Dual Nature of Radiation and Matter — Page 7 | JEE Physics

Q61

An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :

A 10

-

2 nm

B 10

-

1 nm

C 10

-

3 nm

D 10

-

4 nm

Correct Answer

Option C

Solution

Given, V = 1.24 million volt = 1.24 $\times$ 106 volt Since, energy (E) = eV where, e is the charge of electron = 1.6 $\times$ 10 $-$ 19 C $\therefore$ E = 1.6 $\times$ 10 $-$ 19 $\times$ 1.24 $\times$ 106 ..... (i) As we know that, Energy of photon,

E = {{hc} \over \lambda }

.... (ii) Here, Planck's constant, h = 6.67 $\times$ 10 $-$ 34 J-s, c = speed of light in free space, c = 3 $\times$ 108 ms $-$ 1 Equating Eqs. (i) and (ii), we get

1.6 \times {10^{ - 19}} \times 1.24 \times {10^6} = {{6.67 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over \lambda }

\Rightarrow \lambda = {{20.01 \times {{10}^{ - 13}}} \over {1.6 \times 1.24}} = 10.09 \times {10^{ - 13}}

= 1.009 \times {10^{ - 12}} \simeq {10^{ - 3}} \times {10^{ - 9}}

= 10 $-$ 3 nm

Q62

A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.878

\times

10–4. The mass of the particle is close to

A 1.2

\times

10–28 kg

B 9.1

\times

10–31 kg

C 4.8

\times

10–27 kg

D 9.7

\times

10–28 kg

Correct Answer

Option D

Solution

Let mass of particle = m Let speed of e– = V $\therefore$ speed of particle = 5V de-broglie wavelength $\lambda$ d =

{h \over P} = {h \over {mv}}

$\therefore$ ( $\lambda$ d)P =

{h \over {m\left( {5V} \right)}}

....(1) and ( $\lambda$ d)e =

{h \over {{m_e}\left( V \right)}}

....(1) According to question

{{{{\left( {{\lambda _d}} \right)}_P}} \over {{{\left( {{\lambda _d}} \right)}_e}}} = {{{m_e}} \over {5m}}

= 1.878 $\times$ 10–4 $\Rightarrow$ m =

{{{m_e}} \over {5 \times 1.874 \times {{10}^{ - 4}}}}

=

{{9.1 \times {{10}^{ - 31}}} \over {5 \times 1.874 \times {{10}^{ - 4}}}}

= 9.7 $\times$ 10–28 kg

Q63

When a metal surface is illuminated by light of wavelength

\lambda

, the stopping potential is

8 \mathrm{~V}

. When the same surface is illuminated by light of wavelength

3 \lambda

, stopping potential is

2 \mathrm{~V}

. The threshold wavelength for this surface is:

A 3

\lambda

B 9

\lambda

C 5

\lambda

D 4.5

\lambda

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{E}=\phi+\mathrm{K}_{\max } \\ & \phi=\frac{\mathrm{hc}}{\lambda_0} \\ & \mathrm{~K}_{\max }=\mathrm{eV}_0 \\ & 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text{ (i) } \\ & 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text{ (ii) } \\ & \text{ on solving (i) & (ii) } \\ & \lambda_0=9 \lambda \end{aligned}

Q64

Formation of covalent bonds in compounds exhibits

A wave nature of electron

B particle nature of electron

C both wave and particle nature of electron

D none of these

Correct Answer

Option A

Solution

Formation of covalent bond is best explained by molecular orbital theory.

Q65

A nucleus A, with a finite de-broglie wavelength

\lambda

A, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths

\lambda

B and

\lambda

C of B and C are respectively :

A

\lambda

A, 2

\lambda

A

B 2

\lambda

A,

\lambda

A

C

\lambda

A,

\lambda

A/2

D

\lambda

A/2,

\lambda

A

Correct Answer

Option D

Solution

Let mass of B and C is m each. By momentum conservation

2m{v_0} = mv - {{mv} \over 2}

v = 4v0 PA = 2mv0 pB = 4mv0 pc = 2mv0 De-Broglie wavelength

\lambda = {h \over p}

{\lambda _A} = {h \over {2m{v_0}}}

;

{\lambda _B} = {h \over {4m{v_0}}}

;

{\lambda _C} = {h \over {2m{v_0}}}

Q66

A photocell is illuminated by a small bright source placed

1

m

away. When the same source of light is placed

{1 \over 2}

m

away, the number of electrons emitted by photo-cathode would

A increases by a factor of

4

B decreases by a factor of

4

C increases by a factor of

2

D decreases by a factor of

2

Correct Answer

Option A

Solution

I \propto {1 \over {{r^2}}};{{{I_1}} \over {{I_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^2} = {1 \over 4}

{I_2} \to 4\,\,

times

{I_1}

When intensity becomes 4 times, no. of photoelectrons emitted would increase by

4

times, since number of electrons emitted per second is directly proportional to intensity.

