same wavelength, frequency, speed and amplitude, but they vibrate in two mutually perpendicular planes It is now well established that there are many types of electromagnetic radiations, which differ from one another in wavelength (or frequency). These constitute what is called electromagnetic spectrum (Fig.6). The small portion around 1015 Hz, is what is ordinarily called visible light. It is only this part which our eyes can see (or detect). Special instruments are required to detect non-visible radiation. P a g e 5 | 15 JELET CHEMISTRY Th is PD F be lo (a) ng s to cz a6 vy (s ub ha m (b) 37 20 @ gm ai l. co Fig.6 (a) The spectrum of electromagnetic radiation. (b) Visible spectrum. The visible m ,6 region is only a small part of the entire spectrum. 29 7 70 40 These radiations are characterized by the properties, namely, frequency (𝜐) and wavelength (𝜆). 09 ) (i) Frequency: Frequency is defined as the number of oscillations of a wave per unit time. The SI unit for frequency (𝜈) is hertz (Hz, s–1) (ii) Wavelength: Wavelength can be defined as the distance between two successive crests or troughs of a wave. Wavelength should have the units of length and the SI units of length is meter (m). (iii) Velocity: Wave velocity is the distance travelled by a wave per unit time. In vacuum all types of electromagnetic radiations, regardless of wavelength, travel at the same speed, i.e., 3.0 × 108 m s–1. This is called speed of light and is given the symbol ‘c’. (iv) Wave number: The number of waves per unit distance of radiant energy of a given wavelength: the reciprocal of the wavelength (𝜆). Its units are reciprocal of wavelength unit, 𝟏 i.e., m–1 or cm-1 ̅= 𝝊 𝝀 The frequency (𝜈), wavelength (𝜆) and velocity of light (c) are related by the equation: 𝝂𝝀 = 𝒄 2.3.3 Particle Nature of Electromagnetic Radiation Following are some of the observations which could not be explained with the help of even the electromagnetic theory of 19th century physics (known as classical physics): P a g e 6 | 15 JELET CHEMISTRY (i) the nature of emission of radiation from hot bodies (black -body radiation) (ii) ejection of electrons from metal surface when radiation strikes it (photoelectric effect) (iii) variation of heat capacity of solids as a function of temperature (iv) Line spectra of atoms with special reference to hydrogen. Planck’s Quantum Theory Th is PD (i) Atoms and molecules could emit or absorb energy only in discrete quantities and not in a F be continuous manner. He gave the name quantum to the smallest quantity of energy that can be lo ng emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of s to radiation is proportional to its frequency (𝜈) and is expressed by equation: E = h𝝂 cz a6 The proportionality constant, ‘h’ is known as Planck’s constant and has the value 6.626×10– vy (s 34 J s. ub ha (ii) The energy can take any one of the values from the following set, but cannot take on any values m 37 between them. E = 0, h𝜈, 2h𝜈, 3h𝜈....nh𝜈..... 20 @ gm ai 2.3.4 Evidence for the quantized Electronic Energy Levels: Atomic spectra l. co m