The ground state energy of hydrogen atom is −13.6 eV.

Question: The ground state energy of hydrogen atom is 13.6 eV. What are the kinetic and potential energies of the electron in this state? Solution: Ground state energy of hydrogen atom,E= 13.6 eV This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy = E= ( 13.6) = 13.6 eV Potential energy is equal to the negative of two times of kinetic energy. Potential energy= 2 (13.6) = 27 .2 eV...

Solve the equations

Question: Solve the equations $\left|\begin{array}{cccc}x+a x x \\ x x+a x \\ x x x+a\end{array}\right|=0, a \neq 0$ Solution: $\left|\begin{array}{ccc}x+a x x \\ x x+a x \\ x x x+a\end{array}\right|=0$ Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$, we get: $\left|\begin{array}{ccc}3 x+a 3 x+a 3 x+a \\ x x+a x \\ x x x+a\end{array}\right|=0$ $\Rightarrow(3 x+a)\left|\begin{array}{ccc}1 1 1 \\ x x+a x \\ x x x+a\end{array}\right|=0$ Applying $\mathrm{C}_{2} \r...

The ground state energy of hydrogen atom is −13.6 eV.

Question: The ground state energy of hydrogen atom is 13.6 eV. What are the kinetic and potential energies of the electron in this state? Solution: Ground state energy of hydrogen atom,E= 13.6 eV This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy = E= ( 13.6) = 13.6 eV Potential energy is equal to the negative of two times of kinetic energy. Potential energy= 2 (13.6) = 27 .2 eV...

A difference of 2.3 eV separates two energy levels in an atom.

Question: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? Solution: Separation of two energy levels in an atom, E= 2.3 eV = 2.3 1.6 1019 = 3.68 1019J Let be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E=hv Where, $h=$ Planck's constant $=6.62 \times 10^{-34} \mathrm{JS}$ $\the...

What is the shortest wavelength present in the Paschen series of spectral lines?

Question: What is the shortest wavelength present in the Paschen series of spectral lines? Solution: Rydbergs formula is given as: $\frac{h c}{\lambda}=21.76 \times 10^{-19}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$ Where, h= Plancks constant = 6.6 1034Js c= Speed of light = 3 108m/s (n1andn2are integers) The shortest wavelength present in the Paschen series of the spectral lines is given for valuesn1= 3 andn2= . $\frac{h c}{\lambda}=21.76 \times 10^{-19}\left[\frac{1}{(3)^{2}}-\frac{...

Write chemical equations for combustion reaction of the following hydrocarbons:

Question: Write chemical equations for combustion reaction of the followinghydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene Solution: Combustion can be defined as a reaction of a compound with oxygen. Toluene...

Suppose you are given a chance to repeat the alpha-particle

Question: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect? Solution: In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 1027kg) is less than the mass of incident partic...

If a, b and c are real numbers, and

Question: If $a, b$ and $c$ are real numbers, and $\Delta=\left|\begin{array}{lll}b+c c+a a+b \\ c+a a+b b+c \\ a+b b+c c+a\end{array}\right|=0$, Show that eithera+b+c= 0 ora=b=c. Solution: $\Delta=\left|\begin{array}{lll}b+c c+a a+b \\ c+a a+b b+c \\ a+b b+c c+a\end{array}\right|$ Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we have: $\Delta=\left|\begin{array}{lcc}2(a+b+c) 2(a+b+c) 2(a+b+c) \\ c+a a+b b+c \\ a+b b+c c+a\end{array}\right|$ $=2(a+b+c)\left|\begin{array}{ccc}1 1 1 \\ c+a a+b b...

Choose the correct alternative from the clues given at the end of the each statement:

Question: Choose the correct alternative from the clues given at the end of the each statement: (a)The size of the atom in Thomsons model is .......... the atomic size in Rutherfords model. (much greater than/no different from/much less than.) (b)In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force.(Thomsons model/ Rutherfords model.) (c)Aclassicalatom based on .......... is doomed to collapse.(Thomsons model/ Rutherfo...

Propanal and pentan-3-one are the ozonolysis products of an alkene?

