Maharashtra State Board Class 12th Physics Question Paper 2022 with Solutions Answers Pdf Download.

## Class 12 Physics Question Paper 2022 Maharashtra State Board with Solutions

**Section – A**

Question 1.

Select and write the correct answers for the following multiple choice type of questions: (10)

(i) The first law of thermodynamics is concerned with the conservation of ……

(a) momentum

(b) energy

(c) temperature

(d) mass

Answer:

(b) energy

(ii) The average value of alternating current over a full cycle is always ……..

[I_{0} = Peak value of current]

(a) zero

(b) \(\frac{I_0}{2}\)

(c) \(\frac{I_0}{\sqrt{2}}\)

(d) 2I_{0}

Answer:

(a) Zero

(iii) The angle at which maximum torque is exerted by the external uniform electric field on the electric dipole is ……..

(a) 0°

(b) 30°

(c) 45°

(d) 90°

Answer:

(d) 90°

(iv) The property of light which does not change, when it travels from one medium to another is ……..

(a) velocity

(b) wavelength

(c) frequency

(d) amplitude

Answer:

(c) frequency

(v) The root mean square speed of the molecules of a gas is proportional, to ……….

[T = Absolute temperature of gas]

(a) \(\sqrt{T}\)

(b) \(\frac{1}{\sqrt{T}}\)

(c) T

(d) \(\frac{1}{\mathrm{~T}}\)

Answer:

(a) \(\sqrt{T}\)

(vi) The unit Wbm^{-2} is equl to

(a) henry

(b) watt

(c) dyne

(d) tesla

Answer:

(d) tesla

(vii) When the bob performs a vertical circular motion and the string rotates in a vertical plane, the difference in the tension in the string at horizontal position and uppermost position is ………

(a) mg

(b) 2 mg

(c) 3 mg

(d) 6 mg

Answer:

(c) 3 mg

(viii) A liquid rises in glass capillary tube upto a height of 2.5 cm at room temperature. If another glass capillary tube having radius half that of the earlier tube is immersed in the same liquid, the rise of liquid in it will be ……..

(a) 1.25 cm

(b) 2.5 cm

(c) 5 cm

(d) 10 cm

Answer:

(c) 5 cm

(ix) In young’s double slit experiment the two coherent sources have different amplitudes. If the ratio of maximum intensity to minimum intensity is 16 : 1, then the ratio of amplitudes of the two source will be

(a) 4 : 1

(b) 5 : 3

(c) 1 : 4

(d) 1 : 16

Answer:

(b) 5 : 3

(x) The equation of a simple harmonic progressive wave travelling on a string is u = 8 sin (0.02x – 4t) cm. The speed of the wave is ……..

(a) 10 cm/s

(b) 20 cm/s

(c) 100 cm/s

(d) 200 cm/s

Answer:

(d) 200 cm/s

y = 8 sin(0.02x – 4t)

y = A sin(\(\frac{2 \pi x}{\lambda}\) – 2πnt)

know that,

∴ \(\frac{2 \pi}{\lambda}\) = 0.02

∴ λ = \(\frac{2 \pi}{0.02}\)

= 100π

And 2πn = 4

n = \(\frac{2}{\pi}\)

∴ v = nλ

= \(\frac{2}{\pi}\) × 100π

= 200 cm/s.

Question 2.

Answer the following questions: (8)

(i) Define potential gradient of the potentiometer wire.

Answer:

The potential gradient is defined as the fall of potential per unit length of the potentiometer wire.

(ii) State the formula for critical velocity in terms of Reynold’s number for a flow of a fluid.

Answer:

Critical velocity of fluid

V_{c} = \(\frac{R_n \eta}{\rho d}\)

Where R_{n} = Reynold number

(iii) Is it always necessary to use red light to get photoelectric effect?

Answer:

No, photons in red light do not have the necessary energy required to rip an electron out of its orbital due to large wavelength and low energy of red light.

(iv) Write the Boolean expression for Exclusive-OR (X – OR) gate.

Answer:

Boolean expression of Ex – OR gate is

Y = A\(\oplus\)B

OR

Y = \(\overline{\mathrm{A}}\).B + A.\(\overline{\mathrm{B}}\)

(v) Write the differential equation for angular S.H.M

Answer:

\(1 \frac{d^2 \theta}{d t^2}\) + cθ = 0

(vi) What is the mathematical formula for third postulate of Bohr’s atomic model?

