Dual Nature of Radiation and Matter — Page 6 | JEE Physics

Q51

Monochromatic light of frequency

6 \times 10^{14} \mathrm{~Hz}

is produced by a laser. The power emitted is

2 \times 10^{-3} \mathrm{~W}

. How many photons per second on an average, are emitted by the source ? (Given

\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}

)

A

5 \times 10^{15}

B

7 \times 10^{16}

C

6 \times 10^{15}

D

9 \times 10^{18}

Correct Answer

Option A

Solution

To find out the number of photons emitted per second by the laser, we can use the relationship between the energy of a single photon, the total energy emitted per second (power), and the number of photons emitted per second.

The energy $E$ of a single photon is given by Planck's equation:

E = hf

where: $h$ is Planck's constant ( $6.63 \times 10^{-34} \mathrm{Js}$ ), and $f$ is the frequency of the light ( $6 \times 10^{14} \mathrm{Hz}$ ).

Let's first calculate the energy of one photon:

E = (6.63 \times 10^{-34} \mathrm{Js}) \times (6 \times 10^{14} \mathrm{Hz})

E = 3.978 \times 10^{-19} \mathrm{J}

The power ( $P$ ) emitted by the laser is the total energy emitted per second,

P = E_{\text{total per second}} = 2 \times 10^{-3} \mathrm{W} = 2 \times 10^{-3} \mathrm{J/s}

The number of photons ( $N$ ) emitted per second can be found by dividing the total energy emitted per second by the energy of one photon:

N = \frac{P}{E}

Substitute the values we have:

N = \frac{2 \times 10^{-3} \mathrm{J/s}}{3.978 \times 10^{-19} \mathrm{J}}

N = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}}

N = 5.03 \times 10^{15} \text{ photons per second}

The number of photons emitted per second is approximately $5 \times 10^{15}$ .

Therefore, the correct answer, rounded to one significant figure, is: Option A: $5 \times 10^{15}$

Q52

Surface of certain metal is first illuminated with light of wavelength

\lambda

1 = 350 nm and then, by light of wavelength

\lambda

2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to : (Energy of photon n =

{{1240} \over {\lambda (in\,mm)}}

eV)

A 1.8

B 2.5

C 5.6

D 1.4

Correct Answer

Option A

Solution

Let speed of photon electron in first case is 2v then in the second case speed is v. For first case

{{hc} \over {{\lambda _1}}} = \phi + {1 \over 2}

m(2v)2 For second case,

{{hc} \over {{\lambda _2}}} = \phi + {1 \over 2}

mv2 $\therefore$

{{{{hc} \over {{\lambda _1}}} - \phi } \over {{{hc} \over {{\lambda _2}}} - \phi }} = {{{1 \over 2}m \times 4{v^2}} \over {{1 \over 2}m{v^2}}}

$\Rightarrow$

{{{hc} \over {{\lambda _1}}} - \phi }

= 4

\left( {{{hc} \over {{\lambda _2}}} - \phi } \right)

$\Rightarrow$

{{4hc} \over {{\lambda _2}}}

$-$

{{hc} \over {{\lambda _1}}}

= 3 $\phi$ $\Rightarrow$ $\phi$ $=$

{{hc} \over 3}\left( {{4 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right)

$\Rightarrow$ $\phi$ $=$

{{1240} \over 3}\left( {{4 \over {540}} - {1 \over {350}}} \right)

$\Rightarrow$ $\phi$ = 1.8 eV

Q53

In a photoelectric experiment, increasing the intensity of incident light :

A increases the number of photons incident and also increases the K.E. of the ejected electrons

B increases the frequency of photons incident and increases the K.E. of the ejected electrons

C increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged

D increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

Correct Answer

Option D

Solution

$\to$ Increasing intensity means number of incident photons are increased.

$\to$ Kinetic energy of ejected electrons depend on the frequency of incident photons, not the intensity.

Q54

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal. Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal. In the light of the above statements, choose the most appropriate answer from the options given below.

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

B Both A and R are correct but R is not the correct explanation of A.

C A is correct but R is not correct.

D A is not correct but R is correct.

Correct Answer

Option B

Solution

When energy of incident radiation is equal to the work function of the metal, then the KE of photoelectrons would be zero.

But this statement does not comment on the situation when energy is less than the work function.

Q55

The equation

\lambda=\dfrac{1.227}{x} \mathrm{~nm}

can be used to find the de-Brogli wavelength of an electron. In this equation

x

stands for : Where

\mathrm{m}=

mass of electron

\mathrm{P}=

momentum of electron

\mathrm{K}=

Kinetic energy of electron

\mathrm{V}=

Accelerating potential in volts for electron

A

\sqrt{\mathrm{mK}}

B

\sqrt{\mathrm{P}}

C

\sqrt{\mathrm{K}}

D

\sqrt{\mathrm{V}}

Correct Answer

Option D

Solution

The de Broglie wavelength of a particle can be expressed as:

\lambda = \frac{h}{p}

where:

h

is Planck's constant, and

p

is the momentum of the particle. For an electron accelerated through a potential difference of

V

volts, its kinetic energy

K

is given by:

K = eV

where

e

is the charge of an electron. The electron's momentum can be expressed in terms of its kinetic energy as:

p = \sqrt{2mK}

where

m

is the mass of the electron. Substituting this into the de Broglie wavelength equation, we get:

\lambda = \frac{h}{\sqrt{2mK}}

Comparing this with the given equation

\lambda = \frac{1.227}{x} \, nm

, we see that

x

must correspond to:

x = \sqrt{2mK}

Since

K = eV

, we can substitute this into our expression for

x

to get:

x = \sqrt{2meV} = \sqrt{V}

where we've used the fact that the mass and charge of an electron are constants.

