Option 4 : Caesium

__Concept:__

**Photoelectric effect:**When the light of a**sufficiently small wavelength is incident**on the metal surface,**electrons are ejected from the metal instantly**. This phenomenon is called the**photoelectric effect.**

__Explanation__

In the given options

**Sodium, lithium, and magnesium**are electropositive and they also lose electrons easily but their**reactivity is much less than Caesium.**-
**Cesium is the most electropositive**element of all so it has the**minimum ionization energy**and so contains the**maximum capacity**to**lose electrons**.

Option 2 : Increases

__CONCEPT__:

Work function (Φo):

- The minimum energy of incident radiation, required to eject the electrons from the metallic surface is defined as the work function of that surface.

\(\Rightarrow {{\rm{\Phi }}_0} = h{\nu _0} = \frac{{hc}}{{{\lambda _0}}}\)

Where h = Plank's constant, c = speed of light, ν0 = threshold frequency and λ0 = threshold wavelength.

__EXPLANATION__:

- From the above equation, it is clear that the
**work function of the metal is directly proportional to the threshold frequency**. - If the t
**hreshold frequency of a metal surface is doubled**, then the**work function of the metal will get double**. Hence option 2 is correct.

Option 1 : Compton effect and Photoelectric effect

Both the **Compton effect and Photoelectric effect **shows the particle nature of **the electromagnetic radiation.**

__Compton Effect__:

- It is the
**scattering of a photon**by a charged particle, usually an**electron.** - When a photon collides with an electron at rest, the photon gives its energy to the electron.
- Therefore the scattered photon (photon after the collision) will have a higher
**wavelength**(lower energy) compared to the wavelength of the incident photon (photon before the collision). - This
**shift in wavelength**is called the Compton shift. - The effect is named after American physicist Arthur Holly Compton, who first discovered this effect.

__Photoelectric Effect__:

- It is the phenomenon of emission of electrons from the surface of metals when light radiations (Electromagnetic radiations) of suitable frequency fall on them.
- The emitted electrons are called photo-electrons and the current so produced is called photoelectric current.

__Important Points__

The observations and results of the experimental study of photoelectric effect are collectively called Laws of photoelectric emission and are as follows:

1) The number of photoelectrons that are emitted from the metal depends on the intensity of the incident light but is independent of its frequency.

2) The photoelectrons are emitted immediately after light falls on the metal.

3) For the emission of electrons, it is important that the frequency of incident light should be more than the critical value of the metal which is called the threshold frequency of the metal.

__Additional Information__

**Pauli's Principle** states that no two electrons in the same atom can have identical values for all four of their quantum numbers.

** Bragg's Law:** When X-rays are incident on an atom, they make the electronic cloud move, as does any electromagnetic wave.

Option 2 : Maximum kinetic energy of electrons

__CONCEPT:__

- When the photons fall on a metal surface then some electrons get ejected from the metal surface. This phenomenon is called the photoelectric effect.
- The minimum energy needed to remove electrons from the metal surface is called work function (φ) of that metal.
- The maximum energy of ejected electrons from the metal surface after ejection is called maximum kinetic energy (KEmax).
- Einstein’s equation of photoelectric equation:

⇒ E = φ + KEmax

Where E is the incident energy of photons, φ is the work function of metal and KE is the maximum kinetic energy of electrons.

⇒ E = h ν

Where h = Planck constant and ν = the frequency of incident radiation

__EXPLANATION:__

- Einstein’s photoelectric equation is

\(⇒ h\nu = {ϕ} + k\) ----(1)

- According to Einstein’s photoelectric equation:

⇒ E = φ + KEmax

⇒ E = h ν

⇒ hν = φ + KE_{max} ----(2)

On comparing equation 1 and 2, we get to know that,

⇒ k = KEmax

- Therefore
**k represents the maximum kinetic energy of electrons**. Hence option 2 is correct.

KEmax = (h ν - φ)

- From the equation, it is clear that the kinetic energy of the electrons emitted is directly proportional to the frequency of radiation. Therefore option 1 is correct.
- The maximum kinetic energy doesn’t depend upon the intensity of incident radiations and the time for which light falls on the metal.
- When we increase the number of photons or intensity of the incident radiations then the number of electrons ejected will increase but the maximum kinetic energy of electrons will not change.