Answer:

The change in an electron’s energy as it makes a quantum jump from one orbit to another is always accompanied by the emission or absorption of a photon.

(vii) Two inductor coils with inductance 10mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?

Answer:

E = E_{n} – E_{m}

L_{1} = 10 mH

L_{2} = 20 mH

L_{eff} = ?

Resultant inductance of series combination is

L_{eff} = L_{1} + L_{2}

= 10 + 20

= 30 mH

∴ The resultant inductance of combination of two coils is 30 mH.

(viii) Calculate the moment of inertia of a uniform disc of mass 10 kg and radius 60 cm about an axis perpendicular to its length and passing through its centre.

Answer:

Given, m = 10 kg

r = 60 cm

= 60 × 10^{-2} m

I = ?

Moment of inertia of a uniform disc is 1.8 kg m^{2}.

**Section – B**

Attempt onu EIGHT questions of the following: (16)

Question 3.

Define moment of inertia of a rotating rigid body. State its SI unit and dimensions.

Answer:

Moment of inertia of a rotating rigid body is the sum of the product of each point mass and square of its distance from the axis of rotations.

S.I. unit = kgm^{2}

Dimensions = [M^{1}L^{2}T^{0}]

Question 4.

What are polar dielectrics and non polar dielectries?

Answer:

(i) Polar Dielectrics: A dielectric molecule in which the centre of mass of positive charges (protons) does not coincide with the centre of mass of negative charges (electrons), because the asymmetric shape of the molecules is called a polar dielectric.

Ex: HCl, water, alcohol, NH_{3} etc.

(ii) Non polar dielectrics: A dielectric in which the centre of mass of positive charges (protons) coincide with the centre of mass of negative charges (electrons) is called a nonpolar dielectric.

Ex: H_{2}, N_{2}, O_{2}, CO_{2}, methane, benzene etc.

Question 5.

What is a thermodynamic process? Give any two types of it.

Answer:

A process by which two or more state variables of a system can be changed is called a thermodynamic process.

Two types of thermodynamic process: Isochoric, Isobaric, Adiabatic, Isothermal, Revisable, Irreversible, cyclic process.

Question 6.

Derive an expression for the radius of the nth Bohr orbit of the electron in hydrogen atom.

Answer:

Let the mass of electron be ‘m_{e}‘, its velocity in the n^{th} orbit be ‘V_{n}‘ and radius of its orbit be ‘r_{n}‘.

∴ The angular momentum

= m_{e}v_{n}r_{n}

According to second postulate,

m_{e}v_{n}r_{n} = \(_{e}\) …. (i)

According to Bohr’s first postulate,

Centripetal force = Electrostatic force

This is the required expression for the radius of n^{th} Bohr orbit of the electron.

Question 7.

What are harmonics and overtones (Two points)?

Answer:

Harmonics:

- The Lowest allowed natural frequency of vibration of a string and all its integral multiples are called harmonics.
- The fundamental frequency, n is called the first harmonic. The second harmonic is 2n, third harmonic is 3n, …… and so on.

Overtones:

- The higher allowed frequencies of vibration above the fundamental are called overtones.
- Above the fundamental, the first allowed frequency is called the first overtone which may be either the second or third harmonic.

Question 8.

Distinguish between potentiometer and voltmeter.

Answer:

Potentiometer | Voltmeter |

1. AA potentiometer is used to determine the emf of a cell, potential difference and internal resistance. | 1. A voltmeter can be used to measure the potential difference and terminal voltage of a cell. It cannot be used to measure emf of a cell. |

2. Its accuracy and sensitivity are very high. | 2. It’s accuracy and sensitivity are less as compared to a potentiometer. |

3. It is not portable device. | 3. It is a portable device. |

4. It does not give a direct reading. | 4. If gives a direct reading. |

5. It does not draw any current. | 5. It It draws some current. |

Question 9.

What are mechanical equilibrium and thermal equilibrium?

Answer:

Mechanical equilibrium: When there are no unbalanced forces within the system and between the system and its surrounding, the system is in mechanical equilibrium.

Also, the pressure in the system should be same, throughout the system and should not change with time.

Thermal equilibrium: For a system to be in thermal equilibrium the temperature of the system should be uniform throughout and should not change with time.

Question 10.

An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10^{-11} m with a speed of 3 × 10^{6} m/s. Find the angular momentum of electron.

