The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail.
Question: The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 1015m or less. This structure was first probed in early 1970s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.) Solution: Wavelength of a proton or a ...
Read More →For the following compounds, write structural formulas and IUPAC names for
Question: For the following compounds, write structural formulas and IUPAC names for allpossible isomers having the number of double or triple bond as indicated: (a)C4H8(one double bond) (b)C5H8(one triple bond) Solution: (a) The following structural isomers are possible for $\mathrm{C}_{4} \mathrm{H}_{8}$ with one double bond: The IUPAC name of Compound (I) is But-1-ene, Compound (II) is But-2-ene, and Compound (III) is 2-Methylprop-1-ene. (b) The following structural isomers are possible for $...
Read More →An electron microscope uses electrons accelerated by a voltage of 50 kV.
Question: An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? Solution: Electrons are accelerated by a voltage,V= 50 kV = 50 103V Charge on an electron,e= 1.6 1019C Mass of an electron,me= 9.11 1031kg Wa...
Read More →(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV.
Question: (a)Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn= 1.675 1027kg) (b)Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neu...
Read More →The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60.The cost of 2 kg onion, 4 kg
Question: The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method. Solution: Let the cost of onions, wheat, and rice per kg be Rsx, Rsy,and Rszrespectively. Then, the given situation can be represented by a system of equations as: $4 x+3 y+2 z=60$ $2 x+4 y+6 z=90$ $6 x+2 y+3 z=70$ This system of equations can be written in the form...
Read More →Write IUPAC names of the following compounds:
Question: Write IUPAC names of the following compounds: a. $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}$ b. $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}$ c. d e f. g. Solution: (a) IUPAC name: 2-Methylbut-2-ene (b) $\stackrel{1}{\mathrm{C}} \mathrm{H}_{2}=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{3}{\mathrm{C}} \equiv \stackrel{4}{\mathrm{C}}-\stackrel{5}{\mathrm{C}} \mathrm{H}_{3}$ IUPAC name: Pen-1-ene-3-yne (c)can be written as: $\ma...
Read More →Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage.
Question: Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me= 9.11 1031kg). Solution: An X-ray probe has a greater energy than an electron probe for the same wavelength. Wavelength of light emitted from the probe,= 1 Å = 1010m Mass of an electron,me= 9.11 103...
Read More →Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Question: Show that the sum of $(m+n)^{\text {th }}$ and $(m-n)^{\text {th }}$ terms of an A.P. is equal to twice the $m^{\text {th }}$ term. Solution: Let $a$ and $d$ be the first term and the common difference of the A.P. respectively. It is known that the $k^{\text {th }}$ term of an $A . P$. is given by $a_{k}=a+(k-1) d$ $\therefore a_{m+n}=a+(m+n-1) d$ $a_{m-n}=a+(m-n-1) d$ $a_{m}=a+(m-1) d$ $\therefore a_{m+n}+a_{m-n}=a+(m+n-1) d+a+(m-n-1) d$ $=2 a+(m+n-1+m-n-1) d$ $=2 a+(2 m-2) d$ $=2 a+2...
Read More →If, find A−1. Using A−1 solve the system of equations
Question: If $A=\left[\begin{array}{ccc}2 -3 5 \\ 3 2 -4 \\ 1 1 -2\end{array}\right]$, find $A^{-1}$. Using $A^{-1}$ solve the system of equations $\begin{aligned} 2 x-3 y+5 z =11 \\ 3 x+2 y-4 z =-5 \\ x+y-2 z =-3 \end{aligned}$ Solution: $A=\left[\begin{array}{ccc}2 -3 5 \\ 3 2 -4 \\ 1 1 -2\end{array}\right]$ $\therefore|A|=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1 \neq 0$ Now, $A_{11}=0, A_{12}=2, A_{13}=1$ $A_{21}=-1, A_{22}=-9, A_{23}=-5$ $A_{31}=2, A_{32}=23, A_{33}=13$ $\therefore A^{-1}=\frac{1}{|A...
