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}...

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...

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 =...