Question: Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene? Solution: As per the given information, propanal and pentan-3-one are the ozonolysis products of an alkene. Let the given alkene be A. Writing the reverse of the ozonolysis reaction, we get: The products are obtained on the cleavage of ozonide X. Hence, X contains both products in the cyclic form. The possible structure of ozonide can be represented as: Now, X is an additio...

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Question: Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder. Solution: The two-digit numbers, which when divided by 4, yield 1 as remainder, are 13, 17, 97. This series forms an A.P. with first term 13 and common difference 4. Letnbe the number of terms of the A.P. It is known that the $n^{\text {th }}$ term of an A.P. is given by, $a_{n}=a+(n-1) d$ $\therefore 97=13+(n-1)(4)$ $\Rightarrow 4(n-1)=84$ $\Rightarrow n-1=21$ $\Rightarrow n=22$ Sum ofnterms of an A....

An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘

Question: An alkene A contains three C C, eight C Hbonds and one C Cbond. A on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of A. Solution: As per the given information, A on ozonolysis gives two moles of an aldehyde of molar mass 44 u. The formation of two moles of an aldehyde indicates the presence of identical structural units on both sides of the double bond containing carbon atoms. Hence, the structure of A can be represented as: XC = CX There are eight CH ...

Evaluate

Question: Evaluate $\left|\begin{array}{ccc}\cos \alpha \cos \beta \cos \alpha \sin \beta -\sin \alpha \\ -\sin \beta \cos \beta 0 \\ \sin \alpha \cos \beta \sin \alpha \sin \beta \cos \alpha\end{array}\right|$ Solution: $\Delta=\left|\begin{array}{ccc}\cos \alpha \cos \beta \cos \alpha \sin \beta -\sin \alpha \\ -\sin \beta \cos \beta 0 \\ \sin \alpha \cos \beta \sin \alpha \sin \beta \cos \alpha\end{array}\right|$ Expanding along $\mathrm{C}_{3}$, we have: $\begin{aligned} \Delta =-\sin \alpha...

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Question: Find the sum of integers from 1 to 100 that are divisible by 2 or 5. Solution: The integers from 1 to 100, which are divisible by 2, are 2, 4, 6 100. This forms an A.P. with both the first term and common difference equal to 2. ⇒100 = 2 + (n1) 2 ⇒n= 50 $\therefore 2+4+6+\ldots+100=\frac{50}{2}[2(2)+(50-1)(2)]$ $=\frac{50}{2}[4+98]$ $=(25)(102)$ $=2550$ The integers from 1 to 100, which are divisible by 5, are 5, 10 100. This forms an A.P. with both the first term and common difference ...

Answer the following questions:

Question: Answer the following questions: (a)Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e; (1/3)e]. Why do they not show up in Millikans oil-drop experiment? (b)What is so special about the combinatione/m? Why do we not simply talk ofeandmseparately? (c)Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d)Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if in...

An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one.

Question: An alkene A on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of A. Solution: During ozonolysis, an ozonide having a cyclic structure is formed as an intermediate which undergoes cleavage to give the final products. Ethanal and pentan-3-one are obtained from the intermediate ozonide. Hence, the expected structure of the ozonide is: This ozonide is formed as an addition of ozone to A. The desired structure of A can be obtained by the removal of oz...

Compute the typical de Broglie wavelength of an electron in a metal at 27 ºC and compare

Question: Compute the typical de Broglie wavelength of an electron in a metal at 27 C and compare it with the mean separation between two electrons in a metal which is given to be about 2 1010m. [Note:Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, e...

Find the sum of all numbers between 200 and 400 which are divisible by 7.

Question: Find the sum of all numbers between 200 and 400 which are divisible by 7. Solution: The numbers lying between 200 and 400, which are divisible by 7, are 203, 210, 217, 399 First term,a= 203 Last term,l= 399 Common difference,d= 7 Let the number of terms of the A.P. ben. $\therefore a_{n}=399=a+(n-1) d$ $\Rightarrow 399=203+(n-1) 7$ $\Rightarrow 7(n-1)=196$ $\Rightarrow n-1=28$ $\Rightarrow n=29$ $\therefore \mathrm{S}_{29}=\frac{29}{2}(203+399)$ $=\frac{29}{2}(602)$ $=(29)(301)$ $=8729...

