Physics M.C.Q (1 Marks)

It's been determined experimentally that when light shines on a metal surface, the surface emits electrons. For example, you can start a current in a circuit just by shining a light on a metal plate. we were saying earlier that light is made up of electromagnetic waves, and that the waves carry energy. So if a wave of light hit an electron in one of the atoms in the metal, it might transfer enough energy to knock the electron out of its atom. Light has sometimes been viewed as a particle $($photon$)$ rather than a wave. If it's waves, the energy contained in one of those waves should depend only on its amplitude$--$that is, on the intensity of the light.

First electron shell which is closest to the orbit to the nucleus, called the $1s$ orbit can hold up two two electrons. This orbit is equivalent to the innermost electron shell of the Bhor model of atom. It is called the orbital because it is spherical around the nucleus.

Line respectrum is characteristic of the element because it obtained of atoms only. Continous spectrum is characteristics of the source of light.
In the spectrum of sodium, there are two prominent yellow lines of wavelength $589.0\ nm$ and $589.6\ nm.$
Absorption line spectrum is not characteristic of the element.

According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplt of $\frac{\text{h}}{2\pi}$
$\therefore\ \text{L}_\text{n}=\frac{\text{nh}}{2\pi}$
Here,
$n =$ Principal quantum number
The minimum of $n$ is $1$
Thus, the minimum value of the orbit angular momentum of the electron in a hydrogen is given by $\text{L}=\frac{\text{h}}{2\pi}$

Key concept: Total energy $(E)$ is the sum of potential energy and kinetic energy, i.e.$ E = K + U$
$\Rightarrow\ \text{E}=-\frac{\text{kZe}^2}{2\text{r}_\text{n}}\text{also r}_\text{n}=\frac{\text{n}^2\text{h}^2\in_0}{\pi\text{mze}^2}$
Hence $\text{E}=-\bigg(\frac{\text{me}^4}{8\in_0^2\text{h}^2}\bigg)\frac{\text{Z}^2}{\text{n}^2}=-\bigg(\frac{\text{me}^4}{8\in_0^2\text{ch}^3}\bigg)\text{ch}\frac{\text{Z}^2}{\text{n}^2}$
$=-\text{R ch}\frac{\text{Z}^2}{\text{n}^2}=-13.6\frac{\text{Z}^2}{\text{n}^2}\text{eV}$
The lowest state of the atom, called the ground state, is that of the lowest energy. The energy of this state $(n = 1), E_1$ is $-13.6\ eV$
Energy level diagram of hydrogen/hydrogen like atom:

Transition from higher states to $n = 2$ lead to emission of radiation with wavelengths $656.3\ nm$ and $365.0\ nm.$
These wavelengths fall in the visible region and constitute the Balmer series.

The wavelength corresponding the transition from $n_2$ to $n_1$ is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$
Here,
$R =$ Rydberg constant.
$Z =$ Atomic number of the ion.
From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.
Therefore, the wavelength corresponding to $n = 2$ to $n = 1$ will be minimum in doubly ionized lithium ion because for lithium, $Z = 3.$

All the photons emitted in the laser move with the speed equal to the speed of light $(c = 3 \times 10^8m/s)$. Ideally, the light wave through the laser must be coherent, but in practical laser tubes, there is some deviation from the ideal result. Thus, the photons emitted by the laser have little variations in their wavelengths and energies as well as the directions, but the velocity of all the photons remains same.

The hydrogen atom is stable when the electron rests at the ground level of energy, i.e. when the principal quantum number, $n = 1. '$
In other words, when the electron is revolving in the first orbit around the nucleus, the hydrogen atom is stable.

Neil Bohr proposed the concept of stationary orbits in $1913$, which is now called the Bohr model of atom. The electron can only orbit stably, without radiating, in certain orbits at a certain discretesct of distance from the nucleus.

A discharge tube is a tube containing charged electrodes and filled with a gas in which ionization is induced by an electric filed. $X-$ray tube, fluroscent tube and cathode rays oscilloscope can be used as a modified discharge tube in which these tubes contain charged electrodes and filled with a gas.