Read More →Light of intensity
Question: Light of intensity 105W m2falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer? Solution: Intensity of incident light,I= 105W m2 Surface area of a sodium photocell,A= 2 cm2= 2 104m2 Incident power of the light,P = I A = 105 2 104 = 2 109W W...
Read More →Find the sum to n terms of the series whose nth terms is given by (2n – 1)2
Question: Find the sum to $n$ terms of the series whose $n^{\text {th }}$ terms is given by $(2 n-1)^{2}$ Solution: $a_{n}=(2 n-1)^{2}=4 n^{2}-4 n+1$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(4 k^{2}-4 k+1\right)$ $=4 \sum_{k=1}^{n} k^{2}-4 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$ $=\frac{4 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n$ $=\frac{2 n(n+1)(2 n+1)}{3}-2 n(n+1)+n$ $=n\left[\frac{2\left(2 n^{2}+3 n+1\right)}{3}-2(n+1)+1\right]$ $=n\left[\frac{4 n^{2}+6 n+2-6 n-6+3}{3}\right]$ $=n\...
Read More →The work function for the following metals is given:
Question: The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away? Solution: Mo and Ni will not show photoelectric emission in both cases Wavelength for a radiation,= 3300 Å = 3300 1010m Speed of light,c= 3 108m/s Plancks constant,h= 6...
Read More →The work function for the following metals is given:
Question: The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away? Solution: Mo and Ni will not show photoelectric emission in both cases Wavelength for a radiation,= 3300 Å = 3300 1010m Speed of light,c= 3 108m/s Plancks constant,h= 6...
Read More →Find the sum to n terms of the series whose nth terms is given by n2 + 2n
Question: Find the sum to $n$ terms of the series whose $n^{\text {th }}$ terms is given by $n^{2}+2^{n}$ Solution: $a_{n}=n^{2}+2^{n}$ $\therefore S_{n}=\sum_{k=1}^{n} k^{2}+2^{k}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 2^{k}$ (1) Consider $\sum_{k=1}^{n} 2^{k}=2^{1}+2^{2}+2^{3}+\ldots$ The above series $2,2^{2}, 2^{3}, \ldots$ is a G.P. with both the first term and common ratio equal to 2 . $\therefore \sum_{k=1}^{n} 2^{k}=\frac{(2)\left[(2)^{n}-1\right]}{2-1}=2\left(2^{n}-1\right)$ (2) Therefore,...
Read More →A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission,
Question: A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used: 1= 3650 Å,2= 4047 Å,3= 4358 Å,4= 5461 Å,5= 6907 Å, The stopping voltages, respectively, were measured to be: V01= 1.28 V,V02= 0.95 V,V03= 0.74 V,V04= 0.16 V,V05= 0 V Determine the value of Pla...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. $x-y+2 z=7$ $3 x+4 y-5 z=-5$ $2 x-y+3 z=12$ Solution: The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ccc}1 -1 2 \\ 3 4 -5 \\ 2 -1 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$ Now, $|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0$ Thus,Ais non-singular. Therefore, its inverse exists. Now, $A_{11}=7...
Read More →Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
Question: Find the sum to $n$ terms of the series whose $n^{\text {th }}$ term is given by $n(n+1)(n+4)$. Solution: $a_{n}=n(n+1)(n+4)=n\left(n^{2}+5 n+4\right)=n^{3}+5 n^{2}+4 n$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$ $=\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+\frac{4 n(n+1)}{2}$ $=\frac{n(n+1)}{2} \mid \frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4$ $=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4\right]$ $=\frac{n(n...
Read More →Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …
Question: Find the sum to $n$ terms of the series $1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{2}\right)+\ldots$ Solution: The given series is $1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{2}\right)+\ldots$ $a_{n}=\left(1^{2}+2^{2}+3^{2}+\ldots \ldots+n^{2}\right)$ $=\frac{n(n+1)(2 n+1)}{6}$ $=\frac{n\left(2 n^{2}+3 n+1\right)}{6}$ $=\frac{2 n^{3}+3 n^{2}+n}{6}$ $=\frac{1}{3} n^{3}+\frac{1}{2} n^{2}+\frac{1}{6} n$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}$ $=\sum_{k=1}^{n}\left(\frac...
