Dual Nature of Radiation and Matter — Page 4 | JEE Physics

Q31

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)

A

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

B

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

C

{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}

D

c{\left( {2mE} \right)^{{1 \over 2}}}

Correct Answer

Option B

Solution

$\lambda$ e =

{h \over {{p_e}}}

=

{h \over {\sqrt {2mE} }}

E =

{{hc} \over {{\lambda _{photon}}}}

$\Rightarrow$

{{\lambda _{photon}}}

=

{{hc} \over E}

$\therefore$

{{{\lambda _e}} \over {{\lambda _{photon}}}}

=

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

Q32

The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is

\lambda_1

. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

A 2

\lambda_1

B

\dfrac{1}{2}

\lambda_1

C

\dfrac{1}{\sqrt2}

\lambda_1

D

\sqrt2~\lambda_1

Correct Answer

Option C

Solution

The de Broglie wavelength of a particle is given by:

\lambda = \frac{h}{p}

where h is Planck's constant and p is the momentum of the particle.

The momentum of a gas molecule can be related to its kinetic energy (which is related to the temperature of the gas) by:

p = \sqrt{2mK}

where m is the mass of the molecule and K is the kinetic energy of the molecule.

At a given temperature T, the average kinetic energy of a molecule in a gas is given by:

K = \frac{3}{2} kT

where k is Boltzmann's constant. Therefore, the de Broglie wavelength of a molecule in a gas is given by:

\lambda = \frac{h}{\sqrt{2m(3/2)kT}} = \frac{h}{\sqrt{3mkT}}

If the temperature of the gas is increased from T = 300 K to T = 600 K, the new de Broglie wavelength becomes:

\lambda' = \frac{h}{\sqrt{3mk(2T)}} = \frac{h}{\sqrt{2} \sqrt{3mkT}} = \frac{1}{\sqrt{2}} \lambda

So, the de Broglie wavelength of the gas molecule decreases by a factor of

\sqrt{2}

when the temperature of the gas is doubled. Therefore, the correct answer is

\lambda' = \frac{1}{\sqrt{2}} \lambda_1

Q33

Which of the following phenomena does not explain by wave nature of light. A. reflection B. diffraction C. photoelectric effect D. interference E. polarization Choose the most appropriate answer from the options given below:

A C only

B B, D only

C A, C only

D E only

Correct Answer

Option A

Solution

The correct answer is Option A: C only.

Reflection, diffraction, interference, and polarization are all phenomena that can be explained by the wave nature of light.

These phenomena are evidence that light behaves as a wave, evident through various experimental observations: Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated.

The laws of reflection can be explained by treating light as waves.

Diffraction is the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle.

This phenomenon is observed when a wave encounters an obstacle or a slit that is comparable in size to its wavelength, which is best explained by wave theory.

Interference occurs when two or more waves superimpose to form a resultant wave of greater, lower, or the same amplitude.

This phenomenon illustrates the wave nature of light through constructive and destructive interference patterns, such as those observed in the double-slit experiment.

Polarization is a property applying to transverse waves that specifies the geometrical orientation of the oscillations.

For light waves, this can be seen as the direction in which the electric field oscillates.

Polarization can only occur with waves that have transverse components, further supporting the wave nature of light.

On the other hand, the photoelectric effect cannot be explained solely by the wave nature of light.

It is the emission of electrons or other free carriers when light shines on a material.

Electrons emitted in this manner can be called photoelectrons.

The phenomenon is best explained by Albert Einstein's quantum theory of light, where light is considered as quanta of energy called photons.

This effect demonstrates the particle aspect of light, wherein each photon has a discrete packet of energy equal to

hf

, where

h

is Planck's constant and

f

is the frequency of the light.

Thus, the photoelectric effect is the correct answer because it specifically requires the particle theory of light for its explanation, unlike reflection, diffraction, interference, and polarization, which are well-explained by the wave nature of light.

Q34

The kinetic energy of an electron,

\alpha

-particle and a proton are given as

4 \mathrm{~K}, 2 \mathrm{~K}

and

\mathrm{K}

respectively. The de-Broglie wavelength associated with electron

(\lambda \mathrm{e}), \alpha

-particle

((\lambda \alpha)

and the proton

(\lambda p)

are as follows:

A

\lambda \alpha<\lambda p<\lambda e

B

\lambda \alpha>\lambda p>\lambda e

C

\lambda \alpha=\lambda p<\lambda e

D

\lambda \alpha=\lambda p>\lambda e

Correct Answer

Option A

Solution

The de Broglie wavelength of a particle is given by the equation: $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mK}},$ where: $h$ is Planck's constant, $p$ is the momentum of the particle, $m$ is the mass of the particle, and $K$ is the kinetic energy of the particle.

This equation shows that the de Broglie wavelength of a particle is inversely proportional to the square root of its mass and its kinetic energy.

This means that the particle with the smallest mass and kinetic energy will have the largest de Broglie wavelength.

Given that the kinetic energies of the electron, alpha particle, and proton are $4K$ , $2K$ , and $K$ , respectively, and knowing that the mass of the electron is less than the mass of the proton and the mass of the proton is less than the mass of the alpha particle, we can infer that the de Broglie wavelength of the electron is greater than the de Broglie wavelength of the proton, which in turn is greater than the de Broglie wavelength of the alpha particle.

Therefore, the correct answer is $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$ .

Q35

A parallel beam of light of wavelength

900 \mathrm{~nm}

and intensity

100 \,\mathrm{Wm}^{-2}

is incident on a surface perpendicular to the beam. The number of photons crossing

1 \mathrm{~cm}^{2}

area perpendicular to the beam in one second is :

A

3 \times 10^{16}

B

4.5 \times 10^{16}

C

4.5 \times 10^{17}

D

4.5 \times 10^{20}

Correct Answer

Option B

Solution

$\lambda$ = 900 nm I = 100 W/m2 A = 10 $-$ 4 $\Rightarrow$ P = 10 $-$ 2 W $\Rightarrow$ Number of photons incident per second

= {{{{10}^{ - 2}}\lambda } \over {hc}}

= {{9 \times {{10}^{ - 11}} \times {{10}^2}} \over {6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \simeq 4.5 \times {10^{16}}

Q36

Statement -

1

: When ultraviolet light is incident on a photocell, its stopping potential is

{V_0}

and the maximum kinetic energy of the photoelectrons is

{K_{\max }}

. When the ultraviolet light is replaced by

X

-rays, both

{V_0}

and

{K_{\max }}

increase. Statement -

2

: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

A Statement -

1

is true, Statement -

2

is true; Statement -

2

is the correct explanation of Statement -

1

B Statement -

1

is true, Statement -

2

is true; Statement -

2

is not the correct explanation of Statement -

1

C Statement -

1

is is false, Statement -

2

is true

D Statement -

1

is is true, Statement -

2

is false

Correct Answer

Option B

Solution

Statement 1 is true.

The energy of an incident photon (from the ultraviolet light or X-rays) on a photocell is given by Planck's equation, $E = h\nu$ , where $h$ is Planck's constant and $\nu$ is the frequency of the light.

X-rays have a higher frequency than ultraviolet light, so they deliver more energy to the photoelectrons.

This results in a higher stopping potential ( $V_0$ ) and maximum kinetic energy ( $K_{\max}$ ) for the photoelectrons.

Statement 2 is also true.

However, while the speeds (and hence kinetic energies) of photoelectrons do vary, this variation is not because of a range of frequencies in the incident light.

Rather, it's due to the interaction of the incident photons with electrons at different energy levels in the metal.

A single frequency of light can produce photoelectrons with a range of speeds because the electrons they encounter can have a variety of binding energies.

Q37

The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation :

A Phase

B Frequency

C Amnplitude

D Intensity

Correct Answer

Option B

Solution

Stopping potential depends on frequency, according to Einstein's photoelectric equation.

hv - h{v_0} = eV

\Rightarrow V = {h \over e}v - {h \over e}{v_0}

Q38

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : The beam of electrons show wave nature and exhibit interference and diffraction. Reason R : Davisson Germer Experimentally verified the wave nature of electrons. In the light of the above statements, choose the most appropriate answer from the options given below :

A A is not correct but R is correct.

B Both A and R are correct and R is the correct explanation of A

C Both A and R are correct but R is Not the correct explanation of A

D A is correct but R is not correct

Correct Answer

Option B

Solution

The assertion A and the reason R are both correct statements, and the reason R is the correct explanation of the assertion A.

Explanation : The assertion A states that the beam of electrons exhibit wave nature and show interference and diffraction.

This statement is correct because electrons exhibit both particle-like and wave-like behavior.

When electrons are accelerated to high speeds, they have a wavelength associated with them, and this wavelength can interfere and diffract just like any other wave.

The reason R states that the wave nature of electrons was experimentally verified by Davisson Germer.

This statement is also correct because Davisson and Germer performed an experiment in 1927 where they observed diffraction patterns in a beam of electrons that were scattered off a nickel crystal.

This observation provided strong evidence for the wave nature of electrons.

Therefore, both assertion A and reason R are correct statements, and reason R is the correct explanation of assertion A.

Q39

In photoelectric experiment energy of

2.48 \mathrm{~eV}

irradiates a photo sensitive material. The stopping potential was measured to be

0.5 \mathrm{~V}

. Work function of the photo sensitive material is :

A 1.98 eV

B 1.68 eV

C 2.48 eV

D 0.5 eV

Correct Answer

Option A

Solution

To find the work function of the photo sensitive material, we can use the photoelectric equation which relates the kinetic energy of the ejected electrons to the photon energy and the work function ( $\phi$ ) of the material: $KE_{\text{max}} = h\nu - \phi$ Where $KE_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h\nu$ is the energy of the incoming photon, and $\phi$ is the work function of the material.

However, the kinetic energy of the ejected electrons can also be related to the stopping potential ( $V_s$ ) by the equation: $KE_{\text{max}} = e \cdot V_s$ Substituting this into the first equation gives: $e \cdot V_s = h\nu - \phi$ Here, $e$ is the charge of an electron ( $1.6 \times 10^{-19} \, \text{C}$ ), but since we are dealing with energies in electronvolts (eV), and $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$ , we can directly use the values given without converting the units: $\phi = h\nu - e \cdot V_s$ Where $h\nu = 2.48 \, \text{eV}$ is the energy of the irradiating photons, and $V_s = 0.5 \, \text{V}$ .

Substituting these values in, we get: $\phi = 2.48 \, \text{eV} - 0.5 \, \text{eV} = 1.98 \, \text{eV}$ Therefore, the work function ( $\phi$ ) of the photo sensitive material is $1.98 \, \text{eV}$ , which corresponds to Option A.

Q40

What should be the order of arrangement of de-Broglie wavelength of electron (

\lambda

e), an

\alpha

-particle (

\lambda

a) and proton (

\lambda

p) given that all have the same kinetic energy?

A

\lambda

e =

\lambda

p =

\lambda

\alpha

B

\lambda

e <

\lambda

p <

\lambda

\alpha

C

\lambda

e >

\lambda

p >

\lambda

\alpha

D

\lambda

e =

\lambda

p >

\lambda

\alpha

Correct Answer

Option C

Solution

\lambda = {h \over p} = {h \over {\sqrt {2mE} }} \propto {1 \over {\sqrt m }}

m $\alpha$ > mp > me So, $\lambda$ e > $\lambda$ p > $\lambda$ $\alpha$