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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357631 Light rays of wavelengths $6000 A^{\circ}$ and of photon intensity $39.6 w a t t s / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be
[Planks constant $= 6.64 \times 10^{- 34} J - s$ ; Velocity of light $= 3 \times 10^{8} m s^{- 1}$ ]

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{15}

Explanation:

$n_{e} = 10^{- 6} \times \frac{I λ}{h c}$
$\begin{aligned} = \frac{10^{- 6} \times 39.6 \times 6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} \\ = 12 \times 10^{13} . \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357632 A metal of work function $4 e V$ is exposed to a radiation of wavelength $140 \times 10^{- 9} m$ . Then the stopping potential required is $(h = 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8} m / s)$

1

6.42 V

2

2.94 V

3

4.86 V

4

3.2 V

Explanation:

$V_{0} = \frac{[\frac{12400}{λ i n \overset{^{\circ}}{A}} - ϕ_{0}]}{e}$
$\Rightarrow V_{0} = [\frac{12400}{1400} - 4]$
$= (\frac{124}{14} - 4) = \frac{124 - 56}{14}$
$= \frac{68}{14} = \frac{34}{7} = 4.86 V o l t s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357633 The variation of photocurrent with collector potential for different frequencies of incident radiation. $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph, then
supporting img

1

v_{1} = v_{2} = v_{3}

2

v_{1} > v_{2} > v_{3}

3

v_{1} < v_{2} < v_{3}

4

v_{3} = \frac{v_{1} + v_{2}}{2}

Explanation:

$A s e V_{s} = K_{m a x} = h v - ϕ_{0} (1)$
Saturation current is same. It means each source of radiation has same intensity. But corresponding stopping potential is different so they have different frequencies.
So, from eqn. (1) and given graph we conclude that $v_{1} < v_{2} < v_{3} (∵ | V_{01} | < | V_{02} | < | V_{03} |)$

KCET - 2016

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357634 Sodium surface is illuminated with ultraviolet light and visible radiation successively and the stopping potentials are determined. Then the potential

1 Is equal in both the cases

2 Greater for ultraviolet light

3 More for visible light

4 Varies randomly

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357635 Silver has a work function of $4.7 e V$ . When ultraviolet light of wavelength $100 n m$ is incident on it a potential of $7.7 V$ is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when light of wavelength $200 n m$ is incident on it?

1

15.4 V

2

2.35 V

3

3.85 V

4

1.5 V

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 incident wavelength and $ϕ_{0}$ is the work function of the metal.
or $\frac{h c}{λ} = e V_{0} + ϕ_{0}$
For wavelengths $λ_{1}$ and $λ_{2}$ ,
$\frac{h c}{λ_{1}} = e V_{01} + ϕ_{0} (1)$
and $\frac{h c}{λ_{2}} = e V_{02} + ϕ_{0} (2)$
Dividing eqn. (1) by eqn. (2), we get
$\frac{λ_{2}}{λ_{1}} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
Here, $λ_{1} = 100 n m, λ_{2} = 200 n m$
$∴ \frac{200 n m}{100 n m} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
$2 = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}} \Rightarrow e V_{02} = \frac{e V_{01} - ϕ_{0}}{2}$
As $V_{01} = 7.7 V$ and $ϕ_{0} = 4.7 e V$ (given)
$∴ e V_{02} = \frac{e (7.7 V) - 4.7 e V}{2}$
$= \frac{7.7 e V - 4.7 e V}{2} = \frac{3 e V}{2} = 1.5 e V$
$V_{02} = 1.5 V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357631 Light rays of wavelengths $6000 A^{\circ}$ and of photon intensity $39.6 w a t t s / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be
[Planks constant $= 6.64 \times 10^{- 34} J - s$ ; Velocity of light $= 3 \times 10^{8} m s^{- 1}$ ]

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{15}

Explanation:

$n_{e} = 10^{- 6} \times \frac{I λ}{h c}$
$\begin{aligned} = \frac{10^{- 6} \times 39.6 \times 6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} \\ = 12 \times 10^{13} . \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357632 A metal of work function $4 e V$ is exposed to a radiation of wavelength $140 \times 10^{- 9} m$ . Then the stopping potential required is $(h = 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8} m / s)$

1

6.42 V

2

2.94 V

3

4.86 V

4

3.2 V

Explanation:

$V_{0} = \frac{[\frac{12400}{λ i n \overset{^{\circ}}{A}} - ϕ_{0}]}{e}$
$\Rightarrow V_{0} = [\frac{12400}{1400} - 4]$
$= (\frac{124}{14} - 4) = \frac{124 - 56}{14}$
$= \frac{68}{14} = \frac{34}{7} = 4.86 V o l t s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357633 The variation of photocurrent with collector potential for different frequencies of incident radiation. $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph, then
supporting img

1

v_{1} = v_{2} = v_{3}

2

v_{1} > v_{2} > v_{3}

3

v_{1} < v_{2} < v_{3}

4

v_{3} = \frac{v_{1} + v_{2}}{2}

Explanation:

$A s e V_{s} = K_{m a x} = h v - ϕ_{0} (1)$
Saturation current is same. It means each source of radiation has same intensity. But corresponding stopping potential is different so they have different frequencies.
So, from eqn. (1) and given graph we conclude that $v_{1} < v_{2} < v_{3} (∵ | V_{01} | < | V_{02} | < | V_{03} |)$

KCET - 2016

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357634 Sodium surface is illuminated with ultraviolet light and visible radiation successively and the stopping potentials are determined. Then the potential

1 Is equal in both the cases

2 Greater for ultraviolet light

3 More for visible light

4 Varies randomly

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357635 Silver has a work function of $4.7 e V$ . When ultraviolet light of wavelength $100 n m$ is incident on it a potential of $7.7 V$ is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when light of wavelength $200 n m$ is incident on it?

1

15.4 V

2

2.35 V

3

3.85 V

4

1.5 V

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 incident wavelength and $ϕ_{0}$ is the work function of the metal.
or $\frac{h c}{λ} = e V_{0} + ϕ_{0}$
For wavelengths $λ_{1}$ and $λ_{2}$ ,
$\frac{h c}{λ_{1}} = e V_{01} + ϕ_{0} (1)$
and $\frac{h c}{λ_{2}} = e V_{02} + ϕ_{0} (2)$
Dividing eqn. (1) by eqn. (2), we get
$\frac{λ_{2}}{λ_{1}} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
Here, $λ_{1} = 100 n m, λ_{2} = 200 n m$
$∴ \frac{200 n m}{100 n m} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
$2 = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}} \Rightarrow e V_{02} = \frac{e V_{01} - ϕ_{0}}{2}$
As $V_{01} = 7.7 V$ and $ϕ_{0} = 4.7 e V$ (given)
$∴ e V_{02} = \frac{e (7.7 V) - 4.7 e V}{2}$
$= \frac{7.7 e V - 4.7 e V}{2} = \frac{3 e V}{2} = 1.5 e V$
$V_{02} = 1.5 V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357631 Light rays of wavelengths $6000 A^{\circ}$ and of photon intensity $39.6 w a t t s / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be
[Planks constant $= 6.64 \times 10^{- 34} J - s$ ; Velocity of light $= 3 \times 10^{8} m s^{- 1}$ ]

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{15}

Explanation:

$n_{e} = 10^{- 6} \times \frac{I λ}{h c}$
$\begin{aligned} = \frac{10^{- 6} \times 39.6 \times 6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} \\ = 12 \times 10^{13} . \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357632 A metal of work function $4 e V$ is exposed to a radiation of wavelength $140 \times 10^{- 9} m$ . Then the stopping potential required is $(h = 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8} m / s)$

1

6.42 V

2

2.94 V

3

4.86 V

4

3.2 V

Explanation:

$V_{0} = \frac{[\frac{12400}{λ i n \overset{^{\circ}}{A}} - ϕ_{0}]}{e}$
$\Rightarrow V_{0} = [\frac{12400}{1400} - 4]$
$= (\frac{124}{14} - 4) = \frac{124 - 56}{14}$
$= \frac{68}{14} = \frac{34}{7} = 4.86 V o l t s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357633 The variation of photocurrent with collector potential for different frequencies of incident radiation. $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph, then
supporting img

1

v_{1} = v_{2} = v_{3}

2

v_{1} > v_{2} > v_{3}

3

v_{1} < v_{2} < v_{3}

4

v_{3} = \frac{v_{1} + v_{2}}{2}

Explanation:

$A s e V_{s} = K_{m a x} = h v - ϕ_{0} (1)$
Saturation current is same. It means each source of radiation has same intensity. But corresponding stopping potential is different so they have different frequencies.
So, from eqn. (1) and given graph we conclude that $v_{1} < v_{2} < v_{3} (∵ | V_{01} | < | V_{02} | < | V_{03} |)$

KCET - 2016

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357634 Sodium surface is illuminated with ultraviolet light and visible radiation successively and the stopping potentials are determined. Then the potential

1 Is equal in both the cases

2 Greater for ultraviolet light

3 More for visible light

4 Varies randomly

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357635 Silver has a work function of $4.7 e V$ . When ultraviolet light of wavelength $100 n m$ is incident on it a potential of $7.7 V$ is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when light of wavelength $200 n m$ is incident on it?

1

15.4 V

2

2.35 V

3

3.85 V

4

1.5 V

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 incident wavelength and $ϕ_{0}$ is the work function of the metal.
or $\frac{h c}{λ} = e V_{0} + ϕ_{0}$
For wavelengths $λ_{1}$ and $λ_{2}$ ,
$\frac{h c}{λ_{1}} = e V_{01} + ϕ_{0} (1)$
and $\frac{h c}{λ_{2}} = e V_{02} + ϕ_{0} (2)$
Dividing eqn. (1) by eqn. (2), we get
$\frac{λ_{2}}{λ_{1}} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
Here, $λ_{1} = 100 n m, λ_{2} = 200 n m$
$∴ \frac{200 n m}{100 n m} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
$2 = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}} \Rightarrow e V_{02} = \frac{e V_{01} - ϕ_{0}}{2}$
As $V_{01} = 7.7 V$ and $ϕ_{0} = 4.7 e V$ (given)
$∴ e V_{02} = \frac{e (7.7 V) - 4.7 e V}{2}$
$= \frac{7.7 e V - 4.7 e V}{2} = \frac{3 e V}{2} = 1.5 e V$
$V_{02} = 1.5 V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357631 Light rays of wavelengths $6000 A^{\circ}$ and of photon intensity $39.6 w a t t s / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be
[Planks constant $= 6.64 \times 10^{- 34} J - s$ ; Velocity of light $= 3 \times 10^{8} m s^{- 1}$ ]

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{15}

Explanation:

$n_{e} = 10^{- 6} \times \frac{I λ}{h c}$
$\begin{aligned} = \frac{10^{- 6} \times 39.6 \times 6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} \\ = 12 \times 10^{13} . \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357632 A metal of work function $4 e V$ is exposed to a radiation of wavelength $140 \times 10^{- 9} m$ . Then the stopping potential required is $(h = 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8} m / s)$

1

6.42 V

2

2.94 V

3

4.86 V

4

3.2 V

Explanation:

$V_{0} = \frac{[\frac{12400}{λ i n \overset{^{\circ}}{A}} - ϕ_{0}]}{e}$
$\Rightarrow V_{0} = [\frac{12400}{1400} - 4]$
$= (\frac{124}{14} - 4) = \frac{124 - 56}{14}$
$= \frac{68}{14} = \frac{34}{7} = 4.86 V o l t s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357633 The variation of photocurrent with collector potential for different frequencies of incident radiation. $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph, then
supporting img

1

v_{1} = v_{2} = v_{3}

2

v_{1} > v_{2} > v_{3}

3

v_{1} < v_{2} < v_{3}

4

v_{3} = \frac{v_{1} + v_{2}}{2}

Explanation:

$A s e V_{s} = K_{m a x} = h v - ϕ_{0} (1)$
Saturation current is same. It means each source of radiation has same intensity. But corresponding stopping potential is different so they have different frequencies.
So, from eqn. (1) and given graph we conclude that $v_{1} < v_{2} < v_{3} (∵ | V_{01} | < | V_{02} | < | V_{03} |)$

KCET - 2016

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357634 Sodium surface is illuminated with ultraviolet light and visible radiation successively and the stopping potentials are determined. Then the potential

1 Is equal in both the cases

2 Greater for ultraviolet light

3 More for visible light

4 Varies randomly

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357635 Silver has a work function of $4.7 e V$ . When ultraviolet light of wavelength $100 n m$ is incident on it a potential of $7.7 V$ is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when light of wavelength $200 n m$ is incident on it?