Option 3 : 2eV

**Concept:**

Photoelectric effect:

When a light of sufficiently small wavelength is incident on the metal surface, electrons are ejected from the metal instantly. This phenomenon is called the photoelectric effect.

- Mathematically it is (h ν) = ϕ + K.E
- h = planks constant = 6.6 × 10-34 = 4.14 × 10-15 eV-s, ν = incident frequency, ϕ = work function.

ν = h/λ

[h = planks constant = 6.6 × 10-34 = 4.14 × 10-15 eV-s, ν = frequency, λ = wavelength].

Stopping potential -

The photocurrent may be stopped by applying a negative potential to anode w.r.t. cathode. The minimum potential required to stop the electron emitted from metal so that its kinetic energy becomes zero.

Stopping potential (eV) = K.E (Max). = (h c)/λ - ϕ

[h = planks constant = 6.6 × 10-34 = 4.14 × 10-15 eV-s, ν = frequency, λ = wavelength, c = speed of light, e = 1.6× 10-19 coulomb].

Work function: It is the minimum amount of energy required so that metal emits an electron.

- It is represented with ϕ
- Its unit is
__Electron-Volt (eV) or joules.__ - It is having different values for different metals.

In photo electric effect work function (ϕ) = (h ν)

h = planks constant = 6.6 × 10-34 = 4.14 × 10-15 eV-s, ν = incident frequency, ϕ = work function

- Also, Work function (ϕ) = (h c /λ ) - K.E. (electrons) c = speed of light, λ = wavelength

**Explanation:**

Given the work function of the metal is 1 eV so, the minimum amount of energy required to emit the electron is 1 eV or more.

**The only option more than or equal to energy of 1 eV is 2 eV.**

Option 1 : 2.4 × 10^{-19} J

__Concept__:

Photoelectric effect: When the light of a sufficiently small wavelength is incident on the metal surface, electrons are ejected from the metal instantly. This phenomenon is called the** photoelectric effect**.

Mathematically it is

\( h\nu = \phi + K.E \)

\(\nu = \frac{h}{\lambda }\)

Stopping potential: The photocurrent may be stopped by applying a negative potential to anode w.r.t. cathode. The minimum potential required to stop the electron emitted from metal so that its kinetic energy becomes zero.

Mathematically,

The stopping potential (eV) is:

\( K.{E_{\max }} = \frac{{hc}}{\lambda } - \phi = h\nu - \phi \)

Work function: It is the minimum amount of energy required so that metal emits an electron.

- It is represented with ϕ
- Its unit is eV or joules.
- It is having different values for different metals.

In photo electric effect

(ϕ) = (h c /λ ) - K.E. (electrons)

Where h = planks constant = 6.6 × 10-34 = 4.14 × 10-15 eV-s, ν = incident frequency, ϕ = work function, c = speed of light, λ = wavelength,

e = 1.6× 1019 coulomb.

__ Calculation__:

Given that: ϕ = 2.4 eV, hν = 3.9 eV

We have,

\(K.{E_{\max }} = \frac{{hc}}{\lambda } - \phi = h\nu - \phi \)

But, E = hν - ϕ = 3.9 - 2.4 = 1.5 eV = 1.5 × 1.6 × 10^{-19} = 2.4 × 10^{-19}

The maximum kinetic energy of the photoelectrons produced is **2.4 × 10-19**

Option 4 : 10.56 × 10^{-30} J

__CONCEPT__:

Work function (Φo):

- The minimum energy of incident radiation, required to eject the electrons from the metallic surface is defined as the work function of that surface.

\(⇒ {{\rm{Φ }}_0} = h{\nu _0} = \frac{{hc}}{{{\lambda _0}}}\)

Where h = Plank's constant, c = speed of light, ν0 = threshold frequency and λ0 = threshold wavelength.

