If the threshold wavelength is $\lambda_{0} n m\left(=\lambda_{0} \times 10^{-9} \mathrm{~m}\right)$, the K.E. of the radiation would be: $ h\left(\nu-\nu_{0}\right)=\frac{1}{2} m \nu^{2} $ Three...
Given, the work function of caesium $\left(W_{0}\right)=1.9 \mathrm{eV}$ (a)From the $W_{0}=\frac{h c}{\lambda_{0}}$ expression, we get: $\lambda_{0}=\frac{h c}{W_{0}}$ Where, $\lambda_{0}$ is the...
From the expression of energy of one photon (E), $ E=\frac{h c}{\lambda} $ Where, $\lambda$ denotes the wavelength of the radiation $\mathrm{h}$ is Planck's constant c denotes the velocity of the...
No. quanta in $2 \mathrm{~J}$ of energy $\frac{2 J}{32.27 \times 10^{-20} J}$ $ =6.19 \times 10^{18} $ $ =6.2 \times 10^{18} $
Energy of one quantum $(E)=h v$ $ =\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(4.87 \times 10^{14} \mathrm{~s}^{-1}\right) $ Energy of one quantum $(\mathrm{E})=32.27 \times 10^{-20}...
Speed of the radiation, $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ Distance travelled by the radiation in a timespan of $30 \mathrm{~s}$ $ \begin{array}{l} =\left(3 \times 10^{8}...
Wavelength of the emitted radiation $=616 \mathrm{~nm}=616 \times 10^{-9} \mathrm{~m}$ (Given) (a)Frequency of the emission $(\nu)$ $ \nu=\frac{c}{\lambda} $ Where, $c=$ speed of the radiation...
Energy with which it emits photons=Power of laser Power $=E=\frac{N h c}{\lambda}$ Where, $N=$ number of photons emitted $\mathrm{h}=$ Planck's constant $\mathrm{c}=$ velocity of radiation...
Let us consider the total no. electrons present in $A^{3+}$ be $x$. Now, total no. neutrons in it $=x+30.4 \%$ of $x=1.304 x$ Since the ion has a charge of $+3, \Rightarrow$ no. electrons in neutral...
The number of electrons in a negatively charged ion be $x$. Then, no. neutrons present $=x+11.1 \%$ of $x=x+0.111 x=1.111 x$ No. electrons present in the neutral atom $=(x-1)$ (When an ion carries...
Let us consider that the No.of protons in the element be $x$. No. of neutrons $=x+31.7 \%$ of $x$ $=x+0.317 x$ $=1.317 x$ According to the question, Mass number of the element $=81$, which...
The general convention followed while representing elements along with their atomic masses (A), and their atomic numbers $(Z)$ is ${ }_{Z}^{A} \mathrm{X}$. Therefore, ${ }_{35}^{79} \mathrm{Br}$ is...
The findings obtained with a foil made up of heavy atoms will differ from those obtained with a foil made up of comparatively light atoms. The magnitude of positive charge in the nucleus of a...
Charge held by the oil drop $=1.282 \times 10^{-18} C$ Charge held by one electron $=1.6022 \times 10^{-19} C$ Therefore, No. electrons present in the drop of oil $\frac{1.282 \times 10^{-18}...
Charge held by one electron $=1.6022 \times 10^{-19} C \Rightarrow 1.6022 \times 10^{-19} C$ charge is held by one electron. Therefore, electrons carrying charge of $2.5 \times 10^{-16} C \quad...
Radius of carbon atom$ =\frac{2.6}{2} $ $ \begin{array}{l} =1.3 \times 10^{-10} \mathrm{~m} \\ =130 \times 10^{-12} \mathrm{~m}=130 \mathrm{pm} \end{array} $ (b) Length of the arrangement $=1.6...
Length of the arrangement $=2.4 \mathrm{~cm}$ No. carbon atoms present $=2 \times 10^{8}$ The diameter of the carbon atom $=\frac{2.4 \times 10^{-2} m}{2 \times 10^{8} m}$ $=1.2 \times 10^{-10} m...
We know that$ 1 \mathrm{~cm}=10^{-2} \mathrm{~m} $ Length of the scale $=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$ Diameter of one carbon atom $=0.15 \mathrm{~nm}=0.15 \times 10^{-9}...
The energy associated with hydrogen-like species is: $ E_{n}=-2.18 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right) J $ For the ground state of the hydrogen atom, $ \begin{array}{l} \Delta...
The wave number associated with the Balmer transition for the He+ ion (n = 4 to n = 2 ) is given by: $ \bar{\nu}=\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) $...
