QBManager

Dual nature of radiation and Matter

142099 In photoelectric effect, initially when energy of electrons emitted is $E_{0}$ , de-Broglie wavelength associated with them is $λ_{0}$ . Now, energy is doubled then associated de- Broglie wavelength $λ^{'}$ is

1

λ^{'} = \frac{λ_{o}}{\sqrt{2}}

2

λ^{'} = \sqrt{2 λ_{0}}

3

λ^{'} = λ_{o}

4

λ^{'} = \frac{λ_{o}}{2}

Explanation:

A Let, $E$ be the energy and $p$ is momentum of electron,
$E = \frac{p^{2}}{2 m}$
And momentum $(p) = \sqrt{2 mE}$
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
Putting the value of $p$ from equation (i) in equation (ii), we get -
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ^{'}}{λ_{o}} = \sqrt{\frac{E}{E^{'}}}$
$λ^{'} = λ_{o} \sqrt{\frac{E}{2 E}}$
$λ^{'} = \frac{λ_{o}}{\sqrt{2}}$

VITEEE-2016

Dual nature of radiation and Matter

142101 When the momentum of a photon is changed by an amount $p^{'}$ then the corresponding change in the de-Broglie wavelength is found to be $0.20 %$ Then, the original momentum of the photon was

1

300 p^{'}

2

500 p^{'}

3

400 p^{'}

4

100 p^{'}

Explanation:

B Given that,
$\frac{Δ λ}{λ} = 0.2 % = \frac{0.2}{100} = \frac{1}{500}$
We know that,
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
$λ p = h$
$\frac{Δ p}{p} + \frac{Δ λ}{λ} = 0$
$| \frac{Δ p}{p} | = | \frac{- Δ λ}{λ} |$
$\frac{p^{'}}{p} = \frac{1}{500}$
$p = 500 p^{'}$

VITEEE-2015

Dual nature of radiation and Matter

142103 The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

1 double the earlier value

2 unchanged

3 more than doubled

4 less than doubled

Explanation:

C Given that, $v^{'} = 2 v$
We know that,
$Misplaced &$
$Misplaced &$
$= 2 h v - 2 ϕ_{o} + ϕ_{o}$
$= 2 (h v - ϕ_{o}) + ϕ_{o}$
From equation (i) and equation (ii), we get -
$K^{'} = 2 K + ϕ_{o}$
The kinetic energy of the photoelectrons will be more than double when incident frequency is double.

VITEEE-2011

Dual nature of radiation and Matter

142104 The work function of a certain metal is $3.31 \times 10^{- 19} J$ . Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \AA$ is (Given, $h = 6.62 \times 10^{- 34} J$ $s, c = 3 \times 10^{8} {ms}^{- 1}, e = 1.6 \times 10^{- 19} C$ )

1

2.48 eV

2

0.41 eV

3

2.07 eV

4

0.82 eV

Explanation:

B Given that ,
$ϕ_{o} = 3.31 \times 10^{- 19} J$
$λ = 5000 \AA = 5 \times 10^{- 7} m$
$h = 6.62 \times 10^{- 34} J - S$
$c = 3 \times 10^{8} m / s$
$e = 1.6 \times 10^{- 19} C$
According to the Einstein's photoelectric equation -
$(KE)_{max} = E - ϕ_{o}$
$(KE)_{max} = \frac{hc}{λ} - ϕ_{o}$
$(KE)_{max} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{5 \times 10^{- 7}} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 3.972 \times 10^{- 19} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 0.662 \times 10^{- 19} J$
$(KE)_{max} = \frac{0.662 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV$
$(KE)_{max} = 0.41 eV$

VITEEE-2009 - AP EAMCET -2009

Dual nature of radiation and Matter

142105 The variation of photocurrent with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is shown in the graph, then:

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:

C Form the given graph, the magnitude of stopping potential,
$V_{0_{1}} < V_{0_{2}} < V_{0_{3}}$
Stopping potential is directly proportional to the incident frequency i.e. smaller the incident frequency the smaller the stopping potential.
Therefore, $v_{1} < v_{2} < v_{3}$