a. Continuous Spectra: Ordinary white light consists of waves with all the wavelengths in the ,6 29 visible range, a ray of white light is spread out into a series of colored bands called spectrum. Such 7 70 40 a spectrum is called continuous spectrum. Continuous because violet merges into blue, blue into 09 ) green and so on. A similar spectrum is produced when a rainbow forms in the sky. Fig.7 Continuous Spectra b. Emission and Absorption Spectra (Line Spectra of Atom) I. The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. Atoms, molecules or ions that have absorbed radiation are said to be “excited”. II. An absorption spectrum is like the photographic negative of an emission spectrum. A continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelength which corresponds to the radiation absorbed by the matter, leave dark spaces in the bright continuous spectrum. III. The study of emission or absorption spectra is referred to as spectroscopy. Line emission spectra are of great interest in the study of electronic structure. Each element has a unique line emission spectrum. The characteristic lines in atomic spectra can be used in chemical analysis to identify unknown atoms in the same way as fingerprints are used to identify people. P a g e 7 | 15  JELET CHEMISTRY (a) Th is PD F be lo ng s (b) to cz a6 vy (s ub ha m 37 20 @ gm ai l. co m ,6 29 Fig.8 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms (or any other element) can be 7 70 passed through a prism and separated into certain discrete wavelengths. Thus, an emission spectrum, which is a 40 09 photographic recording of the separated wavelengths is called as line spectrum. Any sample of reasonable size contains ) an enormous number of atoms. Although a single atom can be in only one excited state at a time, the collection of atoms contains all possible excited states. The light emitted as these atoms fall to lower energy states is responsible for the spectrum. (b) Atomic absorption. When white light is passed through unexcited atomic hydrogen and then through a slit and prism, the transmitted light is lacking in intensity at the same wavelengths as are emitted in (a) The recorded absorption spectrum is also a line spectrum and the photographic negative of the emission spectrum. 2.3.5 Postulates of Bohr’s Theory According to Bohr’s theory for hydrogen atom: (i) The electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus. (ii) The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state when required amount of energy is absorbed by the electron or energy is emitted when electron moves from higher stationary state to lower stationary state. The energy change does not take place in a continuous manner. (iii) The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ∆E, is given by: ∆𝑬 = 𝑬𝟐 − 𝑬𝟏 = 𝒉𝝊 Where E1 and E2 are the energies of the lower and higher allowed energy states respectively. (iv) The angular momentum of an electron is quantized. In a given stationary state it can be expressed as in equation: 𝒉 𝒎𝒆 𝒗𝒓 = 𝒏 𝟐𝝅 , 𝒏 = 𝟏, 𝟐, 𝟑, … P a g e 8 | 15 JELET CHEMISTRY