Only one electron exists in hydrogen atoms. The angular momentum of this electron, according to Bohr's postulates, is: $m v r=n \frac{h}{2 \pi} \quad \ldots .(1)$ Where, $n=1,2,3, \ldots$ As per de...
(a)The total number of electrons in the atom =$2n^2$ if n is the primary quantum number. Hence, For n = 4, Total no. electrons = 2 (16) = 32 An atom with 32 electrons has the following electron...
(a)n = 1, l = 0 implies a 1s orbital. (b)n = 3 and l = 1 implies a 3p orbital. (c)n = 4 and l = 2 implies a 4d orbital. (d)n = 4 and l = 3 implies a 4f orbital.
(I) The range of potential values for 'l' is 0 to (n – 1). As a result, the possible values of l for n = 3 are 0, 1, and 2. The total number of potential ml = (2l + 1) values. It has a range of...
The number of electrons in H2 is 1 + 1 = 2. 2 – 1 = 1 number of electrons in H2+ H2: The number of electrons in H2 equals 1 + 1 = 2. Number of electrons in O2 = 8 + 8 = 16. In O2+, the number of...
(i)No.protons = no.electrons in a neutral atom. The number of protons in the atoms of the element is 29. (ii)The electronic configuration of this element (atomic number 29) is 1s...
For the 3d orbital: Principal quantum number ,possible values (n) = 3 Azimuthal quantum number, possible values(l) = 2 Magnetic quantum number ,possible values(ml) = – 2, – 1, 0, 1, 2
$(I I I)(a)[H e] 2 s^{1}$ Complete electronic configuration: $1 \mathrm{~s}^{2} 2 \mathrm{~s}{ }^{1}$. $\therefore$ the element's atomic number is 3 . The element is lithium (Li) (b) $[\mathrm{Ne}]...
The electronic configuration of the Hydrogen atom (in its ground state) $1 \mathrm{~s}^{1}$. The single negative charge on this atom indicates that it has gained an electron. Hence, the electronic...
As per de Broglie’s equation, $ \lambda=\frac{h}{m v} $ Given, K.E of electron $=3.0 \times 10^{-25} \mathrm{~J}$ $ \begin{array}{l} \text { Since K.E.= } \frac{1}{2} m v^{2} \therefore...
As per de Broglie’s equation, $ \lambda=\frac{h}{m v} $ Where, $\lambda$ denotes thr wavelength of the moving particle $\mathrm{m}$ is the mass of the particle $v$ denotes the velocity of the...
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as: $E_{5}=\frac{-\left(2.18 \times 10^{-18}\right)}{(5)^{2}}=\frac{-\left(2.18 \times 10^{-18}\right)}{25}=-8.72 \times...
Frequency of radiation $\nu$ $\nu =\frac{1}{2.0\times 10^{-9}s}$ $\nu =5.0\times 10^{8}s^{-1}$ Energy (E) of source = Nhν Where, N is the no. photons emitted h is Planck’s constant ν denotes the...
The energy associated with an incident photon (E) must equal the sum of its kinetic energy and the work function (W0) of the radiation, according to the rule of conservation of energy. E = W0 + K.E...
Energy of incident photon (E) is given by, $E=\frac{hc}{\lambda }$ $E=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(150\times 10^{-12})}=1.3252\times 10^{-15}\, J$ $\simeq 13.252\times 10^{-16}J$...
Wavelength of the transition = 1285 nm =$1285 \times 10^{-9} m$(Given) $\nu =3.29\times 10^{15}(\frac{1}{3^{2}}-\frac{1}{n^{2}})$ Since$ \nu =\frac{c}{\lambda }$ =$\frac{3\times...
The radius of the n th orbit of hydrogen-like particles is given by, $r=\frac{0.529n^{2}}{Z}$ $r=\frac{5.29n^{2}}{Z}pm$ For radius (r1) = 1.3225 nm $=1.32225\times 10^{-9}m$ $=1322.25\times...
As per de Broglie’s equation, $\lambda =\frac{h}{m\nu }$ =$\frac{(6.626\times 10^{-34})}{9.103939\times 10^{-31}kg(1.6\times 10^{6}ms^{-1})}$ =$4.55\times 10^{-10}m\lambda =455pm$ Therefore, de...
From de Broglie’s equation, $\lambda =\frac{h}{m\nu }$ $\nu=\frac{h}{m\lambda}$ Where, v denotes the velocity of the neutron h is Planck’s constant m is the mass of the neutron λ is the wavelength...
As per de Broglie’s equation, $\lambda =\frac{h}{m\nu}$ Where, λ is the wavelength of the electron h is Planck’s constant m is the mass of the electron v denotes the velocity of electron...