1

15.4 V

2

2.35 V

3

3.85 V

4

1.5 V

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 incident wavelength and $ϕ_{0}$ is the work function of the metal.
or $\frac{h c}{λ} = e V_{0} + ϕ_{0}$
For wavelengths $λ_{1}$ and $λ_{2}$ ,
$\frac{h c}{λ_{1}} = e V_{01} + ϕ_{0} (1)$
and $\frac{h c}{λ_{2}} = e V_{02} + ϕ_{0} (2)$
Dividing eqn. (1) by eqn. (2), we get
$\frac{λ_{2}}{λ_{1}} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
Here, $λ_{1} = 100 n m, λ_{2} = 200 n m$
$∴ \frac{200 n m}{100 n m} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
$2 = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}} \Rightarrow e V_{02} = \frac{e V_{01} - ϕ_{0}}{2}$
As $V_{01} = 7.7 V$ and $ϕ_{0} = 4.7 e V$ (given)
$∴ e V_{02} = \frac{e (7.7 V) - 4.7 e V}{2}$
$= \frac{7.7 e V - 4.7 e V}{2} = \frac{3 e V}{2} = 1.5 e V$
$V_{02} = 1.5 V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357631 Light rays of wavelengths $6000 A^{\circ}$ and of photon intensity $39.6 w a t t s / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be
[Planks constant $= 6.64 \times 10^{- 34} J - s$ ; Velocity of light $= 3 \times 10^{8} m s^{- 1}$ ]

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{15}

Explanation:

$n_{e} = 10^{- 6} \times \frac{I λ}{h c}$
$\begin{aligned} = \frac{10^{- 6} \times 39.6 \times 6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} \\ = 12 \times 10^{13} . \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357632 A metal of work function $4 e V$ is exposed to a radiation of wavelength $140 \times 10^{- 9} m$ . Then the stopping potential required is $(h = 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8} m / s)$

1

6.42 V

2

2.94 V

3

4.86 V

4

3.2 V

Explanation:

$V_{0} = \frac{[\frac{12400}{λ i n \overset{^{\circ}}{A}} - ϕ_{0}]}{e}$
$\Rightarrow V_{0} = [\frac{12400}{1400} - 4]$
$= (\frac{124}{14} - 4) = \frac{124 - 56}{14}$
$= \frac{68}{14} = \frac{34}{7} = 4.86 V o l t s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357633 The variation of photocurrent with collector potential for different frequencies of incident radiation. $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph, then
supporting img

1

v_{1} = v_{2} = v_{3}

2

v_{1} > v_{2} > v_{3}

3

v_{1} < v_{2} < v_{3}

4

v_{3} = \frac{v_{1} + v_{2}}{2}

Explanation:

$A s e V_{s} = K_{m a x} = h v - ϕ_{0} (1)$
Saturation current is same. It means each source of radiation has same intensity. But corresponding stopping potential is different so they have different frequencies.
So, from eqn. (1) and given graph we conclude that $v_{1} < v_{2} < v_{3} (∵ | V_{01} | < | V_{02} | < | V_{03} |)$

KCET - 2016

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357634 Sodium surface is illuminated with ultraviolet light and visible radiation successively and the stopping potentials are determined. Then the potential

1 Is equal in both the cases

2 Greater for ultraviolet light

3 More for visible light

4 Varies randomly

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357635 Silver has a work function of $4.7 e V$ . When ultraviolet light of wavelength $100 n m$ is incident on it a potential of $7.7 V$ is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when light of wavelength $200 n m$ is incident on it?

1

15.4 V

2

2.35 V

3

3.85 V

4

1.5 V

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 incident wavelength and $ϕ_{0}$ is the work function of the metal.
or $\frac{h c}{λ} = e V_{0} + ϕ_{0}$
For wavelengths $λ_{1}$ and $λ_{2}$ ,
$\frac{h c}{λ_{1}} = e V_{01} + ϕ_{0} (1)$
and $\frac{h c}{λ_{2}} = e V_{02} + ϕ_{0} (2)$
Dividing eqn. (1) by eqn. (2), we get
$\frac{λ_{2}}{λ_{1}} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
Here, $λ_{1} = 100 n m, λ_{2} = 200 n m$
$∴ \frac{200 n m}{100 n m} = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}}$
$2 = \frac{e V_{01} + ϕ_{0}}{e V_{02} + ϕ_{0}} \Rightarrow e V_{02} = \frac{e V_{01} - ϕ_{0}}{2}$
As $V_{01} = 7.7 V$ and $ϕ_{0} = 4.7 e V$ (given)
$∴ e V_{02} = \frac{e (7.7 V) - 4.7 e V}{2}$
$= \frac{7.7 e V - 4.7 e V}{2} = \frac{3 e V}{2} = 1.5 e V$
$V_{02} = 1.5 V$