Where me is the mass of electron, v is the velocity and r is the radius of the orbit in which electron is moving. Th is PD F be lo ng s to cz a6 vy (s ub ha m 37 20 @ gm Fig.9 Bohr’s Model of Atom ai l. co m ,6 29 2.3.6 Line Spectrum of Hydrogen 7 70 40 When an electric discharge is passed through gaseous hydrogen, the H2 molecules dissociate and the 09 ) energetically excited hydrogen atoms emit electromagnetic radiation of discrete frequencies. The hydrogen spectrum consists of several series of lines named after their discoverers. The Swedish spectroscopist, Johannes Rydberg, noted that all series of lines in the hydrogen spectrum could be described by the following expression: 𝟏 𝟏 𝝂̅ = 𝑹 (𝒏𝟐 − 𝒏𝟐 )𝒁𝟐 [for Hydrogen atom Z=1] 𝟏 𝟐 where n1=1,2........ n2 = n1 + 1, n1 + 2...... R=109,678 cm–1 is called the Rydberg constant Spectral lines of Hydrogen Atom P a g e 9 | 15 JELET CHEMISTRY Th is PD F be lo ng s to cz a6 vy (s ub ha m 37 20 @ gm ai l. co m ,6 29 7 70 40 09 ) Fig.10 Transitions of the electron in the hydrogen atom (The diagram shows the Lyman, Balmer and Paschen series of transitions) 2.3.7 Radius, Velocity and Energy of nth orbital from Bohr’s Theory 𝒏𝟐 𝒉𝟐 𝒏𝟐 𝒏𝟐 i. Radius: 𝒓𝒏 = ⟹ 𝒓𝒏 ∝ ⟹ 𝒓𝒏 = 𝒓𝟏 𝟒𝝅𝟐 𝒎𝒁𝒆 𝟐 𝒁 𝒁 For 1st Bohr radius of Hydrogen, Z=1 and n=1 𝒉𝟐 𝒓𝒏 = = 𝒓𝟏 = 𝟓𝟐. 𝟗 𝒑𝒎 = 𝟎. 𝟓𝟐𝟗 Å 𝟒𝝅𝟐 𝒎𝒆𝟐 P a g e 10 | 15 JELET CHEMISTRY 𝟐𝝅𝒁𝒆𝟐 𝒁 𝒁 ii. Velocity: 𝒗𝒏 = ⟹ 𝒗𝒏 ∝ ⟹ 𝒗𝒏 = 𝒗𝟏 𝒏𝒉 𝒏 𝒏 𝟐𝝅𝒆𝟐 For electron revolving in the 1st Bohr orbital of Hydrogen, Z=1 and n=1 𝒗𝒏 = = 𝒗𝟏 𝒉 Th is 𝟐𝝅𝟐 𝒎𝒁𝟐 𝒆𝟒 𝒁𝟐 𝒁𝟐 PD iii. Energy: 𝑬𝒏 = − ⟹ 𝑬𝒏 ∝ ⟹ 𝑬𝒏 = 𝑬𝟏 F 𝒏𝟐 𝒉𝟐 𝒏𝟐 𝒏𝟐 be lo ng s Energy of 1st Bohr orbital of Hydrogen, Z=1 and n=1 to cz 𝟐𝝅𝟐 𝒎𝒆𝟒 a6 𝑬𝒏 = − = 𝑬𝟏 = −𝟐𝟏𝟕𝟔 × 𝟏𝟎−𝟐𝟏 𝑱(−𝟏𝟑. 𝟔 𝒆𝑽) vy 𝒉𝟐 (s ub ha m What does the negative electronic energy (En) for hydrogen atom mean? 37 20 @ This negative sign means that the energy of the electron in the atom is lower than the energy of a free gm ai electron at rest. A free electron at rest is an electron that is infinitely far away from the nucleus and is l. co m assigned the energy value of zero. Mathematically, this corresponds to setting n equal to infinity in ,6 29 the equation. so that 𝐸∞ = 0. As the electron gets closer to the nucleus (as n decreases), En becomes 7 70 larger in absolute value and more and more negative. The most negative energy value is given by 40 09 n=1 which corresponds to the most stable orbit. We call this the ground state. ) 2.3.8 Drawback of Bohr’s Atomic Model (i) Spectra of atoms or ions having two or more electrons cannot be explained with the help of Bohr’s theory. (ii) When spectroscope with high resolving power used, it was found that each line in spectra split into a number of closely spaced lines called fine structure cannot be explained by Bohr’s model. (iii)Bohr’s theory could not explain the splitting of spectral lines under the influence the magnetic field (Zeeman effect) or electric field (Stark effect) i.e., the formation of fine structure of atomic spectra. (iv) Bohr treated electron only as a particle i.e., ignored the dual nature of an electron. (v) Bohr’s model