Q56

UV light of

4.13 \mathrm{~eV}

is incident on a photosensitive metal surface having work function

3.13 \mathrm{~eV}

. The maximum kinetic energy of ejected photoelectrons will be:

A 4.13 eV

B 1 eV

C 7.26 eV

D 3.13 eV

Correct Answer

Option B

Solution

To find the maximum kinetic energy of the ejected photoelectrons, we'll use the photoelectric effect equation:

KE_{\text{max}} = h\nu - \phi

where

KE_{\text{max}}

is the maximum kinetic energy of the ejected electrons,

h

is Planck's constant, $\nu$ is the frequency of the incident light, and $\phi$ is the work function of the metal.

However, in this problem, we are given the energy of the UV light in electronvolts (eV) directly, which simplifies the problem.

The energy of the UV light in electronvolts also represents the energy of the photons (

h\nu

) hitting the metal surface.

Thus, we can calculate the maximum kinetic energy of the ejected photoelectrons using the given energies directly:

KE_{\text{max}} = E_{\text{photon}} - \phi

Substituting the given values:

KE_{\text{max}} = 4.13 \, \text{eV} - 3.13 \, \text{eV} = 1 \, \text{eV}

Therefore, the maximum kinetic energy of the ejected photoelectrons will be

1 \, \text{eV}

. The correct answer is Option B: 1 eV.

Q57

The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be :

A less than

\sqrt 3

v

B v

C more than

\sqrt 3 \,v

D equal to

\sqrt 3 \,v

Correct Answer

Option C

Solution

The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by

KE = {1 \over 2}m{v^2} = hn - \phi

where h is Planck's constant and $\phi$ is the work function. Therefore, the kinetic energy of the incident light is

{E_1} = hn - \phi

..... (1) When the frequency of the incident light is increased to 3n, then the kinetic energy is given by

{1 \over 2}mv_1^2 = 3hn - \phi

\Rightarrow {E_2} = 3hn - \phi

..... (2) Substituting

hn = {E_1} + \phi

[from Eq. (1)] in Eq. (2), we get

{E_2} = 3({E_1} + \phi ) - \phi

\Rightarrow {E_2} = 3{E_1} + 2\phi

\Rightarrow {1 \over 2}mv_1^2 = 3 \times {1 \over 2}m{v^2} + 2\phi

\Rightarrow v_1^2 = 3{v^2} + 2\phi \times {2 \over m}

\Rightarrow v_1^2 = 3{v^2} + {{4\phi } \over m}

Thus, the velocity is more than

\sqrt 3 v

.

Q58

Particle A of mass mA =

{m \over 2}

moving along the x-axis with velocity v0 collides elastically with another particle B at rest having mass mB =

{m \over 3}

. If both particles move along the x-axis after the collision, the change

\Delta

\lambda

in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength (

\lambda

0) before collision is :

A

\Delta

\lambda

=

{5 \over 2}{\lambda _0}

B

\Delta

\lambda

=

{3 \over 2}{\lambda _0}

C

\Delta

\lambda

= 2

\lambda

0

D

\Delta

\lambda

= 4

\lambda

0

Correct Answer

Option D

Solution

Applying momentum conservation

{m \over 2} \times {V_0} + {m \over 3} \times 0 = {m \over 2}{V_A} + {m \over 3}{V_B}

$\Rightarrow$

{{{V_0}} \over 2} = {{{V_A}} \over 2} + {{{V_B}} \over 3}

.....(1) Since, collision is elastic (e = 1) e = 1 =

{{{V_B} - {V_A}} \over {{V_0}}}

$\Rightarrow$ V0 = VB – VA .....(2) On solving (1) & (2) : VA =

{{{V_0}} \over 5}

Now, De-Broglie wavelength of A before collision : $\lambda$ 0 =

{h \over {{m_A}{V_0}}}

=

{h \over {\left( {{m \over 2}} \right){V_0}}}

=

{{2h} \over {m{V_0}}}

Final De-Broglie wavelength : $\lambda$ f =

{h \over {{m_A}{V_A}}}

=

{h \over {\left( {{m \over 2}} \right)\left( {{{{V_0}} \over 5}} \right)}}

=

{{10h} \over {m{V_0}}}

Now,

\Delta

$\lambda$ = $\lambda$ f - $\lambda$ 0 =

{{10h} \over {m{V_0}}}

-

{{2h} \over {m{V_0}}}

$\Rightarrow$

\Delta

$\lambda$ =

{{8h} \over {m{V_0}}}

$\Rightarrow$

\Delta

$\lambda$ =

4 \times {{2h} \over {m{V_0}}}

$\Rightarrow$

\Delta

$\lambda$ = 4 $\lambda$ 0

Q59

The surface of a metal is illuminated with the light of

400

nm.

The kinetic energy of the ejected photoelectrons was found to be

1.68

eV.

The work function of the metal is :

\left( {hc = 1240eV.nm} \right)

A

1.41

eV

B

1.51

eV

C

1.68

eV

D

3.09

eV

Correct Answer

Option A

Solution

The photoelectric effect equation, which relates the energy of the incident light to the kinetic energy of the ejected photoelectrons and the work function of the metal, is given by: $E = K_{\max} + W$ , where $E$ is the energy of the incident light, $K_{\max}$ is the maximum kinetic energy of the photoelectrons, and $W$ is the work function of the metal.

The energy of the incident light can be calculated using the formula $E = \dfrac{hc}{\lambda}$ , where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light.

However, given that $hc = 1240$ eV⋅nm, we can simplify this to $E = \dfrac{1240}{\lambda}$ .

Substituting the given values into this equation, we have: $E = \dfrac{1240}{400} = 3.1$ eV.

We can then substitute these values into the photoelectric effect equation: $3.1 \text{ eV} = 1.68 \text{ eV} + W$ , which simplifies to: $W = 3.1 \text{ eV} - 1.68 \text{ eV} = 1.42$ eV.

Rounding to two decimal places, the work function of the metal is therefore approximately $1.42$ eV.

Thus, Option A: $1.41$ eV is the closest to the correct answer.

Q60

A metallic surface is illuminated with radiation of wavelength

\lambda

, the stopping potential is

V_{0}

. If the same surface is illuminated with radiation of wavelength

2 \lambda

. the stopping potential becomes

\dfrac{V_{o}}{4}

. The threshold wavelength for this metallic surface will be

A

3 \lambda

B

4 \lambda

C

\dfrac{3}{2} \lambda

D

\dfrac{\lambda}{4}

Correct Answer

Option A

Solution

The photoelectric effect occurs when light (or more generally, electromagnetic radiation) incident on a metallic surface causes the ejection of electrons from the surface.

The energy of the incident photons must be greater than the work function of the metal (denoted by

\phi_0

) for the electrons to be ejected.

The energy of the ejected electrons can be expressed as the difference between the energy of the incident photons and the work function of the metal.

The maximum kinetic energy of the ejected electrons is given by:

K_{max} = h\nu - \phi_0

where

h

is the Planck's constant, $\nu$ is the frequency of the incident light, and

\phi_0

is the work function of the metal. We can also write the maximum kinetic energy in terms of the stopping potential (

V_0

) and the elementary charge (

e

) of an electron:

K_{max} = eV_0

Equating these two expressions for the maximum kinetic energy, we get:

eV_0 = h\nu - \phi_0

We can express the frequency $\nu$ in terms of the speed of light

c

and the wavelength $\lambda$ :

\nu = \frac{c}{\lambda}

Substituting this expression for frequency in the equation, we get:

eV_0 = h\frac{c}{\lambda} - \phi_0

Now, we have two cases: The stopping potential is

V_0

, and the wavelength is $\lambda$ . The stopping potential is

\frac{V_0}{4}

, and the wavelength is

2\lambda

. For the first case, we use the photoelectric effect equation as is:

eV_0 = \frac{hc}{\lambda} - \phi_0

For the second case, we replace

V_0

with

\frac{V_0}{4}

and $\lambda$ with

2\lambda

:

\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0

Now we have two equations: (1)

eV_0 = \frac{hc}{\lambda} - \phi_0

(2)

\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0

We can rewrite equation (1) as:

\frac{eV_0}{4} = \frac{hc}{4\lambda} - \frac{\phi_0}{4}

Now we can equate the two expressions for

\frac{eV_0}{4}

:

\frac{hc}{4\lambda} - \frac{\phi_0}{4} = \frac{hc}{2\lambda} - \phi_0

Now we isolate the terms containing

\phi_0

:

\frac{3\phi_0}{4} = \frac{hc}{4\lambda} - \frac{hc}{2\lambda}

\frac{3\phi_0}{4} = \frac{hc}{4\lambda}

Now we can write the work function

\phi_0

in terms of the threshold wavelength

\lambda_0

:

\phi_0 = \frac{hc}{\lambda_0}

Substituting the expression for the work function

\phi_0

in terms of the threshold wavelength

\lambda_0

into the previous equation, we get:

\frac{3}{4} \frac{hc}{\lambda_0} = \frac{hc}{4\lambda}

Now we can cancel out the common terms

hc

from both sides:

\frac{3}{4\lambda_0} = \frac{1}{4\lambda}

Next, we can cross-multiply to solve for

\lambda_0

:

3\lambda = \lambda_0

Thus, the threshold wavelength for this metallic surface is

3\lambda

.