As per de Broglie’s expression, $\lambda =\frac{h}{m\nu }$ $ =\frac{(6.626\times 10^{-34})}{0.1kg(4.37\times 10^{5}ms^{-1})}$=$1.516\times 10^{-38}m$
As per Heisenberg’s uncertainty principle, ∆x.∆p >= h/4π Where, ∆x = uncertainty in the position of the electron ∆p = uncertainty in the momentum of the electron Substituting the given values in...
The nuclear charge that electrons (which are present in atoms containing multiple electrons) feel is determined by the distance between their orbital and the atom's nucleus. The smaller the...
The net positive charge acting on an electron in an atom's orbital with more than one electron is known as the nuclear charge. The distance between the orbital and the nucleus is inversely...
(a)Phosphorus (P): The atomic number of phosphorus is 15 Electronic configuration of Phosphorus: 1s 2 2s 2 2p 6 3s 2 3p 3 This can be represented as follows: From the diagram, it can be observed...
(a)n = 4 (Given) For some value of ‘n’, the values of ‘l’ range from 0 to (n – 1). Here, the possible values of l are 0, 1, 2, and 3 Therefore, a total of 4 subshells are possible when n=4: the s,...
A total number of 15 lines (5 + 4 + 3 + 2 + 1) will be obtained in this hydrogen emission spectrum. The total number of spectral lines emitted when an electron drops to the ground state from the...
The expression for the ionization energy is given by, $E_{n} =\frac{-(2.18\times 10^{-18})Z^{2}}{n^{2}}$ Where Z denotes the atomic number and n is the principal quantum number For the ionization...
The $n_{i} = 4$ to$ n_{f}$= 2 transition results in a spectral line of the Balmer series. The energy involved in this transition can be calculated using the following expression: $E=2.18\times...
Threshold wavelength of the radiation $(\lambda _{0})= 6800 Å=6800\times 10^{-10}\, m$ Threshold frequency of the metal $(\nu _{0}) =\frac{c}{\lambda _{0}}=$ $\frac{3\times 10^{8}ms^{-1}}{6.8\times...
Power of the bulb,$ P = 25 Watt = 25\, Js^{-1}$ Energy (E) of one photon= $h\nu =\frac{hc}{\nu} $ Substituting these values in the expression for E: $E=\frac{(6.626\times 10^{-34})(3\times...
Ionization energy (E) of sodium =$ \frac{N_{A}hc}{\lambda }$ =$\frac{(6.023\times 10^{23}\, mol^{-1})(6.626\times 10^{-34})Js(3\times 10^{8})ms^{-1}}{242\times 10^{-9}m}$ =$4.947\times 10^{5}\, J\,...
(i) Energy of the photon $(E)= h\nu =\frac{hc}{\lambda }$ Where, h denotes Planck’s constant, whose value is $6.626\times 10^{-34}\,Js$ c denotes the speed of light =$ 3\times 10^{8}\,m/s$...
Energy of one photon (E) =$ h\nu$ Energy of ‘n’ photons $E_{n} = nh\nu \Rightarrow n=\frac{E_{n}\lambda }$ Where, \lambdaλ is the wavelength of the photons = 4000 pm = $4000\times 10^{-12}\, m$ c...
(i) The energy of a photon (E) can be calculated by using the following expression: $E= h\nu$ Where, ‘h’ denotes Planck’s constant, which is equal to $6.626\times 10^{-34}\, Js\nuν$ (frequency of...
Rearranging the expression, $\lambda =\frac{c}{\nu }$ the following expression can be obtained, $\nu =\frac{c}{ \lambda }$ ……….(1) Here, $\nu$ denotes the frequency of the yellow light c denotes...
(i)\({}_{17}^{35}C\) (ii)\({}_{92}^{233}C\) (iii)\({}_{4}^{9}C\)
(i) 1 molecule of methane contains 10 electrons (6 from carbon, 4 from hydrogen) Therefore, 1 mole of methane contains 10*NA = 6.022*1024 electrons. (ii) Number of neutrons in 14g (1 mol) of 14C =...
Ans: 24 is the mass number. The number of protons in an atom equals the number of atoms in the atom. Mass number – Atomic number = 24 – 12 = 12 neutrons Numerical value of mass nubmer = 56 Number...
Ans: Carbon-13 mass number = 13 Carbon atomic number = number of protons in a single carbon atom = 6 As a result, the total number of neutrons in a carbon atom is calculated as follows: mass number...
Ans: (i) 1 molecule of methane contains 10 electrons (6 from carbon, 4 from hydrogen) Therefore, 1 mole of methane contains 10*NA = 6.022*1024 electrons. (ii) Number of neutrons in 14g (1 mol)...