Answer:

Given,

r = 5.3 × 10^{-11}m

v = 3 × 10^{6} m/s

m = 9.1 × 10^{-31} kg

L = ?

Angular momentum of electron is L = m v r

= 9.1 × 10^{-31} × 3 × 10^{6} × 5.3 × 10^{-11}

= 1.447 × 10^{-34} kg m^{2}/s

∴ Angular momentum of electron is 1.447 × 10^{-34} kg m^{2}/s.

Question 11.

Plane wavefront of light of wavelength 6000 A is incident on two slits on a screen perpendicular to the direction of light rays. If the total separation of 10 brights fringes on a screen 2m away is 2 cm, find the distance between the slits.

Answer:

Given,

Distance between 10 fringes

= 2 cm = 0.02 m

λ = 6000 Å

= 6000 × 10^{-10} m

∴ Fringe width, w = \(\frac{0.02}{10}\)

= 0.002 m

D = 2 m

Fringe width w = \(\frac{\lambda D}{d}\)

∴ d = \(\frac{\lambda D}{w}\)

= \(\frac{6000 \times 10^{-10} \times 2}{0.002}\)

= 6 × 10^{-4} m

The distance between the slits is 6 × 10^{-4} m.

Question 12.

Eight droplets of water each of radius 0.2 mm coalesce into a single drop. Find the decrease in the surface area.

Answer:

r = radius of droplet

= 0.2 mm = 2 × 10^{-4}m

R = radius of single drop

Volume of 8 droplets = Volume of a single drop

8 × \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \pi r^3\)

∴ R^{3} = 8r^{3}

∴ R = 2r

Decrease in the surface area

∆A = A_{1} – A_{2}

= 8 × 4πr^{2} – 4πR^{2}

= 32πr^{2} – 4π(2r)^{2}

= 32πr^{2} – 16πr^{2}

= 16πr^{2}

= 16 × 3.142 × (2 × 10^{-4})^{2}

∆A = 2.011 × 10^{-6} m^{2}

∴ Decrease in the surface area is 2.011 × 10^{-6} m^{2}.

Question 13.

A 0.1 H inductor, a 25 × 10^{-6} F capacitor and a 150 resistor are connected in series to a 120 V, 50 Hz AC source. Calculate the resonant frequency.

Answer:

Given,

L = 0.1 H

C = 25 × 10^{-6} F

R = 15 Ω

F_{r} = ?

Resonant frequency

F_{r} = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

∴ F_{r} = \(\frac{1}{2 \times 3.142 \times \sqrt{0.1 \times 25 \times 10^{-6}}}\)

= \(\frac{1}{6.284 \sqrt{2.5 \times 10^{-6}}}\)

= \(\frac{1}{6.284 \times 1.581 \times 10^{-3}}\)

F_{r} = 100.8 Hz

Question 14.

The difference between the two molar specific heats of gas is 9000 J/kg K. If the ratio of the two specific heats is 1.5, calculate the two molar specific heats.

Answer:

Given,

C_{p} – C_{v} = 9000J/kg K

\(\frac{C_p}{C_v}\) = 1.5

C_{v} = ?

C_{p} = ?

\(\frac{C_p}{C_v}\) = 1.5

∴ C_{p} = 1.5 C_{v}

∴ C_{p} – C_{v} = 9000

1.5 C_{v} – C_{v} = 9000

0.5 Ç_{v} = 9000

∴ C_{v} = \(\frac{9000}{0.5}\)

= 18000 J/kg k

∴ C_{p} = 9000 + C_{v}

= 9000 + 18000

= 27000 J/kg k

∴ Two molar specific heats are 18000 J/kg k and 27000 J/kg k.

**Section – C**

Attempt any EIGHT questions of the following: (24)

Question 15.

With the help of a neat diagram, explain the reflection of light on a plane reflecting surface.

Answer:

Consider a plane wavefrorit AB of monochromatic light incident obliquely at an angle on a plane mirror MN as shown in the figure.

MN : plane mirror,

RA and QC : Incident rays,

AP and CS : Normals to MN,

AB : Incident wavefront

CE : Reflected wavefront

i : Angle of incidence

r : Angle of reflection.

The wave front AB touches the reflecting surface MN at A at time t = 0.

Let v be the speed of light in the medium. If T is the time taken by incident wavefront to travel from B to C,

Then BC = vT.

During this time, secondary wave originating from A covers the same distance, so that the secondary spherical wavelet has a radius vT at time T. Draw a tangent CE to the wavelet.

From figure,

∠RAP = ∠i

= angle of incidence

∠PAE = ∠r

= angle of reflection

In ∆ABCand ∆AEC

AC is common side

AE = BE

∠ABC = ∠AEC

= 90°

∴ ∆ABC and ∆AEC are congruent.

∴ ∠ACE = ∠BAC

= ∠i …….. (i)

Also as AE is perpendicular to CE and AP is perpendicular to AC

∠ACE = ∠PAE

= ∠r ……(ii)

∴ From equations (i) and (ii).

Thus, the angle of incidence is equal to angle of reflection. This is the first law of reflection.

Also it can be seen from figure that incident ray, reflected ray lie on the opposite sides of the normal and all of them tie in the some plane. This is second law of reflection.

Thus, the laws of reflection of light are deduced.

Question 16.

What is magnetization, magnetic intensity and magnetic susceptibility?

Answer:

Magnetization: The ratio of magnetic moment to the volume of the material is called magnetization.

M = \(\frac{M_{\text {net }}}{\text { Volume }}\)

Magnetic Intensity: The degree to which a magnetic field can magnetise a substance or the capability of external magnetic field to magnetise the substance is called magnetic intensity.

In vacuum the ratio of magnetic induction (B_{0}) and magnetic permeability (µ_{0}) is called magnetising field(H).

∴ H = \(\frac{B_0}{\mu_0}\)

Magnetic susceptibility: It is a measure of how much a material will become magnetised in an external magnetic field.

It is the ratio of magnetization (M) to the applied magnetising field (H).

∴ χ = \(\frac{M}{H}\)

Question 17.

Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

Answer:

Consider two sound waves of equal amplitude ‘a’ and slightly different frequencies n_{1} and n_{2} (n_{1} > n_{2}) propagating through the same medium in the same direction.

These waves can be represented by the equations.

y_{1} = a sin 2πn, t

y_{2} = a sin 2πn_{2} t

By the principle of superposition of waves,

Y = y_{1} + y_{2}

= a sin 2πn_{1}t + a sin 2π n_{2}t

= 2asin\(\left[2 \pi\left(\frac{n_1-n_2}{2}\right) t\right]\)cos\(\left[2 \pi\left(\frac{n_1-n_2}{2}\right) t\right]\)

By using formula,

This is the equation of a progressive wave having frequency N and amplitude A.

For waxing, amplitude is maximum

Thus, the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

Question 18.

Define:

(a) Inductive reactance

(b) Capacitive reactance

(c) Impedance

Answer:

(a) Inductive reactance: Effective resistance offered by the inductance is called the inductive reactance (X_{L}).

X_{L} = ω_{L} = 2πfl

(b) Capacitive reactance: Effective resistance offered by the capacitor called capacitive reactance (X_{C}).

X_{C} = \(\frac{1}{\omega_C}\)

= \(\frac{1}{2 \pi f C}\)

(c) Impedance: The effective opposition offered by the inductor, capacitor and resistor connected in series to flow of AC current, is called impedance.

Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\)

Question 19.

Derive an expression for the kinetic energy of a body rotating with a uniform angular speed.

Answer:

Consider a rigid body rotating with a constant angular velocity \(\begin{aligned}

&\rightarrow\\

&\omega

\end{aligned}\) about an axis through the point O and perpendicular to the plane of the figure. Suppose that the body consists of N particles of masses m_{1}, m_{2}…….. m_{N} situated at distances r_{1}, r_{2},……….r_{N} respectively from the axis of rotation as shown in figure.

As the body rotates alt, the particles perform uniform circular motion with the same angular velocity ω. However, various particles of the body have different linear speeds depending upon their distances from the

axis of rotation.

The linear speed of the particle with moss

m_{1} is v_{1} = r_{1} ω

Its kinetic energy is

Question 20.

Derive an expression for emf (e) generated in a conductor of length (I) moving in uniform magnetic field (B) with uniform velocity (v) along x-axis.

Answer:

A rectangle frame of wires ABCD of area (lx) is situated in a constant magnetic field \(\vec{B}\). As the wire BC of length l is moved out with velocity \(\vec{v}\) to increase x, the area òf the loop ABCD increases. Thus the flux of \(\vec{B}\) through the loop increasing with time.