The proton, neutron, and electron possess half$-$integer spin states

Coulomb force is responsible for scattering of particles. When the alpha particles $($positive in charge$)$ get closer to the nucleus, which is positive in charge, they get repelled through various angles.

Energy of photon $=\frac{\text{hc}}{\gamma}$
Hence, a photon with longer wavelenght will carry less energy.

Atomic No $=$ no of proton $= 8$
Oxygen has atomic number $8$. So, element $X$

Niels Bohr introduced the atomic Hydrogen model in $1913.$
Neils Bohr developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level $($or orbit$)$ to another.

Radius is directly proportional to square of $n.$
Therefore, ratio of principal quantum number is $1 : 3.$
Hence, ratio of frequencies is $27 : 1.$

The energies of the photons emitted can be expressed as follows,

de $–$ Broglie hypothesis did modify Bohr’s second postulate.
This postulate of Bohr regarding the quantization of the angular momentum of an electron was further explained by Louis de Broglie.
According to de $–$ Broglie, a moving electron in its circular orbit behaves like a particle$-$wave.

The word atom is derived from Greek word, 'atomos'. In Greek, the prefix $"a"$ means "not" and the word "tomos" means cut. Our word atom therefore comes from atomos, a Greek word meaning uncuttable.

For a hydrogen-like ion with $Z$ protons in the nucleus, the radius of the nth state is given by,
$\text{r}_\text{n}=\frac{\text{n}^2\text{a}_0}{\text{Z}}$
Here,
$a_0 = 0.53\ pm$
For lithium, $Z = 3$ Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

The energy of characteristic $X-$ray is a consequence of transition in target atoms.
They cause emission or absorption of electromagnetic radiation.
The other options are not responsible for the energy of characteristic $X-$rays.

The valence electron in an alkali metal is an s-electron.
Generally, they make up Group $1$ of the periodic table.
The different examples that come under this category are lithium, potassium, and francium.

The radius of an electron orbit in a hydrogen atom is of the order of $ 10^{-11} \mathrm{~m} $.
It is equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.

The ionisation energy of a hydrogen like ion of atomic number $Z$ is given by,
$\text{V}=(13.6\text{eV})\times\text{Z}^2$
Thus, the atomic number of ion $A$ is greater than that of B$(Z_A > Z_B)$.
The radius of the orbit is inversely proportional to the atomic number of the ion.
$\therefore\ \text{r}_\text{A}>\text{r}_\text{B}$
The speed of electron is directly proportional to the atomic number.
Therefore, the speed of the electron in the orbit of $A$ will be more than that in $B.$
Thus, $u_A > u_B$ is correct.
The total energy of the atom is given by,
$\text{E}=-\frac{\text{mZ}^2\text{e}^2}{8\in_0\text{h}^2\text{n}^2}$
As the energy is directly proportional to $Z^2$, the energy of $A$ will be less than that of $B,$ i.e. $E_A < E_B$.
The orbital angular momentum of the electron is independent of the atomic number. Therefore, the relation $L_A > L_B$ is invalid.

Molybdenum is used as a target in Coolidge tube for product.
$X-$ ray tube is an energy converter and it’s a device made of cathode and anode.
An electrical current flows through the tube from cathode to anode, when electron undergoes a energy loss, which results in generation of $x -$rays.
Anode is a component in which $x-$rays are produced. It has $2$ primary functions. To convert electricity to $x-$ray
To dissipate heat in the process
For this, material such as molybdenum and tungsten are used with high atomic number which has good heat storage capacity and low rate of evaporation.