Ans: 1 electron weighs 9.109*10-31 kg. Therefore, number of electrons that weigh 1 g (10-3 kg) = 10-3kg/9.109*10-31 kg = 1.098*1027 electrons (ii) Mass of one mole of electrons = NA* mass of one...
(i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true and R is not the correct explanation of A. (iii) A is true and R is false. (iv) Both A and R are false....
(i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) Both A and R are false....
Column 1 Column 2 (i) Cr (a) [Ar]3d84s0 (ii) Fe2+ (b) [Ar]3d104s1 (iii) Ni2+ (c) [Ar]3d64s0 (iv) Cu (d) [Ar] 3d54s1 (e) [Ar]3d64s2 Column 1 Column 2 (i) Cr (d) [Ar]...
Column 1 Column 2 (i) Photon (a)Value is 4 for N shell (ii) Electron (b)Probability density (iii) ψ2 (c)Always a positive value (iv) The principal quantum number n (d)Exhibits both momentum...
Column 1 Column 2 (i) X-rays (a) ν = 100-0 104 Hz (ii) UV (b) ν = 1010Hz (iii) Long radio waves (c) ν = 1016 Hz (iv) Microwave (d) ν = 1018Hz Column 1 Column 2 (i) X-rays (d) ν = 1018Hz (ii) UV...
Rule Principle (i) Hund’s Rule (a) No two electrons in an atom can have the same set of four quantum numbers. (ii) Aufbau Principle (b) Half-filled and filled orbitals have extra stability....
Column 1 Column 2 (i) Principal quantum number (a) orientation of the orbital (ii) Azimuthal quantum number (b) energy and size of orbital (iii) Magnetic quantum number (c) spin of an electron (iv)...
Atom / Ion Electronic configuration (i) Cu (a) 1s2 2s2 2p6 3s2 3p6 3d10 (ii) Cu2+ (b) 1s2 2s2 2p6 3s2 3p6 (iii) Zn2+ (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (iv) Cr3+ (d) 1s2 2s2 2p6 3s2 3p63d9 (e) 1s2...
Hydrogen atom has only one electron, so the mutual repulsion between the electrons is non-existent. However, in multielectron atoms the interaction between electrons is important. This is because,...
Distance between the two identical successive points in a wave is called the wavelength. Given, In the hypothetical wave, Wavelength, λ = 4 × 2.16 pm Hence, Wavelength, λ = 8.64 pm.
(i) 1:2 (ii) 1:1 (iii) 1:3 (iv) 3:1 Correct Answer: (iii) 1:3 Explanation: Ratio of isotope abundance of Chlorine 37 to Chlorine 35 = A1:A2 Ratio of isotope abundance of Chlorine 37...
(i) l (ii) n and l (iii) n and m (iv) m and s Correct Answer: (i) l Explanation: Orbital angular momentum depends on the value of l which is referred to as the azimuthal quantum number.
(i) 2 (ii) 4 (iii) 9 (iv) 3 Correct Answer: (iii) 9 Explanation: The total number of orbitals in nth shell = n2 Hence, the total number of orbitals associated with the third shell will...
(i) 3 (ii) 4 (iii) 2 (iv) 1 Correct Answer: (iv) 1 Explanation: Number of radial nodes for 3p orbital = 3−1−1 Number of radial nodes for 3p orbital = 3−2 Number of radial nodes for 3p...
(i) they have the same atomic number but a different mass number. (ii) they have the same number of electrons but a different number of neutrons. (iii) they have the same number of neutrons but a...
(i) Overall neutrality of atom. (ii) Spectra of a hydrogen atom. (iii) Position of electrons, protons and neutrons in an atom. (iv) Stability of atom. Correct Answer: (i) Overall...
(i) It is a negatively charged particle. (ii) The mass of an electron is equal to the mass of a neutron. (iii) It is a basic constituent of all atoms. (iv) It is a constituent of cathode rays....
(i) They start from the cathode and move towards the anode. (ii) They travel in a straight line in the absence of an external electrical or magnetic field. (iii) Characteristics of cathode rays do...
The density of dots in a region represents the probability density of finding electrons in the region. Based on the above diagram which of the following statements is incorrect? (i) 1s and 2s...
(i) 1s2 2s2 2p6 3s2 3p6 3d8 4s2 (ii) 1s2 2s2 2p6 3s2 3p6 3d9 4s2 (iii) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (iv) 1s2 2s2 2p6 3s2 3p6 3d5 4s1 Correct Answer: (ii) 1s2 2s2 2p6 3s2 3p6 3d9 4s2...
(i) Most of the space in the atom is empty. (ii) The radius of the atom is about 10–10 m while that of a nucleus is 10–15 m. (iii) Electrons move in a circular path of fixed energy called orbits....