Q67

The threshold frequency of a metal with work function

6.63 \mathrm{~eV}

is :

A

16 \times 10^{15} \mathrm{~Hz}

B

16 \times 10^{12} \mathrm{~Hz}

C

1.6 \times 10^{15} \mathrm{~Hz}

D

1.6 \times 10^{12} \mathrm{~Hz}

Correct Answer

Option C

Solution

The threshold frequency,

\nu_0

, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect.

The work function, represented by $\phi$ , is the minimum energy needed to remove an electron from the surface of the metal.

The energy of a photon is given by the equation

E = h\nu

, where:

E

is the energy of the photon,

h

is Planck's constant (

h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}

), and $\nu$ is the frequency of the photon. To find the threshold frequency for a metal with a work function of

6.63 \, \text{eV}

, we must first express the work function in joules (since Planck's constant is in joules per second).

To convert electron volts to joules, use the conversion factor

1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}

:

\phi = 6.63 \, \text{eV} \times 1.602 \times 10^{-19} \, \frac{\text{J}}{\text{eV}} = 1.061 \times 10^{-18} \, \text{J}

The energy of the photon at the threshold frequency is equal to the work function:

h\nu_0 = \phi

Solve for

\nu_0

:

\nu_0 = \frac{\phi}{h} = \frac{1.061 \times 10^{-18} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} = 1.6 \times 10^{15} \, \text{Hz}

Thus, the correct answer is: Option C

1.6 \times 10^{15} \, \text{Hz}

Q68

If the two metals

\mathrm{A}

and

\mathrm{B}

are exposed to radiation of wavelength

350 \mathrm{~nm}

. The work functions of metals

\mathrm{A}

and

\mathrm{B}

are

4.8 \mathrm{eV}

and

2.2 \mathrm{eV}

. Then choose the correct option.

A Metal B will not emit photo-electrons

B Both metals

\mathrm{A}

and

\mathrm{B}

will not emit photo-electrons

C Metal A will not emit photo-electrons

D Both metals A and B will emit photo-electrons

Correct Answer

Option C

Solution

\phi=\frac{h c}{\lambda}=\frac{1240}{350} \mathrm{eV}=3.54 \mathrm{eV}

$\therefore$ Only metal B will emit photoelectron.

Q69

A monochromatic light is incident on a metallic plate having work function

\phi

. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point

B

. The distance between

A

and

B

is : (Given : The magnitude of charge of an electron is e and mass is

\mathrm{m}, \mathrm{h}

is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

A

\sqrt{2 m\left(\dfrac{h c}{\lambda}-\phi\right)} / \mathrm{eB}

B

\sqrt{8 \mathrm{~m}\left(\dfrac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}

C

\sqrt{\mathrm{m}(\mathrm{hc} / \lambda-\phi)} / \mathrm{eB}

D

2 \sqrt{\mathrm{~m}(\mathrm{hc} / \lambda-\phi)} / \mathrm{eB}

Correct Answer

Option B

Solution

To determine the distance between points $A$ and $B$ (where the electron re-enters the metallic plate), we first need to understand the electron's behavior in a magnetic field.

Maximum Kinetic Energy (KE): The energy of the emitted electron can be given by the photoelectric equation: $\text{KE}_{\max} = \dfrac{hc}{\lambda} - \phi$ Here, $\dfrac{hc}{\lambda}$ is the energy of the incident photons, and $\phi$ is the work function of the metal.

Momentum (p): The momentum of the electron is related to its kinetic energy by the equation: $p = \sqrt{2m \cdot \text{KE}_{\max}} = \sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}$ Path of Electron in Magnetic Field: When an electron moves perpendicularly through a magnetic field, it follows a circular path.

The radius $R$ of this path is given by: $R = \dfrac{p}{eB}$ where $e$ is the charge of the electron, and $B$ is the magnetic field strength.

Distance Between $A$ and $B$ : Since the electron travels back to the plate, forming a complete semicircle, the distance between points $A$ and $B$ is twice the radius of this circular path: $d_{A-B} = 2R = 2\left(\dfrac{p}{eB}\right)$ Substituting the expression for momentum, we find: $d_{A-B} = \dfrac{2 \sqrt{2m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB} = \dfrac{\sqrt{8m\left(\dfrac{hc}{\lambda} - \phi\right)}}{eB}$ Therefore, the distance between $A$ and $B$ is $\dfrac{\sqrt{8m(\dfrac{hc}{\lambda} - \phi)}}{eB}$ .

Q70

A small object at rest, absorbs a light pulse of power

20 \mathrm{~mW}

and duration

300 \mathrm{~ns}

. Assuming speed of light as

3 \times 10^{8} \mathrm{~m} / \mathrm{s}

, the momentum of the object becomes equal to :

A

1 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

B

0.5 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

C

3 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

D

2 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

Correct Answer

Option D

Solution

Assuming the small object as photon. Momentum

(p)=\frac{E}{C}

=\frac{20\times10^{-3}\times300\times10^{-9}}{3\times10^8}

=2\times10^{-17}

kg m/s