__EXPLANATION__:

Threshold frequency (ν0) = 1.6 × 104 Hz

- The work function of sodium is

⇒ Φ_{o} = hν0

⇒ Φo = 6.6 × 10-34 × 1.6 × 104 = 10.56 × 10^{-30} J

Option 3 : Photoelectric effect

__ CONCEPT__:

**Photoelectric effect**: The phenomenon in which the light energy forces a metal surface to release electrons is called the photoelectric effect.- When the
**light hits, it shows the particle theory of light, and light is defined as a stream of photons or energy packets**. - The other phenomena such as
**interference, diffraction, and polarization**can only be**explained when the light is treated as a wave**wherein the photoelectric effect, line spectra, and the production and scattering of x rays demonstrate the particle nature of light.

__ EXPLANATION__:

- From the above discussion, the
**photoelectric effect shows the particle nature of light**. So option 3 is correct.

X-rays of wavelength of 0.4500 nm are scattered from free electrons in a target. What is the wavelength of photons scattered at 60° relative to the incident rays? (h = 6.63 × 10^{-34} Js, m_{e} = 9.11 × 10^{-31} kg, c = 3 × 10^{8} m/s)

Option 2 : 0.4512 nm

__Concept:__

**Compton Scattering:** It is the scattering of a photon by a charged particle usually an electron. It results in a decrease in energy (or increase in wavelength) of the photon (which is usually an X-ray or gamma-ray photon).

**Compton relation** is given by:

\({\rm{λ' }} - {\rm{λ }} = \frac{{\rm{h}}}{{{m_e}c}}\;\left( {1 - \cos \theta } \right)\)

where, λ = Initial wavelength, λ’ = Wavelength after & Scattering, h = Planck Constant, m_{e} = Electron rest mass, c = speed of light,θ = Scattering angle

\(\frac{h}{{{m_e}c}}\) is known as **Compton wavelength of the electron and is equal to 2.43 × 10 ^{-12}m.**

__Calculation:__

__Given:__

λ = 0.4500 nm, θ = 60°, h = 6.63 × 10^{-34} J.S, m_{e} = 9.11 × 10^{-31} kg, c = 3 × 10^{8} m/s

We know that

\(λ ' = λ + \frac{h}{{{m_e}c}}\;\left( {1 - \cos \theta } \right)\)

\(λ'= 0.45 + \frac{{6.63 \times {{10}^{ - 34}}}}{{9.11 \times {{10}^{ - 31}} \times 3 \times {{10}^8}}}\;\left( {1 - \cos 60^\circ } \right) \times {10^9}\;\)

λ' = 0.45 + 0.0012 = **0.4512 nm**

Option 3 : Diffraction of light

__Explanation:__

**Diffraction of light** cannot convert **unpolarized light into polarized light.**

**Polarization** is defined as a **phenomenon caused due to the wave nature** **of electromagnetic radiation.**

- Sunlight travels through the vacuum to reach the Earth, which is an example of an electromagnetic wave.
- These waves are called electromagnetic waves because they form when an electric field that interacts with a magnetic field.
- There are two types of waves that are
**transverse waves, and longitudinal waves.**

Diffraction of light is the phenomenon of bending of light from the sharp corners of a slit or obstacle and spreading into the region of geometrical shadow.

- Diffraction can occur only when the wavelength of light is comparable to the size of the obstacle or width of the slit.
- Diffraction is of two types:

**Fresnel Diffraction**- It is the type of diffraction which occurs when the light source lies at a finite distance from the slit.**Fraunhofer Diffraction**- It is the type of diffraction which occurs when a plane wavefront is incident on the slit and the wavefront emerging from the slit is also plane.

__Additional Information__

__Scattering of light: __

The incident in which light rays get deviated from its path when it strikes an obstacle like dust, water vapours, or gas molecules, etc.

Rayleigh's law of scattering: According to Rayleigh's law of scattering, the intensity of light of wavelength λ present in the scattered light is inversely proportional to the fourth power of λ, provided the size of the scattering particles is much smaller than λ. Mathematically,

\(I \propto \frac{1}{\lambda }\)

Thus the scattered intensity is maximum for shorter wavelength.

Scattering of InfraRed light is less than visible light.

