QBManager

Dual nature of radiation and Matter

142201 Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is

1

3 : 2

2

2 : 3

3

1 : 3

4

1 : 2

Explanation:

C We know that kinetic energy in photoelectric effect,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{1}^{2} = 2 ϕ_{o} - ϕ_{o} = ϕ_{o}$
$\frac{1}{2} {mv}_{2}^{2} = 10 ϕ_{o} - ϕ_{o} = 9 ϕ_{o}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{9}$
$\frac{v_{1}}{v_{2}} = \frac{1}{3}$
Hence, the maximum velocity ratio -
$v_{1} : v_{2} = 1 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142202 Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $λ_{1} = 360 nm$ and $λ_{2} = 600 nm$ . What is the ratio of work functions for lights of wavelength ' $λ_{1}$ ' to ' $λ_{2}$ '?

1

6 : 1

2

3 : 5

3

5 : 3

4

1 : 6

Explanation:

C Given that, $λ_{1} = 360 nm, λ_{2} = 600 nm$ Work function -
$ϕ = \frac{h c}{λ}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{600}{360}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{5}{3}$
$ϕ_{1} : ϕ_{2} = 5 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142203 Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is

1

1 : 4

2

1 : 2

3

1 : \sqrt{2}

4

1 : 1

Explanation:

C Maximum K.E = Incident photon energy work function
$K_{max} = E - W$
According to question-
$\frac{1}{2} {mv}_{1}^{2} = 3 W - W = 2 W$
$\frac{1}{2} {mv}_{2}^{2} = 5 W - W = 4 W$
Dividing equation (i) by equation (ii), we get-
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$
$\frac{v_{1}}{v_{2}} = \frac{1}{\sqrt{2}}$
$v_{1} : v_{2} = 1 : \sqrt{2}$

MHT-CET 2020

Dual nature of radiation and Matter

142204 The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $X$ axis respectively are

1 Planck's constant, threshold frequency

2 work function, maximum K.E.

3 Planck's constant, work function

4 Maximum K.E, threshold frequency

Explanation:

A
From Einstein's photoelectric equation,
(K.E $)_{max} = h v - ϕ_{o}$
Equation of straight line,
Here $y = mx + c$
Here $m = h$
$y$ -intercept $c = - ϕ_{o}$
So, the slope of graph and intercept on $x$ -axis shown Planck's constant and threshold frequency.

MHT-CET 2020

Dual nature of radiation and Matter

142205 The light of wavelength ' $λ$ '. Incident on the surface of metal having work function $ϕ$ emits the electrons. The maximum velocity of electrons emitted is
$(c =$ velocity of light, $h =$ Planck's constant, $m =$ mass of electron)

1

{[\frac{2 (hc - λ)}{m λ}]}^{\frac{1}{2}}

2

[\frac{2 (hc - ϕ) λ}{mc}]

3

{[\frac{2 (h c - λ ϕ)}{m λ}]}^{\frac{1}{2}}

4

[\frac{2 (hc - ϕ)}{m λ}]

Explanation:

C According to photo-electric equation,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc}{λ} - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc - ϕ_{o} λ}{λ}$
$v_{m}^{2} = \frac{2 (hc - ϕ_{o} λ)}{λ m}$
$v_{m} = \sqrt{\frac{2 (hc - ϕ_{o} λ)}{λ m}}$

ASSAM CEE-2014

Dual nature of radiation and Matter

142201 Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is

1

3 : 2

2

2 : 3

3

1 : 3

4

1 : 2

Explanation:

C We know that kinetic energy in photoelectric effect,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{1}^{2} = 2 ϕ_{o} - ϕ_{o} = ϕ_{o}$
$\frac{1}{2} {mv}_{2}^{2} = 10 ϕ_{o} - ϕ_{o} = 9 ϕ_{o}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{9}$
$\frac{v_{1}}{v_{2}} = \frac{1}{3}$
Hence, the maximum velocity ratio -
$v_{1} : v_{2} = 1 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142202 Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $λ_{1} = 360 nm$ and $λ_{2} = 600 nm$ . What is the ratio of work functions for lights of wavelength ' $λ_{1}$ ' to ' $λ_{2}$ '?

1

6 : 1

2

3 : 5

3

5 : 3

4

1 : 6

Explanation:

C Given that, $λ_{1} = 360 nm, λ_{2} = 600 nm$ Work function -
$ϕ = \frac{h c}{λ}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{600}{360}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{5}{3}$
$ϕ_{1} : ϕ_{2} = 5 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142203 Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is

1

1 : 4

2

1 : 2

3

1 : \sqrt{2}

4

1 : 1

Explanation:

C Maximum K.E = Incident photon energy work function
$K_{max} = E - W$
According to question-
$\frac{1}{2} {mv}_{1}^{2} = 3 W - W = 2 W$
$\frac{1}{2} {mv}_{2}^{2} = 5 W - W = 4 W$
Dividing equation (i) by equation (ii), we get-
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$
$\frac{v_{1}}{v_{2}} = \frac{1}{\sqrt{2}}$
$v_{1} : v_{2} = 1 : \sqrt{2}$

MHT-CET 2020

Dual nature of radiation and Matter

142204 The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $X$ axis respectively are

1 Planck's constant, threshold frequency

2 work function, maximum K.E.

3 Planck's constant, work function

4 Maximum K.E, threshold frequency

Explanation:

A
From Einstein's photoelectric equation,
(K.E $)_{max} = h v - ϕ_{o}$
Equation of straight line,
Here $y = mx + c$
Here $m = h$
$y$ -intercept $c = - ϕ_{o}$
So, the slope of graph and intercept on $x$ -axis shown Planck's constant and threshold frequency.

MHT-CET 2020

Dual nature of radiation and Matter

142205 The light of wavelength ' $λ$ '. Incident on the surface of metal having work function $ϕ$ emits the electrons. The maximum velocity of electrons emitted is
$(c =$ velocity of light, $h =$ Planck's constant, $m =$ mass of electron)

1

{[\frac{2 (hc - λ)}{m λ}]}^{\frac{1}{2}}

2

[\frac{2 (hc - ϕ) λ}{mc}]

3

{[\frac{2 (h c - λ ϕ)}{m λ}]}^{\frac{1}{2}}

4

[\frac{2 (hc - ϕ)}{m λ}]

Explanation:

C According to photo-electric equation,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc}{λ} - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc - ϕ_{o} λ}{λ}$
$v_{m}^{2} = \frac{2 (hc - ϕ_{o} λ)}{λ m}$
$v_{m} = \sqrt{\frac{2 (hc - ϕ_{o} λ)}{λ m}}$

ASSAM CEE-2014

Dual nature of radiation and Matter

142201 Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is

1

3 : 2

2

2 : 3

3

1 : 3

4

1 : 2

Explanation:

C We know that kinetic energy in photoelectric effect,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{1}^{2} = 2 ϕ_{o} - ϕ_{o} = ϕ_{o}$
$\frac{1}{2} {mv}_{2}^{2} = 10 ϕ_{o} - ϕ_{o} = 9 ϕ_{o}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{9}$
$\frac{v_{1}}{v_{2}} = \frac{1}{3}$
Hence, the maximum velocity ratio -
$v_{1} : v_{2} = 1 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142202 Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $λ_{1} = 360 nm$ and $λ_{2} = 600 nm$ . What is the ratio of work functions for lights of wavelength ' $λ_{1}$ ' to ' $λ_{2}$ '?

1

6 : 1

2

3 : 5

3

5 : 3

4

1 : 6

Explanation:

C Given that, $λ_{1} = 360 nm, λ_{2} = 600 nm$ Work function -
$ϕ = \frac{h c}{λ}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{600}{360}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{5}{3}$
$ϕ_{1} : ϕ_{2} = 5 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142203 Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is

1

1 : 4

2

1 : 2

3

1 : \sqrt{2}

4

1 : 1

Explanation:

C Maximum K.E = Incident photon energy work function
$K_{max} = E - W$
According to question-
$\frac{1}{2} {mv}_{1}^{2} = 3 W - W = 2 W$
$\frac{1}{2} {mv}_{2}^{2} = 5 W - W = 4 W$
Dividing equation (i) by equation (ii), we get-
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$
$\frac{v_{1}}{v_{2}} = \frac{1}{\sqrt{2}}$
$v_{1} : v_{2} = 1 : \sqrt{2}$

MHT-CET 2020

Dual nature of radiation and Matter

142204 The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $X$ axis respectively are

1 Planck's constant, threshold frequency

2 work function, maximum K.E.

3 Planck's constant, work function

4 Maximum K.E, threshold frequency

Explanation:

A
From Einstein's photoelectric equation,
(K.E $)_{max} = h v - ϕ_{o}$
Equation of straight line,
Here $y = mx + c$
Here $m = h$
$y$ -intercept $c = - ϕ_{o}$
So, the slope of graph and intercept on $x$ -axis shown Planck's constant and threshold frequency.

MHT-CET 2020

Dual nature of radiation and Matter

142205 The light of wavelength ' $λ$ '. Incident on the surface of metal having work function $ϕ$ emits the electrons. The maximum velocity of electrons emitted is
$(c =$ velocity of light, $h =$ Planck's constant, $m =$ mass of electron)

1

{[\frac{2 (hc - λ)}{m λ}]}^{\frac{1}{2}}

2

[\frac{2 (hc - ϕ) λ}{mc}]

3

{[\frac{2 (h c - λ ϕ)}{m λ}]}^{\frac{1}{2}}

4

[\frac{2 (hc - ϕ)}{m λ}]

Explanation:

C According to photo-electric equation,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc}{λ} - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc - ϕ_{o} λ}{λ}$
$v_{m}^{2} = \frac{2 (hc - ϕ_{o} λ)}{λ m}$
$v_{m} = \sqrt{\frac{2 (hc - ϕ_{o} λ)}{λ m}}$

ASSAM CEE-2014

Dual nature of radiation and Matter

142201 Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is

1

3 : 2

2

2 : 3

3

1 : 3

4

1 : 2

Explanation:

C We know that kinetic energy in photoelectric effect,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{1}^{2} = 2 ϕ_{o} - ϕ_{o} = ϕ_{o}$
$\frac{1}{2} {mv}_{2}^{2} = 10 ϕ_{o} - ϕ_{o} = 9 ϕ_{o}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{9}$
$\frac{v_{1}}{v_{2}} = \frac{1}{3}$
Hence, the maximum velocity ratio -
$v_{1} : v_{2} = 1 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142202 Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $λ_{1} = 360 nm$ and $λ_{2} = 600 nm$ . What is the ratio of work functions for lights of wavelength ' $λ_{1}$ ' to ' $λ_{2}$ '?

1

6 : 1

2

3 : 5

3

5 : 3

4

1 : 6

Explanation:

C Given that, $λ_{1} = 360 nm, λ_{2} = 600 nm$ Work function -
$ϕ = \frac{h c}{λ}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{600}{360}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{5}{3}$
$ϕ_{1} : ϕ_{2} = 5 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142203 Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is

1

1 : 4

2

1 : 2

3

1 : \sqrt{2}

4

1 : 1

Explanation:

C Maximum K.E = Incident photon energy work function
$K_{max} = E - W$
According to question-
$\frac{1}{2} {mv}_{1}^{2} = 3 W - W = 2 W$
$\frac{1}{2} {mv}_{2}^{2} = 5 W - W = 4 W$
Dividing equation (i) by equation (ii), we get-
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$
$\frac{v_{1}}{v_{2}} = \frac{1}{\sqrt{2}}$
$v_{1} : v_{2} = 1 : \sqrt{2}$

MHT-CET 2020

Dual nature of radiation and Matter

142204 The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $X$ axis respectively are

1 Planck's constant, threshold frequency

2 work function, maximum K.E.

3 Planck's constant, work function

4 Maximum K.E, threshold frequency

Explanation:

A
From Einstein's photoelectric equation,
(K.E $)_{max} = h v - ϕ_{o}$
Equation of straight line,
Here $y = mx + c$
Here $m = h$
$y$ -intercept $c = - ϕ_{o}$
So, the slope of graph and intercept on $x$ -axis shown Planck's constant and threshold frequency.

MHT-CET 2020

Dual nature of radiation and Matter

142205 The light of wavelength ' $λ$ '. Incident on the surface of metal having work function $ϕ$ emits the electrons. The maximum velocity of electrons emitted is
$(c =$ velocity of light, $h =$ Planck's constant, $m =$ mass of electron)

1

{[\frac{2 (hc - λ)}{m λ}]}^{\frac{1}{2}}

2

[\frac{2 (hc - ϕ) λ}{mc}]

3

{[\frac{2 (h c - λ ϕ)}{m λ}]}^{\frac{1}{2}}

4

[\frac{2 (hc - ϕ)}{m λ}]

Explanation:

C According to photo-electric equation,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc}{λ} - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc - ϕ_{o} λ}{λ}$
$v_{m}^{2} = \frac{2 (hc - ϕ_{o} λ)}{λ m}$
$v_{m} = \sqrt{\frac{2 (hc - ϕ_{o} λ)}{λ m}}$

ASSAM CEE-2014

Dual nature of radiation and Matter

142201 Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is

1

3 : 2

2

2 : 3

3

1 : 3

4

1 : 2

Explanation:

C We know that kinetic energy in photoelectric effect,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{1}^{2} = 2 ϕ_{o} - ϕ_{o} = ϕ_{o}$
$\frac{1}{2} {mv}_{2}^{2} = 10 ϕ_{o} - ϕ_{o} = 9 ϕ_{o}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{9}$
$\frac{v_{1}}{v_{2}} = \frac{1}{3}$
Hence, the maximum velocity ratio -
$v_{1} : v_{2} = 1 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142202 Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $λ_{1} = 360 nm$ and $λ_{2} = 600 nm$ . What is the ratio of work functions for lights of wavelength ' $λ_{1}$ ' to ' $λ_{2}$ '?

1

6 : 1

2

3 : 5

3

5 : 3

4

1 : 6

Explanation:

C Given that, $λ_{1} = 360 nm, λ_{2} = 600 nm$ Work function -
$ϕ = \frac{h c}{λ}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{600}{360}$
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{5}{3}$
$ϕ_{1} : ϕ_{2} = 5 : 3$

MHT-CET 2020

Dual nature of radiation and Matter

142203 Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is

1

1 : 4

2

1 : 2

3

1 : \sqrt{2}

4

1 : 1

Explanation:

C Maximum K.E = Incident photon energy work function
$K_{max} = E - W$
According to question-
$\frac{1}{2} {mv}_{1}^{2} = 3 W - W = 2 W$
$\frac{1}{2} {mv}_{2}^{2} = 5 W - W = 4 W$
Dividing equation (i) by equation (ii), we get-
$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$
$\frac{v_{1}}{v_{2}} = \frac{1}{\sqrt{2}}$
$v_{1} : v_{2} = 1 : \sqrt{2}$

MHT-CET 2020

Dual nature of radiation and Matter

142204 The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $X$ axis respectively are

1 Planck's constant, threshold frequency

2 work function, maximum K.E.

3 Planck's constant, work function

4 Maximum K.E, threshold frequency

Explanation:

A
From Einstein's photoelectric equation,
(K.E $)_{max} = h v - ϕ_{o}$
Equation of straight line,
Here $y = mx + c$
Here $m = h$
$y$ -intercept $c = - ϕ_{o}$
So, the slope of graph and intercept on $x$ -axis shown Planck's constant and threshold frequency.

MHT-CET 2020

Dual nature of radiation and Matter

142205 The light of wavelength ' $λ$ '. Incident on the surface of metal having work function $ϕ$ emits the electrons. The maximum velocity of electrons emitted is
$(c =$ velocity of light, $h =$ Planck's constant, $m =$ mass of electron)

1

{[\frac{2 (hc - λ)}{m λ}]}^{\frac{1}{2}}

2

[\frac{2 (hc - ϕ) λ}{mc}]

3

{[\frac{2 (h c - λ ϕ)}{m λ}]}^{\frac{1}{2}}

4

[\frac{2 (hc - ϕ)}{m λ}]

Explanation:

C According to photo-electric equation,
$K_{max} = E - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc}{λ} - ϕ_{o}$
$\frac{1}{2} {mv}_{m}^{2} = \frac{hc - ϕ_{o} λ}{λ}$
$v_{m}^{2} = \frac{2 (hc - ϕ_{o} λ)}{λ m}$
$v_{m} = \sqrt{\frac{2 (hc - ϕ_{o} λ)}{λ m}}$

ASSAM CEE-2014