What is the difference between
Question: What is the difference between the terms orbit and orbital? Solution: The orbit stands for a definite circular path for the electrons to revolve around the nucleus. It represents the two-dimensional motion of the electrons around the nucleus, the orbital is not that well-defined path because its a region around the nucleus where the probability of finding an electron is maximum....
Read More →Chlorophyll present in green leaves
Question: Chlorophyll present in green leaves of plants absorbs light at 4.620 1014 Hz. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to? Solution: Relation b/w wavelength and frequency can be expressed as : = c/, where c be the velocity of light and is the frequency of the radiation. For the given problem = 3 x 108 ms-1 / 4.620 x 1014 Hz = 0.6494 times10-6m-1...
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $2(b c \cos A+c a \cos B+a b \cos C)=\left(a^{2}+b^{2}+c^{2}\right)$ Solution: Need to prove: $2(b c \cos A+c a \cos B+a b \cos C)=\left(a^{2}+b^{2}+c^{2}\right)$ Left hand side $2(b c \cos A+c a \cos B+a b \cos C)$ $2\left(\mathrm{bc} \frac{\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{a}^{2}}{2 \mathrm{bc}}+\mathrm{ca} \frac{\mathrm{c}^{2}+\mathrm{a}^{2}-\mathrm{b}^{2}}{2 \mathrm{ca}}+\mathrm{ab} \frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}}{2 \mathrm{ab}}\right...
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $\left(x_{0}, y_{0}\right)$ Solution: finding the slope of the tangent by differentiating the curve $\frac{x}{a^{2}}-\frac{y}{b^{2}} \frac{d y}{d x}=0$ $\frac{d y}{d x}=\frac{b^{2} x}{y a^{2}}$ $\mathrm{m}$ (tangent) at $\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)=\frac{\mathrm{b}^{2} \mathrm{x}_{0}}{\mathrm{y}_{0} \mathrm{a}^{2}}$ normal is pe...
Read More →Out of electron and proton
Question: Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it. Solution: Out of electron and proton, being the lighter particle electron will have a higher velocity and will also produce matter waves of the same wavelength....
Read More →What is the experimental evidence
Question: What is the experimental evidence in support of the idea that electronic energies in an atom are quantized? Solution: The bright-line spectrum shows that the energy levels in an atom are quantized. These lines are obtained as a result of electronic transitions between the energy .and if the electronic energy levels were continuous and not quantized or discrete; the atomic spectra would have shown a continuous absorption(from lower to higher energy level transition) or emission (from hi...
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{c-b \cos A}{b-\cos A}=\frac{\cos B}{\cos C}$ Solution: Need to prove: $\frac{\mathrm{c}-\mathrm{b} \cos \mathrm{A}}{\mathrm{b}-\mathrm{c} \cos \mathrm{A}}=\frac{\cos \mathrm{B}}{\cos \mathrm{C}}$ Left hand side $=\frac{c-b \cos A}{b-c \cos A}$ $=\frac{c-b \frac{b^{2}+c^{2}-a^{2}}{2 b c}}{b-c \frac{b^{2}+c^{2}-a^{2}}{2 b c}}$ $=\frac{\frac{2 c^{2}-b^{2}-c^{2}+a^{2}}{2 c}}{\frac{2 b^{2}-b^{2}-c^{2}+a^{2}}{2 b}}$ $=\frac{\frac{c^{2}+a^{2}-b^{2}}{2 c}}{\frac{...
Read More →According to de Broglie,
Question: According to de Broglie, the matter should exhibit dual behaviour, that is both particle and wave-like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature. Solution: m= 100g or 0.1kg = 100km/h =100*1000/60*60 = 1000/36m/s =h/m = 2.387*10-34m...
