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...
Read More →The reaction:
Question: The reaction: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{KOH}_{(a q)} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{KI}$ is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition Solution: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{KOH}_{(a q)} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{KI}$ It is an example of nucleophilic substitution reaction. The hydroxyl group of $\m...
Read More →Find the sum to n terms of the series
Question: Find the sum tonterms of the series$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots$ Solution: The given series is $\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots$ $n^{\text {th }}$ term, $a_{n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ (By partial fractions) $a_{1}=\frac{1}{1}-\frac{1}{2}$ $a_{2}=\frac{1}{2}-\frac{1}{3}$ $a_{3}=\frac{1}{3}-\frac{1}{4} \ldots$ $a_{n}=\frac{1}{n}-\frac{1}{n+1}$ Adding the above terms column wise, we obtain ...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. $2 x+y+z=1$ $x-2 y-z=\frac{3}{2}$ $3 y-5 z=9$ Solution: The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ccc}2 1 1 \\ 1 -2 -1 \\ 0 3 -5\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ \frac{3}{2} \\ 9\end{array}\right]$. Now, $|A|=2(10+3)-1(-5-3)+0=2(13)-1(-8)=26+8=34 \neq 0$ Thus,Ais non-singular. Therefore, its inverse exists....
Read More →The best and latest technique for isolation, purification and separation of organic compounds is:
Question: The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography Solution: Chromatography is the most useful and the latest technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances....
Read More →Estimating the following two numbers should be interesting.
Question: Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never count photons, even in barely detectable light. (a)The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m. (b)The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of wh...
Read More →Which of the following carbocation is most stable?
Question: Which of the following carbocation is most stable? (a) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} . \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ (b) $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ (d) $\mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H} \mathrm{CH}_{2} \mathrm{CH}_{3}$ Solution: $\left(\mathrm{CH}_{3}\right)_{3}{ }_{\mathrm{C}}^{+}$is a tertiary carbocation. A tertiary carbocation i...
Read More →Find the sum to n terms of each of the series in Exercises 1 to 7.
Question: Find the sum tonterms of each of the series in Exercises 1 to 7. $3 \times 1^{2}+5 \times 2^{2}+7 \times 3^{2}+\ldots$ Solution: The given series is $3 \times 1^{2}+5 \times 2^{2}+7 \times 3^{2}+\ldots$ $n^{\text {th }}$ term, $a_{n}=(2 n+1) n^{2}=2 n^{3}+n^{2}$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}$ $=\sum_{k=1}^{n}=\left(2 k^{3}+k^{2}\right)=2 \sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} k^{2}$ $=2\left[\frac{n(n+1)}{2}\right]^{2}+\frac{n(n+1)(2 n+1)}{6}$ $=\frac{n^{2}(n+1)}{2}+\frac{n(n+1)(...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. 5x+ 2y= 3 3x+ 2y= 5 Solution: The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ll}5 2 \\ 3 2\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$ Now, $|A|=10-6=4 \neq 0$ Thus,Ais non-singular. Therefore, its inverse exists....
Read More →In the Lassaigne’s test for nitrogen in an organic compound, t
Question: In the Lassaignes test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: (a) $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ (b) $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$ (c) $\mathrm{Fe}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ (d) $\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{4}$ Solution: In the Lassaignes test for nitrogen in an organic compound, the sodium fusion extract is boiled wi...
Read More →In an accelerator experiment on high-energy collisions of electrons with positrons,
Question: In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two-rays of equal energy. What is the wavelength associated with each-ray? (1BeV = 109eV) Solution: Total energy of two-rays: E= 10. 2 BeV = 10.2 109eV = 10.2109 1.61010J $E^{\prime}=\frac{E}{2}$ $=\frac{10.2 \times 1.6 \times 10^{-10}}{2}=8.16 \times 10^{-10} \mathrm{~J}$ Planck's constant, $h=6.62...
Read More →In the organic compound CH2=CH–CH2–CH2–C≡CH, the pair of hydridised orbitals involved in the formation
Question: In the organic compound $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C}=\mathrm{CH}$, the pair of hydridised orbitals involved in the formation of: $\mathrm{C}_{2}-\mathrm{C}_{3}$ bond is: (a) $s p-s p^{2}$ (b) $s p-s p^{3}$ (c) $s p^{2}-s p^{3}$ (d) $s p^{3}-s p^{3}$ Solution: $\mathrm{C} \stackrel{6}{\mathrm{H}}_{2}=\stackrel{5}{\mathrm{C}} \mathrm{H}-\mathrm{C} \stackrel{4}{\mathrm{H}}_{2}-\mathrm{C} \stackrel{3}{\mathrm{H}}_{2}-\stackrel{2}{\mathrm{C}} \equi...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. Solution: The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ll}4 -3 \\ 3 -5\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 7\end{array}\right]$ Now, $|A|=-20+9=-11 \neq 0$ Thus,Ais non-singular. Therefore, its inverse exists. Now, $A^{-1}=\frac{1}{|A|}($ adjA $)=-\frac{1}{11}\left[\begin{array}{ll}-5 3 \\ -3 4\end{array}\right]=\frac...
Read More →In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate.