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}_1}-\frac{1}{\text{n}_2}\Big)$
Here, $R$ is the Rydberg constant.
For the transition from $n = 5$ to $n = 4$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{4^2}-\frac{1}{5^2}\Big)$
$\lambda=\frac{400}{9\text{RZ}^2}$
For the transition from $n = 4$ to $n = 3$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{3^2}-\frac{1}{4^2}\Big)$
$\lambda=\frac{144}{7\text{RZ}^2}$
For the transition from $n = 3$ to $n = 2$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$
$\lambda=\frac{36}{5\text{RZ}^2}$
For the transition form $n = 2$ to $n = 1$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$\lambda=\frac{2}{\text{RZ}^2}$
From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from $n = 2$ to $n = 1$ will be minimum.

Atomic number of any element is equal to number of proton or number of electron
Atomic Number $=$ no. of proton $=$ no of electron.

The radius of the $\mathrm{n}^{\text {th }}$ orbit in one electron system is given by,
$\mathrm{r}_{\mathrm{n}}=\frac{\mathrm{n}^2 \mathrm{a}_0}{\mathrm{z}}$
Here, $a_0=53 \mathrm{pm}$
For the shortest orbit, $\mathrm{n}=1$
For hydrogen, $Z=1$
$\therefore$ Radius of the first state of hydrogen atom $=53 \mathrm{pm}$
For deuterium, $Z=1$
$\therefore$ Radius of the first state of deuterium atom $=53 \mathrm{pm}$
For $\mathrm{He}^{+}, \mathrm{Z}=2$
$\therefore$ Radius of $\mathrm{He}^{+}$atom $=\frac{53}{2} \mathrm{pm}=26.5 \mathrm{pm}$
For $\mathrm{Li}^{++}, \mathrm{Z}=3$
$\therefore$ Radius of $\mathrm{Li}^{++}$atom $=\frac{53}{3} \mathrm{pm}=17.66 \approx 18 \mathrm{pm}$
The given one$-$electron system having radius of the shortest orbit to be $18\ pm$ may be $\mathrm{Li}^{++}$

$J.J$. Thomson proposed plum pudding model in $1904$. Ernest Rutherford introduced the planetary model of an atom in $1911$ which was modified by Neils Bohr in $1913$ as Bohr model of an atom.
Order in which the atomic models were advanced: Plum pudding, planetary, Bohr.

In a Cathode ray tube electrons comes out from cathode. After cathode, anode is placed, which as being $+$ ively charged, accelerates electrons and provide them kinetic energy.

When an alpha particle is emitted from a radioactive source or substance, its atomic number decreases by $2$ and its atomic mass decreases by $4$, which is same as that of helium ion.

de $–$ Broglie concluded his modification of Bohr’s second postulate by stating that the wavelengths of matter waves can be quantized.
This implies that the electrons can exist in those orbits which had a complete set of several wavelengths.

The emission spectrum of $\mathrm{CO}_2$​ gas that has been studied by the electron beam excitation method is a band spectrum.

The first postulate of Bhor theory is that the orbital momentum of the electron is quantized ie, $L = mvr = nh$ where $h$ is Drac constant.

The minimum energy required to excite a hydrogen atom from its ground state to $1^{\text {st }}$ excited state is approximately $10\ eV.$ As the incident electron energy is not sufficient for excitation of the hydrogen atom so electron will not get absorbed in the hydrogen atom so it can not be an inelastic collision. Also this collision can not be partially elastic because in an partially elastic collision, there is a net loss on kinetic energy. If the energy is lost then corresponding amount of heat should have been produced but it is not so which implies that the collision is completely elastic.

Since the brightness or intensity of the display depends on the number of electrons that strike the screen, the control grid is used to control the brightness of the $\text{CRT.}$

The total energy of a hydrogen$-$like ion, having $Z$ protons in its nucleus, is given by,
$\text{E}=-\frac{13.6\text{Z}^2}{\text{n}^2}\text{eV}$
Here, $n =$ Principal quantum number.
For ground state, $n = 1$
$\therefore$ Total energy, $E = -13.6Z^2\ eV$
For hydrogen, $Z = 1$
$\therefore$ Total energy, $E = -13.6\ eV$
For deuterium, $Z = 1$
$\therefore$ Total energy, $e = -13.6\ eV$
For $He^+, Z = 2$
$\therefore$ Total energy $E = -13.6 \times 2^2 = -54.4eV$
For $Li^{++}$,
$Z = 3$
$\therefore$ Total energy, $E = -13.6 \times 3^2 = -122.4\ eV$
Hence, the ion having an energy of $-54.4\ eV$ in its ground state may be $He^+$