Question: Compute the typical de Broglie wavelength of an electron in a metal at 27 C and compare it with the mean separation between two electrons in a metal which is given to be about 2 1010m. [Note:Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, e...

Without expanding the determinant, prove that

Question: Without expanding the determinant, prove that $\left|\begin{array}{lll}a a^{2} b c \\ b b^{2} c a \\ c c^{2} a b\end{array}\right|=\left|\begin{array}{lll}1 a^{2} a^{3} \\ 1 b^{2} b^{3} \\ 1 c^{2} c^{3}\end{array}\right|$ Solution: $\left|\begin{array}{ccc}a a^{2} b c \\ b b^{2} c a \\ c c^{2} a b\end{array}\right|=\left|\begin{array}{lll}1 a^{2} a^{3} \\ 1 b^{2} b^{3} \\ 1 c^{2} c^{3}\end{array}\right|$ L.H.S. $=\left|\begin{array}{lll}a a^{2} b c \\ b b^{2} c a \\ c c^{2} a b\end{arr...

Write IUPAC names of the products obtained by the ozonolysis of the following compounds:

Question: Write IUPAC names of the products obtained by the ozonolysis of the followingcompounds: (i) Pent-2-ene (ii) 3,4-Dimethyl-hept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene Solution: (i)Pent-2-ene undergoes ozonolysis as: The IUPAC name of Product (I) is ethanal and Product (II)is propanal. (ii)3, 4-Dimethylhept-3-ene undergoes ozonolysis as: The IUPAC name of Product (I)is butan-2-one and Product (II)is Pentan-2-one. (iii)2-Ethylbut-1-ene undergoes ozonolysis as: The IUPAC name o...

Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively,

Question: Let the sum of $n, 2 n, 3 n$ terms of an A.P. be $S_{1}, S_{2}$ and $S_{3}$, respectively, show that $S_{3}=3\left(S_{2}-S_{1}\right)$ Solution: Letaandbbe the first term and the common difference of the A.P. respectively. Therefore, $\mathrm{S}_{1}=\frac{n}{2}[2 a+(n-1) d]$$\ldots(1)$ $\mathrm{S}_{2}=\frac{2 n}{2}[2 a+(2 n-1) d]=n[2 a+(2 n-1) d]$ $\ldots(2)$ $\mathrm{S}_{3}=\frac{3 n}{2}[2 a+(3 n-1) d]$ $\ldots(3)$ From (1) and (2), we obtain $\mathrm{S}_{2}-\mathrm{S}_{1}=n[2 a+(2 n-...

Find the typical de Broglie wavelength associated

Question: Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions. Solution: De Broglie wavelength associated with $\mathrm{He}$ atom $=0.7268 \times 10^{-10} \mathrm{~m}$ Room temperature,T= 27C = 27 + 273 = 300 K Atmospheric pressure,P= 1 atm = 1.01 105Pa Atomic weight of a He atom = 4 Avogadros number, NA= 6.023 1023 Boltzmann constant,k= 1.38 ...

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Question: If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers. Solution: Let the three numbers in A.P. be $a-d, a$, and $a+d$. According to the given information, $(a-d)+(a)+(a+d)=24 \ldots(1)$ $\Rightarrow 3 a=24$ $\therefore a=8$ $(a-d) a(a+d)=440$ $\Rightarrow(8-d)(8)(8+d)=440$ $\Rightarrow(8-d)(8+d)=55$ $\Rightarrow 64-d^{2}=55$ $\Rightarrow d^{2}=64-55=9$ $\Rightarrow d=\pm 3$ Therefore, whend= 3, the numbers are 5, 8, and 11 and whend= 3, the numbers are 1...

Prove that the determinant is independent of θ.

Question: Prove that the determinant $\left|\begin{array}{ccc}x \sin \theta \cos \theta \theta \\ -\sin \theta -x 1 \\ \cos \theta 1 x\end{array}\right|$ is independent of $\theta$. Solution: $\Delta=\left|\begin{array}{ccc}x \sin \theta \cos \theta \\ -\sin \theta -x 1 \\ \cos \theta 1 x\end{array}\right|$ $=x\left(-x^{2}-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)$ $=-x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ...