QBManager

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357661 In photoelectric effect, which of the following property of incident light will not affect the stopping potential

1 Frequency

2 Wavelength

3 Energy

4 Intensity

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357662 When a certain metallic surface is illuminated with monochromatic light of wavelength $λ$ , the stopping potential for photoelectric current is $3 V_{0}$ . When the same surface is illuminated with a light of wavelength $2 λ$ , the stopping potential is $V_{0}$ . The threshold wavelength for this surface for photoelectric effect is

1

4 λ

2

6 λ

3

8 λ

4

\frac{4}{3} λ

Explanation:

According to Einstein's photoelectric equation
$e V_{0} = \frac{h c}{λ} - ϕ_{0}$
where $V_{0}$ is the stopping potential, $λ$ is the wavelength of incident light and $ϕ_{0}$ is the work function of the surface.
As per question
$e (3 V_{0}) = \frac{h c}{λ} - ϕ_{0} (1)$
and $e V_{0} = \frac{h c}{2 λ} - ϕ_{0} (2)$
Multiplying eqn. (2) by 3 , we get
$e (3 V_{0}) = \frac{3 h c}{2 λ} - 3 ϕ_{0} (3)$
Equating eqns. (1) and (3), we get
$\frac{h c}{λ} - ϕ_{0} = \frac{3 h c}{2 λ} - 3 ϕ_{0}$ or $ϕ_{0} = \frac{h c}{4 λ}$
$∴$ The threshold wavelength is
$λ_{0} = \frac{h c}{ϕ_{0}} = \frac{h c}{(h c / 4 λ)} = 4 λ$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357663 The stopping potential of the photoelectrons, from a photo cell is

1 Directly proportional to intensity of incident light

2 Directly proportional to frequency of incident light

3 Inversely proportional to frequency of incident light

4 Inversely proportional to intensity of incident light

Explanation:

The stopping potential of photoelectrons is potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation. It is directly proportional to it.
This is beacuse, more the frequency of incident light, more will be the energy of the photons incident photons. $(E = h v)$
Thus, more will be the energy (kinetic) acquired the electrons
More the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential)

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357664 Light of wavelength $330 n m$ falling on piece of metal ejects electrons with sufficient energy which requires voltage $V_{0}$ to prevent them from reaching a collector. In the same set up, light of wavelength $220 n m$ , ejects electrons which required twice the voltage $V_{0}$ to stop them in reaching a collector. The numerical value of voltage $V_{0}$ is

1

\frac{16}{15} V

2

\frac{15}{16} V

3

\frac{15}{8} V

4

\frac{8}{15} V

Explanation:

Let $W$ be the work function of metal, then $e V_{0} = \frac{h c}{330 \times 10^{- 9}} - W$
and $e (2 V_{0}) = \frac{h c}{220 \times 10^{- 9}} - W$
Solving these two equations, we get
$V_{0} = \frac{10^{9} \times h \times c}{110 \times e \times 6}$
$= \frac{10^{9} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{110 \times 1.6 \times 10^{- 19} \times 6} = \frac{15}{8} V$

AIIMS - 2009

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357665 The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photo-current is recorded. An electric field is switched on which has a vertically downward direction, then

1 the photo-current will increase

2 the kinetic energy of the electrons will increase

3 the stopping potential will decrease

4 the threshold wavelength will increase

Explanation:

In electric field, photoelectron will experience force and accelerate opposite to the field so its $K E$ increases $(i . e .,$ stopping potential will increase), no change in photoelectric current and threshold wavelength.
supporting img

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357661 In photoelectric effect, which of the following property of incident light will not affect the stopping potential

1 Frequency

2 Wavelength

3 Energy

4 Intensity

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357662 When a certain metallic surface is illuminated with monochromatic light of wavelength $λ$ , the stopping potential for photoelectric current is $3 V_{0}$ . When the same surface is illuminated with a light of wavelength $2 λ$ , the stopping potential is $V_{0}$ . The threshold wavelength for this surface for photoelectric effect is

1

4 λ

2

6 λ

3

8 λ

4

\frac{4}{3} λ

Explanation:

According to Einstein's photoelectric equation
$e V_{0} = \frac{h c}{λ} - ϕ_{0}$
where $V_{0}$ is the stopping potential, $λ$ is the wavelength of incident light and $ϕ_{0}$ is the work function of the surface.
As per question
$e (3 V_{0}) = \frac{h c}{λ} - ϕ_{0} (1)$
and $e V_{0} = \frac{h c}{2 λ} - ϕ_{0} (2)$
Multiplying eqn. (2) by 3 , we get
$e (3 V_{0}) = \frac{3 h c}{2 λ} - 3 ϕ_{0} (3)$
Equating eqns. (1) and (3), we get
$\frac{h c}{λ} - ϕ_{0} = \frac{3 h c}{2 λ} - 3 ϕ_{0}$ or $ϕ_{0} = \frac{h c}{4 λ}$
$∴$ The threshold wavelength is
$λ_{0} = \frac{h c}{ϕ_{0}} = \frac{h c}{(h c / 4 λ)} = 4 λ$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357663 The stopping potential of the photoelectrons, from a photo cell is

1 Directly proportional to intensity of incident light

2 Directly proportional to frequency of incident light

3 Inversely proportional to frequency of incident light

4 Inversely proportional to intensity of incident light

Explanation:

The stopping potential of photoelectrons is potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation. It is directly proportional to it.
This is beacuse, more the frequency of incident light, more will be the energy of the photons incident photons. $(E = h v)$
Thus, more will be the energy (kinetic) acquired the electrons
More the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential)

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357664 Light of wavelength $330 n m$ falling on piece of metal ejects electrons with sufficient energy which requires voltage $V_{0}$ to prevent them from reaching a collector. In the same set up, light of wavelength $220 n m$ , ejects electrons which required twice the voltage $V_{0}$ to stop them in reaching a collector. The numerical value of voltage $V_{0}$ is

1

\frac{16}{15} V

2

\frac{15}{16} V

3

\frac{15}{8} V

4

\frac{8}{15} V

Explanation:

Let $W$ be the work function of metal, then $e V_{0} = \frac{h c}{330 \times 10^{- 9}} - W$
and $e (2 V_{0}) = \frac{h c}{220 \times 10^{- 9}} - W$
Solving these two equations, we get
$V_{0} = \frac{10^{9} \times h \times c}{110 \times e \times 6}$
$= \frac{10^{9} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{110 \times 1.6 \times 10^{- 19} \times 6} = \frac{15}{8} V$

AIIMS - 2009

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357665 The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photo-current is recorded. An electric field is switched on which has a vertically downward direction, then

1 the photo-current will increase

2 the kinetic energy of the electrons will increase

3 the stopping potential will decrease

4 the threshold wavelength will increase

Explanation:

In electric field, photoelectron will experience force and accelerate opposite to the field so its $K E$ increases $(i . e .,$ stopping potential will increase), no change in photoelectric current and threshold wavelength.
supporting img

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357661 In photoelectric effect, which of the following property of incident light will not affect the stopping potential

1 Frequency

2 Wavelength

3 Energy

4 Intensity

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357662 When a certain metallic surface is illuminated with monochromatic light of wavelength $λ$ , the stopping potential for photoelectric current is $3 V_{0}$ . When the same surface is illuminated with a light of wavelength $2 λ$ , the stopping potential is $V_{0}$ . The threshold wavelength for this surface for photoelectric effect is

1

4 λ

2

6 λ

3

8 λ

4

\frac{4}{3} λ

Explanation:

According to Einstein's photoelectric equation
$e V_{0} = \frac{h c}{λ} - ϕ_{0}$
where $V_{0}$ is the stopping potential, $λ$ is the wavelength of incident light and $ϕ_{0}$ is the work function of the surface.
As per question
$e (3 V_{0}) = \frac{h c}{λ} - ϕ_{0} (1)$
and $e V_{0} = \frac{h c}{2 λ} - ϕ_{0} (2)$
Multiplying eqn. (2) by 3 , we get
$e (3 V_{0}) = \frac{3 h c}{2 λ} - 3 ϕ_{0} (3)$
Equating eqns. (1) and (3), we get
$\frac{h c}{λ} - ϕ_{0} = \frac{3 h c}{2 λ} - 3 ϕ_{0}$ or $ϕ_{0} = \frac{h c}{4 λ}$
$∴$ The threshold wavelength is
$λ_{0} = \frac{h c}{ϕ_{0}} = \frac{h c}{(h c / 4 λ)} = 4 λ$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357663 The stopping potential of the photoelectrons, from a photo cell is

1 Directly proportional to intensity of incident light

2 Directly proportional to frequency of incident light

3 Inversely proportional to frequency of incident light

4 Inversely proportional to intensity of incident light

Explanation:

