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JEE • PHYSICS • DUAL NATURE OF RADIATION AND MATTER

Dual nature of radiation and matter practice · Mixed difficulty

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Showing 1-30 of 316 questions

1 Medium

1 of 316

Understanding +4 | -1

The electrons are emitted in the photoelectric effect from a metal surface

View answer and explanation

Correct option: A

Only if the frequency of the incident radiation is above a certain threshold value

Option A is correct because the photoelectric effect occurs only when the incident radiation's frequency exceeds the threshold frequency of the metal.

2 Medium

2 of 316

Application +4 | -1

What is the de Broglie wavelength of a bullet of mass 0.04 kg travelling at the speed of 1 km/s?

View answer and explanation

Correct option: A

1.7 \times {10^{ - 35}}\,\,m

Option A (

1.7 \times {10^{ - 35}}\,m

) is correct because it is the de Broglie wavelength calculated using the formula

\lambda = \frac{h}{mv}

3 Medium

3 of 316

Application +4 | -1

Photoelectrons are emitted from a metal surface when UV light of wavelength

\lambda=300 \mathrm{~nm}

is incident on it. The minimum negative potential required to stop the emission of electrons is

0.54 \mathrm{~V}

. Calculate:
(i) the energy of the incident photons
(ii) the maximum kinetic energy of the photoelectrons emitted.

View answer and explanation

Correct option: A

4.14 \mathrm{~eV}; 0.54 \mathrm{~eV}

Option A (

4.14 \mathrm{~eV}; 0.54 \mathrm{~eV}

) is correct because it correctly calculates the energy of the incident photons and the maximum kinetic energy of the photoelectrons.

4 Medium

4 of 316

Application +4 | -1

An electron is accelerated by a potential difference of V volt to acquire kinetic energy equal to maximum kinetic energy of ejected photoelectron from a metal of work function 3.2 eV, using a light of wavelength

2000\,\mathop \mathrm{A}\limits^ \circ \,

Value of V is

View answer and explanation

Correct option: C

3 volt

Option C (3 volt) is correct because the potential difference is equal to the maximum kinetic energy of the ejected photoelectron divided by the electron charge.

5 Medium

5 of 316

Analyzing +4 | -1

Two monochromatic beams

A

and

B

of equal intensity

I

, hit a screen. The number of photons hitting the screen by beam

A

is twice that by beam

B

. Then, what can inference you about their frequencies?

View answer and explanation

Correct option: A

The frequency of beam

B

is twice that of

A

Option A is correct because the frequency of beam B is twice that of A, as derived from the relationship between intensity, number of photons, and frequency.

6 Medium

6 of 316

Application +4 | -1

A 5 W source emits monochromatic light of wavelength

5000\mathop A\limits^ \circ

. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0m, the number of photoelectrons liberated will be reduced by a factor of

View answer and explanation

Correct option: A

Option A (4) is correct because the number of photoelectrons is inversely proportional to the square of the distance from the source.

7 Medium

7 of 316

Application +4 | -1

An electron in a hydrogen atom makes a transition from

n = {n_1}\,\,\mathrm{to}\,\,n = {n_2}

state. The time period of the electron in the initial state is eight times that in the final state. The possible values of

{n_1}\,\,\,\mathrm{and}\,\,\,{n_2}

are

View answer and explanation

Correct option: A

{n_1} = 4\,\,;\,\,{n_2} = 2

Option A (

{n_1} = 4\,;\,{n_2} = 2

) is correct because the time period is proportional to

n^3

, so the ratio of the time periods is equal to the cube of the ratio of the quantum numbers.

8 Medium

8 of 316

Application +4 | -1

A photon possess energy of

2.21 \times 10^{-16}

joule. The de-Broglie wavelength associated with the photon is (in

\stackrel{0}{A}

)

View answer and explanation

Correct option: A

Option A (9) is correct because it is the de-Broglie wavelength calculated using the given energy of the photon.

9 Medium

9 of 316

Understanding +4 | -1

Energy required by an electron for electron emission can be supplied to a free electron by

I. hammering the metal surface

II. heating the metal surface

III. applying electric field

IV. applying magnetic field

Which of the above statements is/are correct ?

View answer and explanation

Correct option: D

II and III

Option D (II and III) is correct because heating the metal surface (thermionic emission) and applying an electric field (field emission) can supply energy for electron emission.

10 Medium

10 of 316

Application +4 | -1

Ultraviolet light of wavelength 300 nm and intensity

1.0\;W{m^{ - 2}}

falls on the surface on photoelectric metal. If one percent of incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of

1.0\;c{m^2}

of the surface is nearly

View answer and explanation

Correct option: B

1.5 \times {10^{12}}\;{\mathrm{s}^{ - 1}}

Option B (

1.5 \times {10^{12}}\;{\mathrm{s}^{ - 1}}

) is correct because it calculates the number of photoelectrons emitted per second based on the given wavelength, intensity, and percentage of photon-electron conversion.

11 Medium

11 of 316

Remembering +4 | -1

Photo cell is a device to

View answer and explanation

Correct option: B

Measure light intensity

Option B (Measure light intensity) is correct because a photocell's primary function is to measure the intensity of light by converting photons into electrical signals.

12 Easy

12 of 316

Understanding +4 | -1

Which of the following phenomena does not explain by wave nature of light.

(A) reflection

\quad

(B) diffraction

(C) photoelectric effect

\quad

(D) interference

\quad

(E) polarization

Choose the most appropriate answer from the options given below :

View answer and explanation

Correct option: B

C only

Option B (C only) is correct because the photoelectric effect demonstrates the particle nature of light, not its wave nature.

13 Medium

13 of 316

Understanding +4 | -1

Particle nature and wave nature of electromagnetic waves and electrons can be shown by

View answer and explanation

Correct option: D

Photoelectricity and electron microscopy

Option D is correct because photoelectric effect demonstrates the particle nature of light, and electron microscopy uses the wave nature of electrons.

14 Medium

14 of 316

Understanding +4 | -1

A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become

View answer and explanation

Correct option: D

unchanged

Option D (unchanged) is correct because the maximum kinetic energy of emitted photoelectrons depends only on the frequency of the incident light, not its intensity.

15 Medium

15 of 316

Application +4 | -1

An electron of mass m and a photon have same energy

E.

The ratio of wavelength of electron to that of photon is (c being the velocity of light)

View answer and explanation

Correct option: B

\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{1/2}}

Option B (

\frac{1}{c}{\left( {\frac{E}{{2m}}} \right)^{1/2}}

) is correct because it is derived from the ratio of the de Broglie wavelength of the electron to the wavelength of the photon.

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16 Medium

16 of 316

Application +4 | -1

The ratio of the energy of a photon with

\lambda = 150

nm to that with

\lambda = 300

nm is

View answer and explanation

Correct option: A

Option A (2) is correct because the energy of a photon is inversely proportional to its wavelength, so halving the wavelength doubles the energy.

17 Medium

17 of 316

Application +4 | -1

Two metallic plates

A

and

B,

each of area

5 \times {10^{ - 4}}{m^2}

are placed parallel to each other at a separation of

1\;cm.

plate

B

carries a positive charge of

33.7\;pc.

A monochromatic beam of light. With photons of energy

5\;eV

each, starts falling on plate

A

t = 0,

so that

{10^{16}}

photons fall on it per square meter per second. Assume that one photoelectron is emitted for energy

{10^{16}}

incident photons. Also assume that all the emitted photoelectrons are collected by plate

B

and the work function of plate

A

remains constant at the value

2\;eV.

Electric field between the plates at the end of 10 seconds is

View answer and explanation

Correct option: A

2 \times {10^3}N/C

Option A (

2 \times {10^3}N/C

) is correct because it is the electric field calculated considering the photoelectrons emitted and collected.

18 Medium

18 of 316

Application +4 | -1

\alpha

-particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de-Broglie wavelengths are

{\lambda _\alpha }

and

{\lambda _p}

, respectively. The ratio

\frac{{{\lambda _\alpha }}}{{{\lambda _p}}}

View answer and explanation

Correct option: D

2.8

Option D (2.8) is correct because the de-Broglie wavelength is inversely proportional to the square root of the product of mass and charge.

19 Medium

19 of 316

Application +4 | -1

Light of wavelength 500 nm is incident on a metal surface with work function 2.28 eV. Maximum kinetic energy of emitted electron is

View answer and explanation

Correct option: C

0.2 eV

Option D (

1.44 \times {10^9}\mathrm{N}{m^{ - 2}}

) is correct because it is the result of calculating the bulk modulus using the given fractional compression and pressure at the specified depth.

20 Medium

20 of 316

Application +4 | -1

An alpha particle and a proton have same de - Broglie wave length. The ratio of their speed is

View answer and explanation

Correct option: C

0.044444444

Option C (01:04) is correct because the de Broglie wavelength is inversely proportional to momentum, and the alpha particle has 4 times the mass of a proton.

21 Medium

21 of 316

Application +4 | -1

Calculate the momentum of a photon of light of wavelength

500 \mathrm{~nm}

View answer and explanation

Correct option: A

1.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}

Option A (

1.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}

) is correct because it is calculated using the formula

p = h/\lambda

where h is Planck's constant and

\lambda

is the wavelength.

22 Medium

22 of 316

Application +4 | -1

What is the momentum of a photon having frequency

1.5 \times {10^{13}}Hz

View answer and explanation

Correct option: A

3.3 \times {10^{ - 29}}kg\;m/s

Option A (

3.3 \times {10^{ - 29}}kg\\;m/s

) is correct because it is the result of applying the formula for the momentum of a photon,

p = \frac{{hv}}{c}

23 Medium

23 of 316

Understanding +4 | -1

V

be the accelerating voltage, then the maximum frequency of continuous

X

-rays is given by

View answer and explanation

Correct option: C

\frac{{eV}}{h}

Option C (

\frac{{eV}}{h}

) is correct because the maximum frequency of continuous X-rays is directly proportional to the accelerating voltage V and inversely proportional to Planck's constant h.

24 Medium

24 of 316

Application +4 | -1

Photoelectrons are emitted from a metal surface when UV light of wavelength

\lambda=300 \mathrm{~nm}

is incident on it. The minimum negative potential required to stop the emission of electrons is

0.54 \mathrm{~V}

. Calculate the work function of the metal.

View answer and explanation

Correct option: A

3.6 \mathrm{~eV}

Option A (

3.6 \mathrm{~eV}

) is correct because it is calculated by subtracting the stopping potential energy from the incident photon energy.

25 Medium

25 of 316

Application +4 | -1

The work function of tungsten and sodium are

4.5 {eV}

and

2.3 {eV}

respectively. If the threshold wavelength,

\lambda

for sodium is

5460 \mathop {\text{A}}\limits^ \circ

, the value of

\lambda

for tungsten is

View answer and explanation

Correct option: A

2791 \mathop {\text{A}}\limits^ \circ

Option A (

2791 \mathop {\text{A}}\limits^ \circ

) is correct because it is calculated using the inverse relationship between work function and threshold wavelength.

26 Medium

26 of 316

Analyzing +4 | -1

Assertion :The de Broglie equation has significance for any microscopic or sub-microscopic particles.
Reason: The de Broglie wavelength is inversely proportional to the mass of the object if velocity is constant.

View answer and explanation

Correct option: A

If both Assertion & reason are True & the reason is a correct explanation of the assertion

Option A is correct because the de Broglie equation applies to microscopic particles, and the wavelength is inversely proportional to mass at constant velocity, explaining its significance.

27 Medium

27 of 316

Remembering +4 | -1

Momentum of a photon of wavelength

\lambda

View answer and explanation

Correct option: A

h/\lambda

Option A (

h/\lambda

) is correct because the momentum of a photon is given by the equation

p = h/\lambda

, where h is Planck's constant and λ is the wavelength.

28 Medium

28 of 316

Application +4 | -1

Ultraviolet radiations of 6.2 eV falls on an aluminium surface. KE of fastest electron emitted is (work function = 4.2 eV)

View answer and explanation

Correct option: B

3.2 \times {10^{ - 19}}\mathrm{J}

Option B (

3.2 \times {10^{ - 19}}\mathrm{J}

) is correct because it is the kinetic energy calculated using the photoelectric effect equation.

29 Medium

29 of 316

Understanding +4 | -1

Assertion: Photoelectric effect is the phenomenon of emission of photon by metal when illuminated by light of suitable frequency.

Reason: An electron beam carries sufficient energy to release photons from the metal.

View answer and explanation

Correct option: D

Both Assertion and Reason are false

Option D is correct because both the assertion and the reason are false. The photoelectric effect involves the emission of electrons, not photons, and is caused by light, not an electron beam.

30 Medium

30 of 316

Understanding +4 | -1

Which of the following statement are true:

(a) Dual nature of matter is evident in de-Broglie equation

\lambda=\frac{\mathrm{h}}{\mathrm{p}}

(b) Wave nature of macroscopic particles not visible in daily life

(c)

\lambda=\frac{\mathrm{h}}{\mathrm{p}}

is not satisfied by a photon

View answer and explanation

Correct option: B

(a, b)

Option B ((a, b)) is correct because the de-Broglie equation demonstrates the dual nature of matter, and the wave nature of macroscopic particles is not readily observable.

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Key Takeaways

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