As the flux φ through the frame ABCD is Blx, magnitude of the induced emf can be written as:

|e| = \(\frac{d \phi}{d t}\) = \(\frac{d}{d t}\)(Blx)

= Bl\(\left(\frac{d x}{d t}\right)\) = Blv

Question 21.

Derive an expression for terminal velocity of a spherical object falling under gravity through a viscous medium.

Answer:

Consider a spherical object falling through a viscous fluid. Forces experienced by it during its downward motion are:

- Viscous force (F
_{v}), directed upwards. - Gravitational force (Fg). directed downwards.
- Buoyant force on upthrust (Fu), directed upwards.

Net downward force given by

f = F_{g} – (F_{v} + F_{u}), is responsible for initial increase is velocity.

Let the radius of the sphere be r, its mass m and density ρ. Let the density of the medium be a and its coefficient of viscosity be \(\eta\). When the sphere attains the terminal velocity, the total downward force acting on the sphere is balanced by the total upward force acting on the

sphere.

Total downward force = Total upward force

Weight of sphere (mg) = Viscous force + buoyant force

This is the expression for the terminal velocity of the sphere.

Question 22.

Determine the shortest wavelengths of Balmer and Paschen series. Given the limit for Lyman series is 912 Å.

Answer:

The wavelength of lines are given by,

Question 23.

Calculate the value of magnetic field at a distance of 3 cm from a very long, straight wire carrying a current of 6 A.

Answer:

Given

R = 3 cm = 3 × 10^{-2}m

I = 6A

B = ?

Magnetic field B = \(\frac{\mu_0 I}{2 \pi R}\)

= \(\frac{4 \pi \times 10^{-7} \times 6}{2 \pi \times 3 \times 10^{-2}}\)

B = 4 × 10^{-5}T

The value of magnetic field is 4 × 10^{-5}T.

Question 24.

A parallel plate capacitor filled with air has an area of 6 cm^{2} and plate separation of 3 mm. Calculate its capacitance.

Answer:

Given.

A = 6 cm^{2} = 6 × 10^{-4} m^{2}

d = 3 mm = 3 × 10^{-3} m

k = 1

= 8.85 × 10^{-12} c^{2}/Nm^{2}

C = ?

Capacitance of parallel plate capacitor

C = \(\frac{A \varepsilon_0 k}{d}\)

= \(\frac{6 \times 10^{-4} \times 8.85 \times 10^{-12} \times 1}{3 \times 10^{-3}}\)

= 17.7 × 10^{-13}

= 1.77 × 10^{-12} F

∴ Capacitance of a parallel plate capacitor is 1.77 × 10^{-12} F.

Question 25.

An emf of 91 mV is induced in the windings of a coil, when the current in a nearby coil is increasing at the rate of 1.3 A/s, what is the mutual inductance (M) of the two coils in mH?

Answer:

Question 26.

Two cells of emf 4 V and 2 V having respective internal resistance of 1Ω and 2Ω are connected in parallel, so as to send current in the same direction through an external resistance of 5Ω. Find the current through the external resistance.

Answer:

Applying Kirchhoff’s voltage low to loop ABCDEFA.

We get,

-5(I_{1} + I_{2}) – I_{1} + 4 = 0

∴ -5I_{1} – 5I_{2} – I_{1} = -4

6I_{1} + 5I_{2} = 4 …… (i)

Applying Kirchhoffs voltage Law to loop BCDEB,

We get

-5(I_{1} + I_{2}) – 2I_{2} + 2 = 0

-5I_{1} – 5I_{2} – 2I_{2} = -2

-5I_{1} + 7I_{2} = 2 ….. (ii)

Multiplying eqn (i) by S and equation (ii) by 6 we get

30I_{1} + 25I_{2} = 20 ……. (iii)

And 30I_{1} + 42I_{2} = 12 ….. (iv)

Subtracting eqn (iii) from eqn (iv) we get,

17I_{2} = -8

I_{2} = \(\frac{-8}{17}\)A

Substituting this value of I_{2} in equation (i), we get

**Section – D**

Attempt any THREE questions of the following: (12)

Question 27.

Derive an expression for a pressure exerted by a gas on the basis of kinetic theory of gases.

Answer:

Let there be n moles of an ideal gas enclosed in a cubical box of volume V = L^{3} and constant temperature T.