The electrons are accelerated by a second anode at high potential, more than $500V$

Niels Bohr calculated the energies that a hydrogen atom would have in each of its energy levels, based on the wavelength of the spectral lines.
Then he found out that there are four spectral lines for hydrogen, namely, Lyman, Balmer, Paschen, and Brackett series.
The Lyman series lies in the $UV$ region, whereas the Balmer series lies in the visible region, and the last two lie in the infrared region.

Values of Principle quantum number are $1, 2, 3, 4.....8. 0$ is not a Principle quantum number.
Only odd numbers are not Principle quantum numbers either.
Odd numbers, as well as even numbers, are Principle quantum number except $0$ and negative integers.

1. Mainly there are five series and each series is named after its discoverser as Lyman series, Balmer series, Paschen series, Bracked series and pfumd series.
2. According to the Bohr's theory, the wavelength of the rediations emitted from hydrogen atom is given by

Values for the quantum number ml​ are $ -1, 0, + 1$

Bohr showed that the electron in the hydrogen atom could only be found in certain selected $($quantized$)$ orbits, and no others.

Spins are paired means clockwise and anticlock wise rotation cancel each other, so it does not have any free electron.
$\therefore$ it is in dimagnetic in nature.

In atomic physics, hyperfine structure is the different effects leading to small shifts and splitting in the energy levels of atoms, molecules and ions. The name is a reference to the fine structure which results from the interaction between the magnetic moments associated with electron spin and the electrons' orbital angular momentum. Hyperfine structure, with energy shifts is typically orders of magnitude smaller than the fine structure, results from the interactions of the nucleus $($or nuclei, in molecules$)$ with internally generated electric and magnetic fields.

Atomic number of any element is equal no of electron.
Atomic no. $=$ no. of proton $=$ no of electron.

John Dalton $(1766 – 1844)$, an English chemist and physicist, is the man credited today with the development of atomic theory. However, a theory of atoms was actually formulated $2,500$ years before Dalton by an Indian sage and philosopher, known as Acharya Kanad.

The time by a photoeletron to come out after the photon strikes is approximately $10^{-10}$ seconds. This is a fact.

Key concept: Bohr's redius of orbit $($for Hydrogen and $H_2$-like atoms$)$: For an electron around a stationary nucleus, the electrostatics force of attraction provides the necessary centripetal force.

i.e., $\frac{1}{4\pi\in_0}\frac{(\text{Ze})\text{e}}{\text{r}^2}=\frac{\text{mv^2}}{\text{r}}\ .....(\text{i})$
Also $\text{mvr}=\frac{\text{nh}}{2\pi}\ .....(\text{ii})$
From equation $(i)$ and $(ii),$ radius of $n^{th}$ orbit
$\text{r}_\text{n}=\frac{\text{n}^2\text{h}^2}{4\pi^2\text{kZme}^2}=\frac{\text{n}^2\text{h}^2\in_0}{\pi\text{mZe}^2}=0.53\frac{\text{n}^2}{\text{Z}}\mathring{\text{A}}\ \Big(\text{k}=\frac{1}{4\pi\in_0}\Big)$
$\Rightarrow\ \text{r}_\text{n}\propto\frac{\text{n}^2}{\text{Z}}\text{ or }\text{r}_\text{n}\propto\frac{1}{\text{Z}}$
$\text{r}_\text{n}=\text{a}_0\frac{\text{n}^2}{\text{Z}},$ where $a_0 =$ the Bohr radius $= 53\ pm$
The atomic number $(Z)$ of lithium is $3.$
As $\text{r}_\text{n}=\text{a}_0\frac{\text{n}^2}{\text{Z}},$
Therefore, the radius of $Li^{++}$ ion in its ground statr, on the basis of Bohr's model, will be about $\frac{1}{3}$ times to that of Bohr radius.
Therefore, the radius of lithium ion is near $\text{r}=\frac{53}{3}\approx18\text{pm}.$