directly contradicts the Heisenberg’s uncertainty principle. (vi) It fails to give the actual 3D electronic model of atom. 3. Towards Quantum Mechanical Model of the Atom In view of the shortcoming of the Bohr’s model, attempts were made to develop a more suitable and general model for atoms. Two important developments which contributed significantly in the formulation of such a model were: 1. Dual behavior of matter 2. Heisenberg uncertainty principle P a g e 11 | 15 JELET CHEMISTRY 3.1 Dual Behavior of Matter The French physicist, de Broglie, in 1924 proposed that matter, like radiation, should also exhibit dual behavior i.e., both particle and wavelike properties. de Broglie gave the following relation between wavelength (𝜆) and momentum (p) of a material particle Th is 𝒉 𝒉 𝝀= = PD F 𝒑 𝒎𝒗 be lo where m is the mass of the particle, v its velocity and p its momentum ng s The wavelengths associated with ordinary objects are so short (because of their large masses) that to cz their wave properties cannot be detected. The wavelengths associated with electrons and other a6 vy subatomic particles (with very small mass) can however be detected experimentally. (s ub 1 If K.E.=E then 𝐸 = 2 𝑚𝑣 2 ⟹ 2𝑚𝐸 = 𝑚2 𝑣 2 ⟹ 𝑚𝑣 = √2𝑚𝐸 ha m 37 𝒉 𝒉 𝒉 𝝀= = = 20 @ 𝒑 𝒎𝒗 √𝟐𝒎𝑬 gm de Broglie wavelength of an electron after being accelerated by a potential difference of V volt from ai l. co rest: m ,6 29 1 2𝑒𝑉 𝑒𝑉 = 𝐾. 𝐸. = 2 𝑚𝑣 2 ⟹ 𝑣 = √ 7 70 𝑚 40 09 𝒉 𝒉 𝒉 𝝀= = = ) 𝒑 𝒎𝒗 √𝟐𝒎𝒆𝑽 3.2 Heisenberg’s Uncertainty Principle Werner Heisenberg a German physicist in 1927, stated uncertainty principle which is the consequence of dual behavior of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as in equation 𝒉 𝚫𝒙 𝐗 𝚫𝒑𝑿 ≥ 𝟒𝝅 𝒉 ⟹ 𝚫𝒙 𝐗 𝚫(𝒎𝒗𝑿 ) ≥ 𝟒𝝅 𝒉 ⟹ 𝚫𝒙 𝐗 𝚫𝒗𝑿 ≥ 𝟒𝝅𝒎 Where ∆𝑥 is the uncertainty in position and ∆𝑝𝑥 (or ∆𝑣𝑥 ) is the uncertainty in momentum (or velocity) of the particle. 𝒉 Energy-Time uncertainty: 𝚫𝒕 𝐗 𝚫𝐄 ≥ 𝟒𝝅 3.2.1 Significance of Uncertainty Principle One of the important implications of the Heisenberg Uncertainty Principle is that it rules out existence of definite paths or trajectories of electrons and other similar particles. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. i. If uncertainty principle is applied to an object of mass, say about a milligram (10–6 kg), then 𝒉 ∆𝒗. ∆𝒙 = 𝟒𝝅𝒎 P a g e 12 | 15 JELET CHEMISTRY 𝟔.𝟔𝟐𝟔 𝑿 𝟏𝟎−𝟑𝟒 𝑱𝒔 = = 𝟏𝟎−𝟐𝟖 𝒎−𝟐 𝒔−𝟏 𝟒 𝑿 𝟑.𝟏𝟒𝟏𝟔 𝑿 𝟏𝟎−𝟔 𝒌𝒈 The value of ∆𝑣. ∆𝑥 obtained is extremely small and is insignificant. Therefore, one may say that in dealing with milligram-sized or heavier objects, the associated uncertainties are hardly of any real consequence. ii. In the case of a microscopic object like an electron on the other hand, ∆𝑣. ∆𝑥 obtained is much Th is larger and such uncertainties are of real consequence. For example, for an electron PD F whose mass is 9.11×10-31 kg., according to Heisenberg uncertainty principle be lo 𝒉 ng ∆𝒗. ∆𝒙 = s to 𝟒𝝅𝒎 cz 𝟔.𝟔𝟐𝟔 𝑿 𝟏𝟎−𝟑𝟒 𝑱𝒔 a6 = = 𝟏𝟎−𝟒 𝒎−𝟐 𝒔−𝟏 vy 𝟒 𝑿 𝟑.𝟏𝟒𝟏𝟔 𝑿 𝟗.