__Reflection: __

The phenomena in which light ray is sent back into the same medium from which it is coming, on interaction with boundary, is called reflection

Laws of reflection:

The angle of incidence (θ i ) = Angle o f reflection (θ r )

The incident ray, the reflected ray, and normal to the surface of incidence always lies in the same plane.

**Double refraction** is an optical property in which a single ray of **unpolarized light **entering an anisotropic medium is **split into two rays**, each traveling in a different direction.

Option 2 : Cathode ray particles start from anode and move towards cathode.

__CONCEPT:__

**Cathode rays:**Cathode rays are the stream of**high speed negatively charged particles**moving from**cathode to anode**in a discharge tube.

**Properties of Cathode Rays**:

**Cathode rays**travel in a**straight line**.- They are streams off
**fast-moving electrons**. - Cathode rays
**heat up the material**on which they fall. - The cathode rays in the discharge tube are the
**electrons**produced due to the**ionization of gas**and that**emitted by cathode**due to the**collision of positive ions**. - Cathode rays are emitted normally from the
**cathode surface**. Their**direction**is**independent of the position of the anode**. - Cathode rays are
**deflected**by an**electric field**and also by a**magnetic field**. - Cathode rays
**ionise**the gases through which they are**passed**. - Cathode rays can
**penetrate through thin foils of metal**.

__EXPLANATION:__

- From above it is clear that
**Cathode ray**particles**start from cathode and move towards anode**. Thus option 2 is incorrect.

Option 1 : -1.51

__Concept-__

- The atoms have protons and neutrons in the nucleous of the atom and electrons revolve around the nucleous in the orbits.
- For each orbits there is a certain amount of energy for the electrons.

The energy of electrons in any orbit is given by:

\({\bf{Energy}}\;{\bf{in}}\;{\bf{nth}}\;{\bf{orbit}}\left( {{E_{n\;}}} \right) = - 13.6\;\frac{{{Z^2}}}{{{n^2}}}eV\)

Where n is principle quantum number and Z is atomic number

__CALCULATION:__

For hydrogen atom:

Atomic number (Z) = 1

Given that

Energy of electron in first orbit , E0 = -16.6

Principle quantum number of 3rd orbit, n = 3

\({\bf{Energy}}\;{\bf{in}}\;{\bf{nth}}\;{\bf{orbit}}{\rm{\;}}\left( {{E_{n\;}}} \right) = E_0\;\frac{{{Z^2}}}{{{n^2}}}\)

\({\bf{Total}}\;{\bf{energy}}\;{\bf{of}}\;{\bf{an}}\;{\bf{electron}}\;{\bf{in}}\;3{\bf{nd}}\;{\bf{orbit}}\left( {{E_{3\;}}} \right) = - 13.6\;\frac{{{Z^2}}}{{{n^2}}} = - 13.6\;\frac{{{1^2}}}{{{3^2}}} = - 1.51\;eV\)

Which radiation has the highest penetrating power?

Option 3 : Gamma radiation

__CONCEPT__:

Radioactivity:

- The phenomenon of spontaneous emission of radiations by heavy elements is called radioactivity.
- The elements which show this phenomenon is called radioactive elements.

__EXPLANATION__:

Nuclear Radiations:

- According to Rutherford's experiment when a sample of a radioactive substance is put in a lead box and allows the emission of radiation through a small hole only.
- When the radiation enters into the external electric field, they split into three parts (α – rays, β – rays, and γ – gamma rays).

- The penetration power of β – rays is 100 times that of α – rays. They are absorbed by the aluminum foil of 5 mm thickness.
- The penetration power of γ – rays is 1000 times that of α – rays. They can penetrate a 30 cm thick iron block.
- The increasing order of penetrating powers is γ – rays > β – rays > α – gamma. Therefore option 4 is correct.

- Thus,
**Gamma rays**have the**highest penetrating power**. Therefore option 3 is correct.

Option 4 : Photon

A quantum is defined as the** minimum amount of any physical entity** involved in an interaction. For example, **a photon** is a single quantum of light.

__Photon__:

1) A photon is a **tiny light particle** that comprises waves of electromagnetic radiation.

2) James Maxwell observed photons are just electric fields traveling through space.

3) Photons have no charge and no resting mass.

4) It travels at the speed of light.