Read More →How do you account for the formation of ethane during chlorination of methane?
Question: How do you account for the formation of ethane during chlorination of methane? Solution: Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps. Step 1:Initiation: The reaction begins with the homolytic cleavage of Cl Cl bond as: $\mathrm{Cl}-\mathrm{Cl} \stackrel{h \nu}{\longrightarrow} \dot{\mathrm{C}} 1+\dot{\mathrm{C}}$ Chlorine free radicals Step 2:Propagation: In the second step, chlorine free radicals attack m...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. $2 x+3 y+3 z=5$ $x-2 y+z=-4$ $3 x-y-2 z=3$ Solution: The given system of equations can be written in the formAX=B, where $A=\left[\begin{array}{ccc}2 3 3 \\ 1 -2 1 \\ 3 -1 -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}5 \\ -4 \\ 3\end{array}\right]$ Now, $|A|=2(4+1)-3(-2-3)+3(-1+6)=2(5)-3(-5)+3(5)=10+15+15=40 \neq 0$ Thus,Ais non-singular. Therefore, its inverse exists. No...
Read More →Monochromatic radiation of wavelength 640.2 nm (1nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten.
Question: Monochromatic radiation of wavelength 640.2 nm (1nm = 109m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage. Solution: Wavelength of the monochromatic radiation,= 640.2 nm = 640.2 109m Stopping potential of the neon lamp,V0= 0.54 V Charge on an electron,e= 1.6 1019C Plancks constan...
Read More →Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…
Question: Find the sum to $n$ terms of the series $3 \times 8+6 \times 11+9 \times 14+\ldots$ Solution: The given series is $3 \times 8+6 \times 11+9 \times 14+\ldots$ $a_{n}=\left(n^{\text {th }}\right.$ term of $\left.3,6,9 \ldots\right) \times\left(n^{\text {th }}\right.$ term of $\left.8,11,14, \ldots\right)$ $=(3 n)(3 n+5)$ $=9 n^{2}+15 n$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(9 k^{2}+15 k\right)$ $=9 \sum_{k=1}^{n} k^{2}+15 \sum_{k=1}^{n} k$ $=9 \times \frac{n(n+1)(2 n...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. xy+z= 4 2x+y 3z= 0 x+y+z= 2 Solution: The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ccc}1 -1 1 \\ 2 1 -3 \\ 1 1 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$ Now, $|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10 \neq 0$ Thus,Ais non-singular. Therefore, its inverse exists. Now, $A_{4}=4, A_{12}=-5, A_{13}=1$ $...
Read More →Find the sum to n terms of the series
Question: Find the sum tonterms of the series$5^{2}+6^{2}+7^{2}+\ldots+20^{2}$ Solution: The given series is $5^{2}+6^{2}+7^{2}+\ldots+20^{2}$ $n^{\text {th }}$ term, $a_{n}=(n+4)^{2}=n^{2}+8 n+16$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(k^{2}+8 k+16\right)$ $=\sum_{k=1}^{n} k^{2}+8 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 16$ $=\frac{n(n+1)(2 n+1)}{6}+\frac{8 n(n+1)}{2}+16 n$ $16^{\text {th }}$ term is $(16+4)^{2}=20^{2} 2$ $\therefore S_{16}=\frac{16(16+1)(2 \times 16+1)}{6}+\frac{8 ...
Read More →Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal.
Question: Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is 1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (105W m2) red light of wavelength 6328 Å produced by a He-Ne laser? Solution: Wavelength of ultraviolet light,= 2271 Å = 2271 1010m Stopping potential of the metal,V0= 1.3 V Plancks constant,h= 6.6 1034J Charge on an electron,e= 1.6 1019C Work...
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