Karnataka CET-2016

Dual nature of radiation and Matter

142099 In photoelectric effect, initially when energy of electrons emitted is $E_{0}$ , de-Broglie wavelength associated with them is $λ_{0}$ . Now, energy is doubled then associated de- Broglie wavelength $λ^{'}$ is

1

λ^{'} = \frac{λ_{o}}{\sqrt{2}}

2

λ^{'} = \sqrt{2 λ_{0}}

3

λ^{'} = λ_{o}

4

λ^{'} = \frac{λ_{o}}{2}

Explanation:

A Let, $E$ be the energy and $p$ is momentum of electron,
$E = \frac{p^{2}}{2 m}$
And momentum $(p) = \sqrt{2 mE}$
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
Putting the value of $p$ from equation (i) in equation (ii), we get -
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ^{'}}{λ_{o}} = \sqrt{\frac{E}{E^{'}}}$
$λ^{'} = λ_{o} \sqrt{\frac{E}{2 E}}$
$λ^{'} = \frac{λ_{o}}{\sqrt{2}}$

VITEEE-2016

Dual nature of radiation and Matter

142101 When the momentum of a photon is changed by an amount $p^{'}$ then the corresponding change in the de-Broglie wavelength is found to be $0.20 %$ Then, the original momentum of the photon was

1

300 p^{'}

2

500 p^{'}

3

400 p^{'}

4

100 p^{'}

Explanation:

B Given that,
$\frac{Δ λ}{λ} = 0.2 % = \frac{0.2}{100} = \frac{1}{500}$
We know that,
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
$λ p = h$
$\frac{Δ p}{p} + \frac{Δ λ}{λ} = 0$
$| \frac{Δ p}{p} | = | \frac{- Δ λ}{λ} |$
$\frac{p^{'}}{p} = \frac{1}{500}$
$p = 500 p^{'}$

VITEEE-2015

Dual nature of radiation and Matter

142103 The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

1 double the earlier value

2 unchanged

3 more than doubled

4 less than doubled

Explanation:

C Given that, $v^{'} = 2 v$
We know that,
$Misplaced &$
$Misplaced &$
$= 2 h v - 2 ϕ_{o} + ϕ_{o}$
$= 2 (h v - ϕ_{o}) + ϕ_{o}$
From equation (i) and equation (ii), we get -
$K^{'} = 2 K + ϕ_{o}$
The kinetic energy of the photoelectrons will be more than double when incident frequency is double.

VITEEE-2011

Dual nature of radiation and Matter

142104 The work function of a certain metal is $3.31 \times 10^{- 19} J$ . Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \AA$ is (Given, $h = 6.62 \times 10^{- 34} J$ $s, c = 3 \times 10^{8} {ms}^{- 1}, e = 1.6 \times 10^{- 19} C$ )

1

2.48 eV

2

0.41 eV

3

2.07 eV

4

0.82 eV

Explanation:

B Given that ,
$ϕ_{o} = 3.31 \times 10^{- 19} J$
$λ = 5000 \AA = 5 \times 10^{- 7} m$
$h = 6.62 \times 10^{- 34} J - S$
$c = 3 \times 10^{8} m / s$
$e = 1.6 \times 10^{- 19} C$
According to the Einstein's photoelectric equation -
$(KE)_{max} = E - ϕ_{o}$
$(KE)_{max} = \frac{hc}{λ} - ϕ_{o}$
$(KE)_{max} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{5 \times 10^{- 7}} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 3.972 \times 10^{- 19} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 0.662 \times 10^{- 19} J$
$(KE)_{max} = \frac{0.662 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV$
$(KE)_{max} = 0.41 eV$

VITEEE-2009 - AP EAMCET -2009

Dual nature of radiation and Matter

142105 The variation of photocurrent with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is shown in the graph, then:

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:

