Kishan Sinha

The Trailing Zeros Suppose you measure length of an object using a metre scale with least count of \(0.1 \, cm\). The length comes out to be \(35 \, cm\). Which of the following, in your opinion, is the best way to state this data? \(35.000 \, cm\) \(35 \, cm\) \(35.0 \, cm\) \(35.00 \, cm\) Are these numbers equivalent? Mathematically, there is no difference between these numbers. However, to an experimentalist, each of the numbers above tell a different story. The number \(35.0 \, cm\) conveys that we are certain about the first two digits \(3 \, \mbox{and} \, 5\) of this measurement and we can say with a 'good degree of confidence' that the number in the tenth place is 'close' to zero, if not exactly zero. However, we are not able to say anything about numbers in hundredth position and onward. The number \(35.00 \, cm\), on the other hand, conveys that we are certain about the first three digits — \(3, \, 5\) and the first \(0\) a

PREVIOUS: Measurement and Significant Figures NEXT: Combining Uncertainties Rules to identify signficant figures are: Zeros between non-zero digits are significant. 1.023 has 4 significant figures, 30405 has five significant figures. Leading zeros (zeros to the left of non-zero digits) are never significant. 00509 has three significant figures, 0.00345 has three significant figures (3,4 and 5). In numbers with decimal point, zeros to the right of the last non-zero digit after the decimal point (trailing zeros) are significant. 509.0 has four significant figures, 0.0034500 has five significant figures. In numbers without decimal point, trailing zeros (zeros to the right of the last non-zero digit) may or may not be significant. Rules for rounding off If the digit immediately to the right of the last significant digit is 4 or less, the last significant digit is retained and all the digits to the right of it are dropped. In

PREVIOUS: Combining Uncertainties I: Propagation of Errors DEFINITIONS: If \(\Delta h\) is the absolute uncertainty in a physical quantity \(h\), then: a) the quantity \(h\) is properly expressed as \(h \pm \Delta h\), b) the fractional uncertainty in \(h\) is given by \(\frac{\Delta h}{h}\), c) percent uncertainty in \(h\) is given by \(\frac{\Delta h}{h} \times 100\). Suppose you make following measurements of lengths in a lab experiment using a meter scale with the least count of \(0.1 \, cm\): \[p = (80.0 \pm 0.1) \, cm \]\[ q = (12.0 \pm 0.1) \, cm \]\[ r = (5.3 \pm 0.1) \, cm \] Here, \(0.1 \, cm\) is referred to as absolute uncertainty in \(p,q,r\). According to the above definitions, fractional uncertainties in these measurements are: \[\frac{\Delta p}{p} = \frac{0.1}{80} = 0.00125\] \[\frac{\Delta q}{q} = \frac{0.1}{12} = 0.00833\] \[\frac{\Delta r}{r} = \frac{0.1}{5.3} = 0.01887\] Results from the previous post \[ \begin{array}{|c|c|} \hline \

The idea of direct exchange interaction was originally derived in Heitler-London model which explained the binding energy of the hydrogen molecule. This model inspired Heisenberg's theory of ferromagnetism which based on an exchange interaction between the spins of two neighboring atoms. Consider two hydrogen atoms \(a\) and \(b\) with their isolated atomic orbitals represented as \(\phi_{a}(\vec{r})\) and \(\phi_{b}(\vec{r})\), respectively. The hamiltonian of an isolated hydrogen atom is given as: \[ H = - \frac{\hbar^{2}}{2m}\nabla^{2} - \frac{e^{2}}{r}\] The hamiltonian of the hydrogen molecule is given by the sum of those for the isolated atoms: \[ H = - \frac{\hbar^{2}}{2m}\nabla_{1}^{2} - \frac{\hbar^{2}}{2m}\nabla_{2}^{2} - \frac{e^{2}}{r_{a1}} - \frac{e^{2}}{r_{b2}} + \left( \frac{e^{2}}{r_{12}} + \frac{e^{2}}{r_{ab}} - \frac{e^{2}}{r_{b1}} - \frac{e^{2}}{r_{a2}} \right)\] where \(r_{1}\) and \(r_{2}\) represent coordinates of the two electrons in the m