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $\left(x_{1}, y_{1}\right)$ Solution: finding the slope of the tangent by differentiating the curve $\frac{x}{a^{2}}-\frac{y}{b^{2}} \frac{d y}{d x}=0$ $\frac{d y}{d x}=\frac{b^{2} x}{y a^{2}}$ $\mathrm{m}($ tangent $)$ at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\frac{\mathrm{b}^{2} \mathrm{x}_{1}}{\mathrm{y}_{1} \mathrm{a}^{2}}$ normal is...
Read More →The Balmer series in the hydrogen spectrum
Question: The Balmer series in the hydrogen spectrum corresponds to the transition from n1 = 2 to n2 = 3,4,. This series lies in the visible region. Calculate the wavenumber of the line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (RH= 109677 cm-1) Solution: According to Bohrs model for the hydrogen atom = RH(1/n12-1/ n22)cm-1 here, n1 = 2 and n2 = 4 and H = Rydbergs constant = 109677 Hence, wave number -v= 109677 ( -1/16) = 20564.44cm-1...
Read More →The electronic configuration of the valence
Question: The electronic configuration of the valence shell of Cu is 3d10 4s1 and not 3d94s2. How is this configuration explained? Solution: Configuration with filled and half-filled orbitals has extra stability. In 3d104s1, d orbitals are filled and s orbitals are half-filled....
Read More →Wavelengths of different radiations are given below :
Question: Wavelengths of different radiations are given below : (A) = 300 nm (B) = 300 m (c) = 3 nm (D) 30 A Arrange these radiations in the increasing order of their energies. Solution: According to Plancks quantum theory, energy is related to the frequency of radiation by : E = h Frequency So, E is proportional to 1/ Hence, the relation b/w energy and wavelength are inversely proportional, therefore lesser the wavelength higher will be the energy of the radiation. For the given wavelengths (A)...
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=\frac{\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)}{2 \mathrm{abc}}$ Solution: Need to prove: $\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=\frac{\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)}{2 \mathrm{abc}}$ Left hand side $=\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \m...
Read More →An atom having atomic mass number 13 has 7
Question: An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom? Solution: The mass number of an atom = number of protons + number of neutrons Therefore atomic number ( number of protons ) = mass number no. Of neutrons. The atomic number of atom: 13 7 = 6....
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $x y=c^{2}$ at $(c t, c / t)$ Solution: finding slope of the tangent by differentiating the curve $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}$ $m($ tangent $)$ at $\left(c t, \frac{c}{t}\right)=-\frac{1}{t^{2}}$ normal is perpendicular to tangent so, $m_{1} m_{2}=-1$ $\mathrm{m}$ (normal) at $\left(\mathrm{ct}, \frac{\mathrm{c}}{\mathrm{t}}\right)=\mathrm{t}^{2}$ equatio...
Read More →Which of the following will not show deflection
Question: Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron Solution: Neutron will not show deflection from the path on passing through an electric field. This is due to the neutral nature of the neutron particles. Therefore, it has no charge and does not get affected by any electric field. Among other 3 particles proton (positive ), electron (negative), cathode rays (the beam of electrons, negatively charg...
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $a c \cos B-b c \cos A=\left(a^{2}-b^{2}\right)$ Solution: Left hand side, $a c \cos B-b c \cos A$ $=a c \frac{a^{2}+c^{2}-b^{2}}{2 a c}-b c \frac{b^{2}+c^{2}-a^{2}}{2 b c}\left[A s, \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right]$ $=\frac{a^{2}+c^{2}-b^{2}}{2}-\frac{b^{2}+c^{2}-a^{2}}{2}$ $=\frac{a^{2}+c^{2}-b^{2}-b^{2}-c^{2}+a^{2}}{2}$ $=\frac{2\left(a^{2}-b^{2}\right)}{2}$ $=a^{2}-b^{2}$ $=$ Right hand side. [Proved]...