𝟏𝟏 𝑿 𝟏𝟎−𝟑𝟏 𝒌𝒈 (s ub ha m 37 20 Numerical @ gm ai l. co m Q1. The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). ,6 29 Express these wavelengths in frequencies (Hz). (1nm = 10–9 m) 7 70 40 09 𝒄 𝟑 𝑿 𝟏𝟎𝟖 𝒎𝒔−𝟏 ) Solution: 𝝂 = = = 𝟕. 𝟓𝟎 𝑿 𝟏𝟎𝟏𝟒 𝑯𝒛 (Violet) 𝝀 𝟒𝟎𝟎 𝑿 𝟏𝟎−𝟗 𝒎 𝒄 𝟑 𝑿 𝟏𝟎𝟖 𝒎𝒔−𝟏 𝝂= = = 𝟒 𝑿 𝟏𝟎𝟏𝟒 𝑯𝒛 (Red) 𝝀 𝟕𝟓𝟎 𝑿 𝟏𝟎−𝟗 𝒎 The range of visible spectrum is from 4.0 × 1014 to 7.5 × 1014 Hz Q2. Calculate energy of one mole of photons of radiation whose frequency is 5 ×1014 Hz. Solution: Energy (E) of one photon is given by the expression 𝑬 = 𝒉𝝂 = (6.626 × 10−34 𝐽 𝑠) × (5 × 1014 𝑠 −1 ) = 3.313 × 10−19 𝐽 Energy of one mole of photons = (3.313 ×10-19 J) × (6.022 × 1023 mol-1) = 199.51 kJ mol-1 Q3. A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb. Solution: Power of the bulb = 100 watt =100 J s–1 6.626 ×1034 J s × 3 ×108 ms −1 Energy of one photon E = h𝝂 = hc/𝝀 = = 4.969 × 10−19 J 400 × 10−9 𝑚 100 𝐽𝑠 −1 Number of photons emitted= 4.969 ×10−19 𝐽 = 2.012 × 1020 𝑠 −1 P a g e 13 | 15 JELET CHEMISTRY Q4. What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s–1? Solution: According to de Broglie equation 𝒉 6.626 𝑋 10−34 𝐽𝑠 Th 𝝀= = (0.1 = 6.626 × 10−34 m (J=kg m2 s-2) is 𝑘𝑔)(10 𝑚𝑠 −1 ) PD 𝒎𝒗 F be lo ng s to Q5. The mass of an electron is 9.1×10–31 kg. If its K.E. is 3.0×10–25 J, calculate its wavelength. cz a6 vy (s 𝟐𝑲𝑬 𝟏 ub 𝟏 𝟐×𝟑×𝟏𝟎−𝟐𝟓 𝒌𝒈𝒎𝟐 𝒔−𝟐 1/2 Solution: Since 𝑲. 𝑬. = 𝒎𝒗𝟐 ⟹ 𝒗 = ( )𝟐 = ( ) ha m 𝟐 𝒎 𝟗.𝟏×𝟏𝟎−𝟑𝟏 𝒌𝒈 37 20 = 𝟖𝟏𝟐 𝒎 𝒔−𝟏 @ gm ai 6.626 𝑋 10−34 𝐽𝑠 l. 𝒉 co 𝝀= = = 8967 × 10−10 m = 897.6 nm m ,6 𝒎𝒗 (0.1 𝑘𝑔)(812 𝑚𝑠 −1 ) 29 7 70 40 09 ) Q6. Determine the wavelengths of 𝑯𝜶 line of the Balmer series. 𝟏 𝟏 ̅ =𝑹( Solution: We know, 𝝂 − )𝒁𝟐 𝒏𝟐𝟏 𝒏𝟐𝟐 For Balmer series n1=2 and for the line 𝑯𝜶 , n2=3 𝟏 𝟏 𝟏 𝝂̅ = = 𝑹 ( − ) (For Hydrogen Z=1) 𝝀 𝒏𝟐𝟏 𝒏𝟐𝟐 𝟏 𝟏 = 𝟏𝟎𝟗𝟔𝟕𝟖 ( − ) 𝟐𝟐 𝟑𝟐 ∴ 𝝀 = 𝟔. 𝟓𝟔𝟒 × 𝟏𝟎−𝟓 𝒄𝒎 Q7. What is the value of the highest and lowest wavelength for Lyman series of hydrogen spectra? 𝟏 𝟏 𝟏 ̅ =𝑹( Solution: 𝝂 − )𝒁𝟐 and 𝝂̅ = 𝒏𝟐𝟏 𝒏𝟐𝟐 𝝀 For Lyman series n1=1 When wavelength is maximum, wavenumber is minimum. This is satisfied when n2=2 𝟏 𝟏 𝝂̅ = 𝟏𝟎𝟗𝟔𝟕𝟖 ( − ) ⟹ 𝝀 = 𝟏𝟐𝟏𝟓. 𝟔𝟕 × 𝟏𝟎−𝟖 𝒄𝒎 = 𝟏𝟐𝟏𝟓. 𝟔𝟕Å 𝟏𝟐 𝟐𝟐 When wavelength is minimum, wavenumber is maximum. This is satisfied when n2=∞ 𝟏 𝟏 𝝂̅ = 𝟏𝟎𝟗𝟔𝟕𝟖 ( 𝟐 − 𝟐 ) ⟹ 𝝀 = 𝟗𝟏𝟏. 𝟕𝟓 × 𝟏𝟎−𝟖 𝒄𝒎 = 𝟗𝟏𝟏. 𝟕𝟓Å 𝟏 ∞ P a g e 14 | 15 JELET CHEMISTRY Q8. If the energy of the 1st Bohr orbit of hydrogen is -13.6eV. What will be the energy of the 3rd Bohr orbit? 𝒁𝟐 Th Solution: We know, 𝑬𝒏 = 𝑬𝟏 is 𝒏𝟐 PD F 𝟏𝟐 be 𝑬𝒏 = −𝟏𝟑. 𝟔 ( 𝟐) 𝒆𝑽 = −𝟏. 𝟓𝟏𝟏 𝒆 lo 𝟑 ng s to cz a6 vy Q9. Uncertainty in position of an electron moving with velocity of 𝟑 × 𝟏𝟎𝟒 𝒎/𝒔 accurate upto (s ub ha 0.001%. What will be the uncertainty in position? m 37 20 @ 0.001 Solution: Accuracy= 0.001%= 100 gm ai l. 0.001 co Δ𝑣 = 3 × 104 × = 0.3 𝑚/𝑠 m ,6 100 29 𝒉 6.626 × 10−27 7 70 𝚫𝒙 = = = 1.93 𝑐𝑚 40 𝟒𝝅𝒎𝚫𝒗 4𝜋(9.1 × 10−28 ) × 0.3 09 ) Q10. The difference between the radii of 3rd and 4th orbitals of Li+2 is ∆𝑹𝟏. The difference between the radii of 3rd and 4th orbitals of He+ is ∆𝑹𝟐. What is the ratio of ∆𝑹𝟏 and ∆𝑹𝟐 ? 𝒏𝟐 Solution: 𝒓𝒏 = 𝒓𝟏 𝒁 For Li 2+ 𝒓𝟒 − 𝒓𝟑 = 𝒓𝟏 (𝟏𝟔 − 𝟗)/𝟑 = ∆𝑹𝟏 For He+ 𝒓𝟒 − 𝒓𝟑 = 𝒓𝟏 (𝟏𝟔 − 𝟗)/𝟐 = ∆𝑹𝟐 ∆𝑹𝟏 𝒓𝟏 (𝟏𝟔−𝟗)/𝟑 2 Now, = = ∆𝑹𝟐 𝒓𝟏 (𝟏𝟔−𝟗)/𝟐 3 P a g e 15 | 15