C Form the given graph, the magnitude of stopping potential,
$V_{0_{1}} < V_{0_{2}} < V_{0_{3}}$
Stopping potential is directly proportional to the incident frequency i.e. smaller the incident frequency the smaller the stopping potential.
Therefore, $v_{1} < v_{2} < v_{3}$

Karnataka CET-2016

Dual nature of radiation and Matter

142099 In photoelectric effect, initially when energy of electrons emitted is $E_{0}$ , de-Broglie wavelength associated with them is $λ_{0}$ . Now, energy is doubled then associated de- Broglie wavelength $λ^{'}$ is

1

λ^{'} = \frac{λ_{o}}{\sqrt{2}}

2

λ^{'} = \sqrt{2 λ_{0}}

3

λ^{'} = λ_{o}

4

λ^{'} = \frac{λ_{o}}{2}

Explanation:

A Let, $E$ be the energy and $p$ is momentum of electron,
$E = \frac{p^{2}}{2 m}$
And momentum $(p) = \sqrt{2 mE}$
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
Putting the value of $p$ from equation (i) in equation (ii), we get -
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ^{'}}{λ_{o}} = \sqrt{\frac{E}{E^{'}}}$
$λ^{'} = λ_{o} \sqrt{\frac{E}{2 E}}$
$λ^{'} = \frac{λ_{o}}{\sqrt{2}}$

VITEEE-2016

Dual nature of radiation and Matter

142101 When the momentum of a photon is changed by an amount $p^{'}$ then the corresponding change in the de-Broglie wavelength is found to be $0.20 %$ Then, the original momentum of the photon was

1

300 p^{'}

2

500 p^{'}

3

400 p^{'}

4

100 p^{'}

Explanation:

B Given that,
$\frac{Δ λ}{λ} = 0.2 % = \frac{0.2}{100} = \frac{1}{500}$
We know that,
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
$λ p = h$
$\frac{Δ p}{p} + \frac{Δ λ}{λ} = 0$
$| \frac{Δ p}{p} | = | \frac{- Δ λ}{λ} |$
$\frac{p^{'}}{p} = \frac{1}{500}$
$p = 500 p^{'}$

VITEEE-2015

Dual nature of radiation and Matter

142103 The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

1 double the earlier value

2 unchanged

3 more than doubled

4 less than doubled

Explanation:

C Given that, $v^{'} = 2 v$
We know that,
$Misplaced &$
$Misplaced &$
$= 2 h v - 2 ϕ_{o} + ϕ_{o}$
$= 2 (h v - ϕ_{o}) + ϕ_{o}$
From equation (i) and equation (ii), we get -
$K^{'} = 2 K + ϕ_{o}$
The kinetic energy of the photoelectrons will be more than double when incident frequency is double.

VITEEE-2011

Dual nature of radiation and Matter

142104 The work function of a certain metal is $3.31 \times 10^{- 19} J$ . Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \AA$ is (Given, $h = 6.62 \times 10^{- 34} J$ $s, c = 3 \times 10^{8} {ms}^{- 1}, e = 1.6 \times 10^{- 19} C$ )

1

2.48 eV

2

0.41 eV

3

2.07 eV

4

0.82 eV

Explanation:

B Given that ,
$ϕ_{o} = 3.31 \times 10^{- 19} J$
$λ = 5000 \AA = 5 \times 10^{- 7} m$
$h = 6.62 \times 10^{- 34} J - S$
$c = 3 \times 10^{8} m / s$
$e = 1.6 \times 10^{- 19} C$
According to the Einstein's photoelectric equation -
$(KE)_{max} = E - ϕ_{o}$
$(KE)_{max} = \frac{hc}{λ} - ϕ_{o}$
$(KE)_{max} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{5 \times 10^{- 7}} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 3.972 \times 10^{- 19} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 0.662 \times 10^{- 19} J$
$(KE)_{max} = \frac{0.662 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV$
$(KE)_{max} = 0.41 eV$