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $c^{2}\left(x^{2}+y^{2}\right)=x^{2} y^{2}$ at $\left(\frac{c}{\cos \theta}, \frac{c}{\sin \theta}\right)$ Solution: finding the slope of the tangent by differentiating the curve $c^{2}\left(2 x+2 y \frac{d y}{d x}\right)=2 x y^{2}+2 x^{2} y \frac{d y}{d x}$ $2 x c^{2}-2 x y^{2}=2 x^{2} y \frac{d y}{d x}-2 y c^{2} \frac{d y}{d x}$ $\frac{d y}{d x}=\frac{x c^{2}-x y^{2}}{x^{2} y-y c^{2}}$ $\...
Read More →The arrangement of orbitals based on energy
Question: The arrangement of orbitals based on energy is based upon their (n+l ) value. Lower the value of (n+l ), lower is the energy. For orbitals having the same values of (n+l), the orbital with a lower value of n will have lower energy. I. Based upon the above information, arrange the following orbitals in the increasing order of energy (a) 1s, 2s, 3s, 2p (b) 4s, 3s, 3p, 4d (c) 5p, 4d, 5d, 4f, 6s (d) 5f, 6d, 7s, 7p II. Based upon the above information, solve the questions given below : (a) ...
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $a(b \cos C-c \cos B)=\left(b^{2}-c^{2}\right)$ Solution: Left hand side, $a(b \cos C-c \cos B)$ $=a b \cos C-a c \cos B$ $=a b \frac{a^{2}+b^{2}-c^{2}}{2 a b}-a c \frac{a^{2}+c^{2}-b^{2}}{2 a c}\left[A s, \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right]$ $=\frac{a^{2}+b^{2}-c^{2}}{2}-\frac{a^{2}+c^{2}-b^{2}}{2}$ $=\frac{a^{2}+b^{2}-c^{2}-a^{2}-c^{2}+b^{2}}{2}$ $=\frac{2\left(b^{2}-c^{2}\right)}{2}$ $=b^{2}-c^{2}$ $=$ Right ...
Read More →Calculate the total number of angular
Question: Calculate the total number of angular nodes and radial nodes present in 3p orbital. Solution: Nodes are the region present among the orbitals where the probability density of finding electrons will be zero. In case of np orbitals , radial nodes = n l 1 = 3 1 1 = 1 Angular nodes = l = 1....
Read More →Which of the following orbitals
Question: Which of the following orbitals are degenerate? 3dxy, 4dxy, 3dz2 , 3dyx, 4dyx, 4dzz Solution: The energy of orbitals depends on the principal quantum number or the main shell to a large extent. Hence, orbitals with an equal value of n will have the same levels of energy and will be called degenerate orbitals. Degenerate orbitals are 3dxy, 3dz2, 3dyx because they have the same main shell n = 3. And 4dxy, 4dyx, 4dzz because they have the same value of n=4....
Read More →Nickel atom can lose two electrons
Question: Nickel atom can lose two electrons to form Ni2+ ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons. Solution: One Ni atom has 28 electrons and its electronic configuration is : [Ar] 4s2 3d8 It becomes Ni2+ by losing 2 electrons, hence configuration of Ni2+ is : [Ar] 4s0 3d8 So, nickel loses two electrons from the 4s orbital, not the 3d orbital as per the Aufbau principle...
Read More →Arrange s, p and d sub-shells of a
Question: Arrange s, p and d sub-shells of a shell in the increasing order of effective nuclear charge (Zeff) experienced by the electron present in them. Solution: dps S orbitals shield the electrons from the nucleus more than p-orbitals which shield more in d....
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $y^{2}=4 a \times a t\left(a / m^{2}, 2 a / m\right)$ Solution: finding the slope of the tangent by differentiating the curve $2 y \frac{d y}{d x}=4 a$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{\mathrm{y}}$ $\mathrm{m}$ (tangent) at $\left(\frac{\mathrm{a}}{\mathrm{m}^{2}}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)$ $\mathrm{m}$ (tangent) $=\mathrm{m}$ normal is perpendicular to tan...
Read More →