Question: In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound. Solution: Total mass of organic compound = 0.468 g [Given] Mass of barium sulphate formed = 0.668 g [Given] 1 mol of BaSO4= 233 g of BaSO4= 32 g of sulphur Thus, $0.668 \mathrm{~g}$ of $\mathrm{BaSO}_{4}$ contains $\frac{32 \times 0.668}{233} \mathrm{~g}$ of sulphur $=0.0917 \mathrm{~g}$ of sulphur Therefore...
Read More →(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å.
Question: (a)An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation? (b)From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube? Solution: (a)Wavelength produced by an X-ray tube, Plancks constant,h= 6.6261034Js Speed of light,c= 3108m/s The maximum energy of a photon is given as: $E=\frac{h c}{\lambda}$ $=\frac{6.626 \times 10^{-34} \times 3 \t...
Read More →0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation.
Question: $0.3780 \mathrm{~g}$ of an organic chloro compound gave $0.5740 \mathrm{~g}$ of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. Solution: Given that, Mass of organic compound is 0.3780 g. Mass of AgCl formed = 0.5740 g 1 mol of AgCl contains 1 mol of Cl. Thus, mass of chlorine in 0.5740 g of AgCl $=\frac{35.5 \times 0.5740}{143.32}$ $=0.1421 \mathrm{~g}$ $\therefore$ Percentage of chlorine $=\frac{0.1421}{0.3780} \times 100=37.59 \%$ ...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. Solution: The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{cc}2 -1 \\ 3 4\end{array}\right], X=\left[\begin{array}{c}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}-2 \\ 3\end{array}\right]$. Now, $|A|=8+3=11 \neq 0$ Thus,Ais non-singular. Therefore, its inverse exists. Now, $A^{-1}=\frac{1}{|A|}$ adj $A=\frac{1}{11}\left[\begin{array}{cc}4 1 \\ -3 2\end{array}\right]$ $\therefor...
Read More →An electron gun with its collector at a potential of 100 V fires
Question: An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (102mm of Hg). A magnetic field of 2.83 104T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the fine beam tube method. Determinee/mfrom the data. Solution: Potential...
Read More →A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method.
Question: A sample of $0.50 \mathrm{~g}$ of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in $50 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$. The residual acid required $60 \mathrm{~mL}$ of $0.5 \mathrm{M}$ solution of $\mathrm{NaOH}$ for neutralisation. Find the percentage composition of nitrogen in the compound. Solution: Given that, total mass of organic compound = 0.50 g 60 mL of 0.5 M solution of NaOH was required by re...
Read More →Solve system of linear equations, using matrix method.
Question: Solve system of linear equations, using matrix method. $5 x+2 y=4$ $7 x+3 y=5$ Solution: The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ll}5 2 \\ 7 3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$ Now, $|A|=15-14=1 \neq 0$. Thus,Ais non-singular. Therefore, its inverse exists. Now, $A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$ $\therefore A^{-1}=\left[\begin{array}...
Read More →https://www.esaral.com/q/find-the-sum-to-n-terms-of-the-series-1-2-2-3-3-4-4-5-32672/
Question: Find the sum to $n$ terms of the series $1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\ldots$ Solution: The given series is $1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\ldots$ $n^{\text {th }}$ term, $a_{n}=n(n+1)(n+2)$ $=\left(n^{2}+n\right)(n+2)$ $=n^{3}+3 n^{2}+2 n$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}$ $=\sum_{k=1}^{n} k^{3}+3 \sum_{k=1}^{n} k^{2}+2 \sum_{k=1}^{n} k$ $=\left[\frac{n(n+1)}{2}\right]^{2}+\frac{3 n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2}$ ...
Read More →(a) A monoenergetic electron beam with electron speed of
Question: (a)A monoenergetic electron beam with electron speed of 5.20 106m s1is subject to a magnetic field of 1.30 104T normal to the beam velocity. What is the radius of the circle traced by the beam, givene/mfor electron equals 1.76 1011C kg1. (b)Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified? [Note:Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book....
Read More →Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Question: Find the sum to $n$ terms of the series $1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots$ Solution: The given series is $1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots$ $n^{\text {th }}$ term, $a_{n}=n(n+1)$ $\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k(k+1)$ $=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$ $=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$ $=\frac{n(n+1)}{2}\left(\frac{2 n+1}{3}+1\right)$ $=\frac{n(n+1)}{2}\left(\frac{2 n+4}{3}\right)$ $=\frac{n(n+1)(n+2)}{3}$...
Read More →An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen.
Question: An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion. Solution: Percentage of carbon in organic compound = 69 % That is, 100 g of organic compound contains 69 g of carbon. $\therefore 0.2 \mathrm{~g}$ of organic compound will contain $=\frac{69 \times 0.2}{100}=0.138 \mathrm{~g}$ of $\mathrm{C}$ Molecular mass of carbon dioxide, $...
Read More →Examine the consistency of the system of equations.
Question: Examine the consistency of the system of equations. 5xy+ 4z= 5 2x+ 3y+ 5z= 2 5x 2y+ 6z= 1 Solution: The given system of equations is: 5xy+ 4z= 5 2x+ 3y+ 5z= 2 5x 2y+ 6z= 1 This system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ccc}5 -1 4 \\ 2 3 5 \\ 5 -2 6\end{array}\right], X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ 2 \\ -1\end{array}\right]$ Now, $\begin{aligned}|A| =5(18+10)+1(12-25)+4(-4-15) \\ =5(28)+1...
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