The energy of Photon is given by:

E = hν

Where,

h is Planck's constant, and

ν is the frequency of the Electromagnetic wave associated with the photon

h = 6.626 070 15 x 10-34 J Hz-1

__Additional Information__

__Positrons__:

- Positron, also called
**positive electron**, positively charged subatomic particle having the same mass and magnitude of charge as the electron and constituting the antiparticle of a negative electron. - The first of the antiparticles to be detected, positrons were discovered by Carl David Anderson.

__Phonons__**:**

- The thermal vibrations resulting from the thermal energy of the substances are called phonons.
- They can't absorb or emit electromagnetic radiation.

__Optical phonons__**:**

- The quantum of the lattice vibration in opposite directions are called optical phonons.
- The optical phonons are observed for the crystals containing multiple atoms as a basis such as NaCl, GaAs, ZnO, etc.

**Phantom** __Loading__:

- It is the phenomenon in which the appliances
**consume electricity even when they turn off.** **The phantom loading is used for examining the current rating ability of the energy meter.**- The actual loading arrangement will waste a lot of power.
- The phantom loading consumes very less power as compared to real loading, and because of this reason, it is used for testing the meter.

Option 3 : finite, single valued and continuous

__Concept:__

In one dimension, wave functions are often **denoted by the symbol ψ(x,t)**.

They are functions of the coordinate x and the time t. But **ψ(x,t) is not a real, but a ****complex function**, the Schroedinger equation does not have real, but complex solutions.

**The wave function of a particle, at a particular time, contains all the information that anybody at that time can have about the particle**.

But the wave function itself has no physical interpretation. It is not measurable. However, the square of the absolute value of the wave function has a physical interpretation.

In one dimension, we interpret **|ψ(x,t)|**^{2} as a probability density, a probability per unit length of finding the particle at a time t at position x.

**The probability of finding the particle at time t in an interval ∆x about the position x is proportional to |ψ(x,t)| ^{2}∆x.**

This interpretation is possible because the square of the magnitude of a complex number is real.

For the probability interpretation to make sense, the wave function must satisfy certain conditions. We should be able to find the particle somewhere, we should only find it at one place at a particular instant, and the total probability of finding it anywhere should be one. This leads to the requirements listed below.

**The wave function must be single valued and continuous. The probability of finding the particle at time t in an interval ∆x must be some number between 0 and 1.**- We must be able to normalize the wave function. We must be able to choose an arbitrary multiplicative constant in such a way, so that if we sum up all possible values ∑|ψ(x
_{i},t)|^{2}∆x_{i}we must obtain 1.

Option 1 : E/c

Relation between Relativistic E and \(\vec{P}\)

1) E = K + m_{o}c^{2}

2) \(E = \dfrac{m_oc^2}{\sqrt{1-\dfrac{r^2}{c^2}}}\)

3) \(P = \dfrac{m_o}{\sqrt{1-\dfrac{r^2}{c^2}}}\)

**Derivation:**

We begin by squaring equation (2) on both sides, i.e.

\(E^2 = \dfrac{m_o^2c^4}{1-r^2/c^2}\)

Next, we insert the term (V^{2} - V^{2}) as shown:

\(E^2 = \dfrac{m_o^2 c^2 (V^2 - V^2 +c^2)}{1-r^2/c^2} \)

\(=\dfrac{m_o^2c^2V^2}{1-r^2/c^2} - \dfrac{m_o^2 c^2 V^2}{1-r^2 /c^2} + \dfrac{m_o^2 c^4}{1-r^2/c^2}\)

\(E^2=\underset{= \ p}{\underbrace{\left(\dfrac{m_oV}{\sqrt{1-r^2/c^2}}\right)^2}}c^2 + \dfrac{m_oc^2 c^4 - m^2 c^2 r^2}{1-r^2/c^2}\)

\(E^2 =p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - r^2}{1-r^2/c^2}\right)\)

\(E^2 = p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - V^2}{\dfrac{c^2 - V^2}{C^2}}\right)\)

\(E^2 = p^2 c^2 + m_o^2 c^4\) where E is the total energy of the particle

\(m_o^2 c^4 = E^2 - p^2 c^2\) ← will be identical in all inertial reference frames.