Molecule is moving with the velocity V collide with the shaded wall of the cube. The wall is Parallel to yz plane.

The change in momentum of X-component is

∆P_{x} = final momentum – initial momentum

= (-mv_{x}) – (mv_{x})

= 2 mv_{x}

The momentum transferred to the wall is + 2mv_{x} in magnitude. The rebounded molecule then goes to the opposite walt and collides with it.

As L is the length of the cubical box, the time for the molecule to travel back and forth to the shaded walL is

∆t = \(\frac{2 L}{V_x} .\)

Average force exerted on the shaded wall by molecule 1 is given as:

Average force = Average rote of change of momentum

= \(\frac{2 m v_{x_1}}{\Delta L N_{x_1}}\) = \(\frac{m v_{x_1}^2}{\mathrm{~L}}\)

Total average force on the wall from all molecule is

Question 28.

What is a rectifier? With the help of a neat circuit diagram, explain the working of a halfwave rectifier.

Answer:

Rectifier: A rectifier is a device that converts an oscillating two – directional alternating current (AC) into a single – directional direct current (DC).”

The conversion of AC voltage into DC voltage is called rectification.

There are two types of rectifier circuit

- Halfwave rectifier
- Full wave rectifier

Halfwave rectifier:

Construction: The AC input voltage is connected to step up or step down transformer. The secondary coil AB of a transformer is connected in series with a diode D and the load resistance R_{L}.

Working: When the positive half cycle begins, the voltage at the point A is higher with respect to that at point B, therefore, the diode (D) is forward biased. It conducts (Works as a closed switch) and current flows through the circuit.

When the negative half cycle begins, the potential at the point A is lower with respect to that at the point B and the diode is reverse biased, therefore, it does not conduct (work as an open switch).

In this way, current always flows through the load R_{L} in the same direction for alternate positive half cycles.

The wave form for input and output voltages are shown below.

Question 29.

Draw a neat, labelled diagram of a suspended coil type moving coil galvanometer.

The initial pressure and volume of a gas enclosed in a cylinder are 2 × 10^{5} N/m^{2} and 6 × 10^{-3} m^{3} respectively. If the work done in compressing the gas

at constant pressure is 150 J, find the final volume of the gas.

Answer:

Given data,

ρ = 2 × 10^{5} N/m^{2}

V_{i} = 6 × 10^{-3}

W = -15oJ

V_{f} = ?

W = P(V_{f} – V_{i})

∴ V_{f} = V_{i} + \(\frac{W}{p}\)

Question 30.

Define second’s pendulum. Derive a formula for the length of second’s pendulum.

A particle performing linear S.H.M. has maximum velocity 25 cm/s and maximum acceleration 100 cm/ s^{2}. Find period of oscillations.

Answer:

Seconds Pendulum: A simple pendulum whose period is two seconds is called seconds pendulum.

Period T = \(2 p \sqrt{\frac{L}{g}}\)

∴ For seconds pendulum

2 = \(2 p \sqrt{\frac{L_s}{g}}\)

Where L_{s} is the length of seconds pendulum having period

T = 2s

∴ L_{s} = \(\frac{g}{p^2}\)

This is a formula for the length of second’s pendulum. Given data,

Question 31.

Explain de-Broglie wavelength. Obtain an expression for de-Broglie wavelength of wave associated with material particles. The photoelectric work function for a metal is 4.2 eV. Find the threshold wavelength.

Answer:

The wavelength that is associated with an object in relation to its momentum and mass is known as the de-Broglie wavelength.

de-Broglie equated the energy equation of Plank (wave nature) and Einstein (particle nature) such that

E = hv (Plank energy relation)

E = mc^{2} (Einsteins mass-energy relation)

Where,

E = energy associated with the particle

h = planks constant

v = frequency associated with the particle

m = mass of the particle

c = speed of light

After equating equations (1) and (2) we get

hv = mc^{2}

h\(\frac{c}{\lambda}\) = mc^{2} (∵ v = \(\frac{c}{\lambda}\))

λ = \(\frac{h c}{m c^2}\) = \(\frac{h}{m c}\)

If the particle is moving with velocity ‘V’ then equation (3) becomes,

λ = \(\frac{h}{m v}\)

The wavelength of the particle if energy is in electron volt is,

Where E should be in eV.

After substituting the value of E i.e., 4.2 eV in equation

Hence, the threshold wevelegth of the particle is 2952.38Å.