1 . An excess of \(Zn\) is added to \(100 \, cm^{3}\) of \(1.0 \, mol/dm^{3}\) of hydrochloric acid. The equation for the reaction is: \[Zn \,\,\, + \,\,\, HCl \,\,\, \rightarrow \,\,\, ZnCl_{2} \,\,\, + \,\,\, H_{2} \uparrow \] What is the maximum volume of hydrogen evolved at room temperature and pressure? a) \(1.2 \, dm^{3}\) b) \(2.0 \, dm^{3}\) c) \(2.4 \, dm^{3}\) d) \(24 \, dm^{3}\) Observe that one \(H_{2}\) molecule is evolved when one \(Zn\) atom reacts with one \(HCl\) molecule. So, the total number of \(H_{2}\) molecules evolved must be the same as the molecules of \(HCl\) or \(Zn\) that participate in the reaction. Therefore, we will adopt the following strategy to solve this problem: First , find out the number of moles of \(H_{2}\) gas evolved by finding out how many \(HCl\) or \(Zn\) that participate in the reaction. Second , we use the relationship \(24 \, dm^{3} \, \equiv \, 1 \, mole\) for any gas at r.t.p. t

PREVIOUS: Quantifying Compounds in Bulk Having found a way to find number of molecules in solids in the previous post, its time to turn our attention toward solutions. Solutions are obtained by dissolving solute (such as salts, polymers, etc.) in a solvent. Such solutions are used in myriad of applications in research, industry as well as our homes. Consider the example of hypothetical lemonade we made in the last post wherein we had dissolved \(100 \, gm\) of sugar in \(1 \, liter\) of water. Since lemonade was prepared for four individuals, each individual would have got \(25 \, gm\) of sugar in their share along with \(0.25 \, liter\) or \(250 \, ml\) of water. Now, it is not too hard to imagine that if we add some more sugar, say \(10 \, gm\), to the lemonade of one individual, she would find it sweeter compared to lemonades of her friends. This is because though the amount of sugar has increased from \(25 \, gm\) to \(35 \, gm\), the amount of water in her le

PREVIOUS: Concept of Mole: From Atomic Scale to Large Scale In the previous post, we devised a technique to estimate number of atoms of an element in a given mass. In brief, if an element \(^{A}_{Z}X\) has mass number \(A\), then \(1 \, mole\) of atoms of that element will have mass \(A \, gm\). The question now is: How do we extend this concept to find number of molecules in compounds? After all, much of the world we see around us and materials we use in daily life are compounds that consist of multiple elements. Concepts of \(mole\) and \(amu\) would be of limited use if we cannot apply them to analyze compounds. So, let's invest some time in generalizing concepts developed in previous posts to include compounds. Say you are given \(50 \, gm\) of common salt (\(NaCl\)) and are asked to find the number of formula units (or molecules) in it. How will you solve this problem? To begin with, note that one formula unit (or molecule) of salt contains one \(Na\) atom

A proton and a neutron have nearly equal masses – \( 1.66 \times 10^{-27} kg \) or \( 1.66 \times 10^{-24} gm \). Compare this with an electron which has mass of about \( 9.31 \times 10^{-31} kg \) or \( 9.31 \times 10^{-28} gm \), nearly 1837 times less than the mass of a proton. Therefore, while estimating the mass of an atom, we may safely ignore masses of electrons and consider contributions from protons and neutron only. Of course this will not yield the most accurate estimate of mass of an atom but it will get us a good enough estimate to work with on large scale. We thus assume that mass of an atom is, to a good approximation, almost equal to the mass of its nucleus. Because an atomic nucleus consists of nucleons - proton and neutron - the mass of any nucleus must be an integral multiple of \( 1.66 \times 10^{-27} kg \) or \( 1.66 \times 10^{-24} gm \). For instance, if a nucleus has 14 protons and 13 neutrons, its mass is mass \(27 \times 1.66 \times 10^{-24} gm

PREVIOUS: What is amu ? Using the concept of amu from the previous post, and a bit of creative thinking, we try to devise a technique to count the number of atoms on large scale. Let’s consider hydrogen gas. One hydrogen atom measures \(1 \, amu = 1.66 \times 10^{-24} gm\), same as mass of a proton. Now we ask: If we took \(1 \, gm\) of hydrogen atom, exactly how many atoms would be in it? This is a straightforward calculation: \[ \begin{array}{l} \mbox{Number of hydrogen atoms in 1 gm} &=& \frac{\mbox{1 gm}}{\mbox{mass of one hydrogen atom in grams}} \\ & = & \frac{1 gm}{1.66 \times 10^{-24} gm} \\ & = & 6.023 \times 10^{23} \end{array} \] Note that this is number of atoms and has no units. Consider the case of helium gas, \(^{4}_{2}He\). An atom of helium has 4 nucleons; therefore, its atomic mass is 4 amu. The mass of \(6.023 \times 10^{23} \) helium atoms will be \[\begin{array}{l} 6.023