Option 2 : Linear differential equation

__CONCEPT__:

**Schroedinger wave equation:**

**Schrodinger wave equation**is a mathematical expression describing the**energy**,**momentum**, and**position of the electron**in space and time.- The
**Schrödinger equation**(also known as**Schrödinger’s wave equation**) is a partial differential equation that describes the dynamics of quantum mechanical systems via the wave function.

__EXPLANATION__:

- The one - dimensional time-dependent Schrödinger equation for a free particle is

\(\Rightarrow ih\frac{dΨ(x,t)}{dt}=\frac{h^2}{2m}\frac{d^2Ψ(x,t)}{dx{^2}}\)

Where Ψ(x,t) = Wave function, h = Planck constant and m = mass

- From the above, it is clear that the
**Schrödinger equation**is a**linear partial differential equation**. Therefore option 1 is incorrect and 2 is correct. - In the above equation, the
**left-hand side**depends only on the**'t'**and the**right-hand side**only on**'x'**. Therefore, both sides must be equal to a constant E which is equal to the energy of the particle, we thus obtain the two equations

\(\Rightarrow ih=\frac{df(t)}{dt}=Ef(t)\)

The equation depends on **time**.

\(\Rightarrow [-\frac{h^2}{2m}\Delta^2+V(x)\Psi(r)=E\Psi(r)]\)

- The second equation depends only on the space coordinates. Therefore the
**Schrödinger equation**is a**second-order**differential equation in time. Hence option 3 and 4 is incorrect.

Option 2 : 345 nm

__Concept:__

Stopping potential is defined as the potential necessary to stop any electron from reaching the other side. It is a measure of the maximum kinetic energy of the electrons emitted as a result of the photoelectric effect.

**work function** (ϕ_{o}) is the minimum energy required to remove on electron form the surface of the material.

stopping potential \(\left( {{V_o}} \right) = \frac{{{{\left( {K.E} \right)}_{max}}}}{e}\)

\(\begin{array}{l} {\left( {K.E} \right)_{max}} = hv - {\phi _o} = \frac{{{h_c}}}{λ } - {\phi _o}\\ \Rightarrow {V_o}e = \frac{{hc}}{λ } - {\phi _o} \end{array}\)

__Calculation:__

__Given:__

ϕ_{o} = 2.3 eV, V_{o} = 1.3 V

\(\Rightarrow \frac{{hc}}{λ } = {V_o}e + {\phi _o}\)

\(\Rightarrow \frac{{hc}}{λ } = 1.3eV + 2.3eV = 3.6\;eV\)

\(\Rightarrow λ = \frac{{hc}}{{3.6eV}} = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.6 \times 1.6 \times {{10}^{ - 19}}}}\)

λ = 3.45 × 10^{-7} m

A nuclear reaction gives off a total of 10^{17} J energy. How much mass is spent in the process?

Option 2 : 1.11 kg

__Concept:__

Einstein gave the famous mass-energy equivalence relation

**E = mc ^{2}**

Here the energy equivalent of mass m is related by the above equation and c is the velocity of light in vacuum and is approximately equal to 3×10^{8} ms^{-1}.

__Calculation:__

__Given:__

E = 1017 J

We know that,

E = mc2

10^{17} = m(3 × 10^{8})^{2}

**m = 1.1 kg**

Option 2 : T + V

__CONCEPT__:

**Hamiltonian operator:**

- In classical physics, the total energy of the particle i.e., the
**sum of the kinetic energy and potential energy**is called it's Hamiltonian - It is denoted by the letter
**"H"**. - Mathematically, the Hamiltonian operator is written as

\(\Rightarrow H = -\frac{h^2\Delta^2}{2m}+V(r,t)\)

Where V (r, t) = potential energy of the particle, m = mass of the particle

**EXPLANATION:**

- From the above, it is clear that if
**T and V represent the kinetic and potential energies**of a body respectively the**Hamiltonian**is given by T + V.

**NOTE:**

- In classical physics, the
**difference**between the kinetic energy and potential energy of a body is called it's**Lagrangian**. - It is denoted by the letter "L".