The stopping potential of photoelectrons is potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation. It is directly proportional to it.
This is beacuse, more the frequency of incident light, more will be the energy of the photons incident photons. $(E = h v)$
Thus, more will be the energy (kinetic) acquired the electrons
More the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential)

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357664 Light of wavelength $330 n m$ falling on piece of metal ejects electrons with sufficient energy which requires voltage $V_{0}$ to prevent them from reaching a collector. In the same set up, light of wavelength $220 n m$ , ejects electrons which required twice the voltage $V_{0}$ to stop them in reaching a collector. The numerical value of voltage $V_{0}$ is

1

\frac{16}{15} V

2

\frac{15}{16} V

3

\frac{15}{8} V

4

\frac{8}{15} V

Explanation:

Let $W$ be the work function of metal, then $e V_{0} = \frac{h c}{330 \times 10^{- 9}} - W$
and $e (2 V_{0}) = \frac{h c}{220 \times 10^{- 9}} - W$
Solving these two equations, we get
$V_{0} = \frac{10^{9} \times h \times c}{110 \times e \times 6}$
$= \frac{10^{9} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{110 \times 1.6 \times 10^{- 19} \times 6} = \frac{15}{8} V$

AIIMS - 2009

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357665 The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photo-current is recorded. An electric field is switched on which has a vertically downward direction, then

1 the photo-current will increase

2 the kinetic energy of the electrons will increase

3 the stopping potential will decrease

4 the threshold wavelength will increase

Explanation:

In electric field, photoelectron will experience force and accelerate opposite to the field so its $K E$ increases $(i . e .,$ stopping potential will increase), no change in photoelectric current and threshold wavelength.
supporting img

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357661 In photoelectric effect, which of the following property of incident light will not affect the stopping potential

1 Frequency

2 Wavelength

3 Energy

4 Intensity

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357662 When a certain metallic surface is illuminated with monochromatic light of wavelength $λ$ , the stopping potential for photoelectric current is $3 V_{0}$ . When the same surface is illuminated with a light of wavelength $2 λ$ , the stopping potential is $V_{0}$ . The threshold wavelength for this surface for photoelectric effect is

1

4 λ

2

6 λ

3

8 λ

4

\frac{4}{3} λ

Explanation:

According to Einstein's photoelectric equation
$e V_{0} = \frac{h c}{λ} - ϕ_{0}$
where $V_{0}$ is the stopping potential, $λ$ is the wavelength of incident light and $ϕ_{0}$ is the work function of the surface.
As per question
$e (3 V_{0}) = \frac{h c}{λ} - ϕ_{0} (1)$
and $e V_{0} = \frac{h c}{2 λ} - ϕ_{0} (2)$
Multiplying eqn. (2) by 3 , we get
$e (3 V_{0}) = \frac{3 h c}{2 λ} - 3 ϕ_{0} (3)$
Equating eqns. (1) and (3), we get
$\frac{h c}{λ} - ϕ_{0} = \frac{3 h c}{2 λ} - 3 ϕ_{0}$ or $ϕ_{0} = \frac{h c}{4 λ}$
$∴$ The threshold wavelength is
$λ_{0} = \frac{h c}{ϕ_{0}} = \frac{h c}{(h c / 4 λ)} = 4 λ$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357663 The stopping potential of the photoelectrons, from a photo cell is

1 Directly proportional to intensity of incident light

2 Directly proportional to frequency of incident light

3 Inversely proportional to frequency of incident light

4 Inversely proportional to intensity of incident light

Explanation:

The stopping potential of photoelectrons is potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation. It is directly proportional to it.
This is beacuse, more the frequency of incident light, more will be the energy of the photons incident photons. $(E = h v)$
Thus, more will be the energy (kinetic) acquired the electrons
More the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential)

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357664 Light of wavelength $330 n m$ falling on piece of metal ejects electrons with sufficient energy which requires voltage $V_{0}$ to prevent them from reaching a collector. In the same set up, light of wavelength $220 n m$ , ejects electrons which required twice the voltage $V_{0}$ to stop them in reaching a collector. The numerical value of voltage $V_{0}$ is

1

\frac{16}{15} V

2

\frac{15}{16} V

3

\frac{15}{8} V

4

\frac{8}{15} V

Explanation:

Let $W$ be the work function of metal, then $e V_{0} = \frac{h c}{330 \times 10^{- 9}} - W$
and $e (2 V_{0}) = \frac{h c}{220 \times 10^{- 9}} - W$
Solving these two equations, we get
$V_{0} = \frac{10^{9} \times h \times c}{110 \times e \times 6}$
$= \frac{10^{9} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{110 \times 1.6 \times 10^{- 19} \times 6} = \frac{15}{8} V$

AIIMS - 2009

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357665 The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photo-current is recorded. An electric field is switched on which has a vertically downward direction, then

1 the photo-current will increase

2 the kinetic energy of the electrons will increase

3 the stopping potential will decrease

4 the threshold wavelength will increase

Explanation:

In electric field, photoelectron will experience force and accelerate opposite to the field so its $K E$ increases $(i . e .,$ stopping potential will increase), no change in photoelectric current and threshold wavelength.
supporting img

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357661 In photoelectric effect, which of the following property of incident light will not affect the stopping potential

1 Frequency

2 Wavelength

3 Energy

4 Intensity

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357662 When a certain metallic surface is illuminated with monochromatic light of wavelength $λ$ , the stopping potential for photoelectric current is $3 V_{0}$ . When the same surface is illuminated with a light of wavelength $2 λ$ , the stopping potential is $V_{0}$ . The threshold wavelength for this surface for photoelectric effect is

1

4 λ

2

6 λ

3

8 λ

4

\frac{4}{3} λ

Explanation:

According to Einstein's photoelectric equation
$e V_{0} = \frac{h c}{λ} - ϕ_{0}$
where $V_{0}$ is the stopping potential, $λ$ is the wavelength of incident light and $ϕ_{0}$ is the work function of the surface.
As per question
$e (3 V_{0}) = \frac{h c}{λ} - ϕ_{0} (1)$
and $e V_{0} = \frac{h c}{2 λ} - ϕ_{0} (2)$
Multiplying eqn. (2) by 3 , we get
$e (3 V_{0}) = \frac{3 h c}{2 λ} - 3 ϕ_{0} (3)$
Equating eqns. (1) and (3), we get
$\frac{h c}{λ} - ϕ_{0} = \frac{3 h c}{2 λ} - 3 ϕ_{0}$ or $ϕ_{0} = \frac{h c}{4 λ}$
$∴$ The threshold wavelength is
$λ_{0} = \frac{h c}{ϕ_{0}} = \frac{h c}{(h c / 4 λ)} = 4 λ$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357663 The stopping potential of the photoelectrons, from a photo cell is

1 Directly proportional to intensity of incident light

2 Directly proportional to frequency of incident light

3 Inversely proportional to frequency of incident light

4 Inversely proportional to intensity of incident light

Explanation:

The stopping potential of photoelectrons is potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation. It is directly proportional to it.
This is beacuse, more the frequency of incident light, more will be the energy of the photons incident photons. $(E = h v)$
Thus, more will be the energy (kinetic) acquired the electrons
More the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential)

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357664 Light of wavelength $330 n m$ falling on piece of metal ejects electrons with sufficient energy which requires voltage $V_{0}$ to prevent them from reaching a collector. In the same set up, light of wavelength $220 n m$ , ejects electrons which required twice the voltage $V_{0}$ to stop them in reaching a collector. The numerical value of voltage $V_{0}$ is

1

\frac{16}{15} V

2

\frac{15}{16} V

3

\frac{15}{8} V

4

\frac{8}{15} V

Explanation:

Let $W$ be the work function of metal, then $e V_{0} = \frac{h c}{330 \times 10^{- 9}} - W$
and $e (2 V_{0}) = \frac{h c}{220 \times 10^{- 9}} - W$
Solving these two equations, we get
$V_{0} = \frac{10^{9} \times h \times c}{110 \times e \times 6}$
$= \frac{10^{9} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{110 \times 1.6 \times 10^{- 19} \times 6} = \frac{15}{8} V$

AIIMS - 2009

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357665 The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photo-current is recorded. An electric field is switched on which has a vertically downward direction, then

1 the photo-current will increase

2 the kinetic energy of the electrons will increase

3 the stopping potential will decrease

4 the threshold wavelength will increase

Explanation:

In electric field, photoelectron will experience force and accelerate opposite to the field so its $K E$ increases $(i . e .,$ stopping potential will increase), no change in photoelectric current and threshold wavelength.
supporting img