The angular momentum of the electron for the $n^{th}$ state is given by,
$\text{L}_\text{n}=\frac{\text{nh}}{2\pi}$
Angular momentum of the electron for $\text{n}=3,\ \text{L}_\text{i}=\frac{3\text{h}}{2\pi}$
Angular momentum of the electron for $\text{n}=2,\ \text{L}_\text{f}=\frac{2\text{h}}{2\pi}$
The torque is the time rate of change of the angular momentum.
Torque, $\tau=\frac{\text{L}_\text{f}-\text{L}_\text{i}}{\text{t}}$
$=\frac{\big(\frac{2\text{h}}{2\pi}\big)-\big(\frac{3\text{h}}{2\pi}\big)}{10^{-8}}$
$=\frac{-\big(\frac{\text{h}}{2\pi}\big)}{10^{-8}}$
$=\frac{-10^{-34}}{10^{-8}}$ $\Big[\because\frac{\text{h}}{2\pi}\approx10^{-34}\text{J}-\text{s}\Big]$
$=-10^{-42}\text{N}-\text{m}$
The magnitude of the torque is $10^{-42}$N-m.

The shape of orbitals i.e. number of orbitals depends on the subshell in which they are found.
The maximum possible number of orbitals i.e. the allowed orientations in space are denoted by magnetic quantum number and is given by.$ml ​= 2l + 1$
$7 = 2l + 1$
$2l = 6$
$l = 3$

According to the uncertainty principle,
$\triangle\text{E}.\triangle\text{t}>=\frac{\text{h}}{2\pi}.$
Thus the time measured will become uncertain if $\triangle\text{E}$ is measured with high certainty.

Infinitely large transition are possible $($in principle$)$ for the hydrogen atom

Number of spectral lines in hydrogen atom $\infty .$

Cathode$-$ray tubes that are used to convert electric signals into visible images may be divided into, for example, the following categories: television picture tubes, or kinescopes; cathode$-$ray oscillograph tubes; numerical indicator tubes; the indicator tubes used in radar sets

Bohr's hypothesis
Electrons revolves round the nucleus in water orbits.
Orbit of the electron around the nucleus can take only some special values of radius.
The energy of the atom as a definite value in these orbits.
In this Orbits, Angular momentum $(e)$ of the electron is integral multiple of the plank's constant h divided by $2n$
$\text{i.e. l}=\text{n}\frac{\text{h}}{2\text{n}}.$

Ionization energy is given as:
$E = 13.6 Z2 eV$
For $Li^+, Z = 3$
$E = 13.6 \times 9$
$E = 122.4 eV$

$1 : 4 : 9$, In Bohr’s atomic model, $r_n n^2$

In a hydrogen atom, electron revolving around a fixed proton nucleus have some centripetal acceleration. Therefore its frame of reference is non$-$inertial. If the frame of reference, where the electron is at rest, the given expression is not true as it forms the non$-$inertial frame of reference.

The orbital quantum number, $l,$ divides the shells up into smaller groups of subshells called orbitals. The orbital quantum number describes shape of subshells. The principal quantum number$(n)$ determines the possible values of l. For $n = 1$ i.e.$ K-$shell there is no subshell so, $l = 0$ i.e. $(n − 1)$. Hence, The possible values of orbital quantum numbers are from $0$ to $(n − 1).$

In transition from $n_1 = 3$ and $n_2 = 4, 5, 6,….$
Infrared radiation of Paschen spectral is emitted.

Phosphorus has atomic no$-17.$ so its $k$ shell has $2$ electron, $L$ shell has $8$ electron and $M$ shell has $5$ electron.