VITEEE-2009 - AP EAMCET -2009

Dual nature of radiation and Matter

142105 The variation of photocurrent with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is shown in the graph, then:

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:

C Form the given graph, the magnitude of stopping potential,
$V_{0_{1}} < V_{0_{2}} < V_{0_{3}}$
Stopping potential is directly proportional to the incident frequency i.e. smaller the incident frequency the smaller the stopping potential.
Therefore, $v_{1} < v_{2} < v_{3}$

Karnataka CET-2016

Dual nature of radiation and Matter

142099 In photoelectric effect, initially when energy of electrons emitted is $E_{0}$ , de-Broglie wavelength associated with them is $λ_{0}$ . Now, energy is doubled then associated de- Broglie wavelength $λ^{'}$ is

1

λ^{'} = \frac{λ_{o}}{\sqrt{2}}

2

λ^{'} = \sqrt{2 λ_{0}}

3

λ^{'} = λ_{o}

4

λ^{'} = \frac{λ_{o}}{2}

Explanation:

A Let, $E$ be the energy and $p$ is momentum of electron,
$E = \frac{p^{2}}{2 m}$
And momentum $(p) = \sqrt{2 mE}$
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
Putting the value of $p$ from equation (i) in equation (ii), we get -
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ^{'}}{λ_{o}} = \sqrt{\frac{E}{E^{'}}}$
$λ^{'} = λ_{o} \sqrt{\frac{E}{2 E}}$
$λ^{'} = \frac{λ_{o}}{\sqrt{2}}$

VITEEE-2016

Dual nature of radiation and Matter

142101 When the momentum of a photon is changed by an amount $p^{'}$ then the corresponding change in the de-Broglie wavelength is found to be $0.20 %$ Then, the original momentum of the photon was

1

300 p^{'}

2

500 p^{'}

3

400 p^{'}

4

100 p^{'}

Explanation:

B Given that,
$\frac{Δ λ}{λ} = 0.2 % = \frac{0.2}{100} = \frac{1}{500}$
We know that,
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
$λ p = h$
$\frac{Δ p}{p} + \frac{Δ λ}{λ} = 0$
$| \frac{Δ p}{p} | = | \frac{- Δ λ}{λ} |$
$\frac{p^{'}}{p} = \frac{1}{500}$
$p = 500 p^{'}$

VITEEE-2015

Dual nature of radiation and Matter

142103 The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

1 double the earlier value

2 unchanged

3 more than doubled

4 less than doubled

Explanation:

C Given that, $v^{'} = 2 v$
We know that,
$Misplaced &$
$Misplaced &$
$= 2 h v - 2 ϕ_{o} + ϕ_{o}$
$= 2 (h v - ϕ_{o}) + ϕ_{o}$
From equation (i) and equation (ii), we get -
$K^{'} = 2 K + ϕ_{o}$
The kinetic energy of the photoelectrons will be more than double when incident frequency is double.

VITEEE-2011

Dual nature of radiation and Matter

142104 The work function of a certain metal is $3.31 \times 10^{- 19} J$ . Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \AA$ is (Given, $h = 6.62 \times 10^{- 34} J$ $s, c = 3 \times 10^{8} {ms}^{- 1}, e = 1.6 \times 10^{- 19} C$ )

1

2.48 eV

2

0.41 eV

3

2.07 eV

4

0.82 eV

Explanation:

B Given that ,
$ϕ_{o} = 3.31 \times 10^{- 19} J$
$λ = 5000 \AA = 5 \times 10^{- 7} m$
$h = 6.62 \times 10^{- 34} J - S$
$c = 3 \times 10^{8} m / s$
$e = 1.6 \times 10^{- 19} C$
According to the Einstein's photoelectric equation -
$(KE)_{max} = E - ϕ_{o}$
$(KE)_{max} = \frac{hc}{λ} - ϕ_{o}$
$(KE)_{max} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{5 \times 10^{- 7}} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 3.972 \times 10^{- 19} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 0.662 \times 10^{- 19} J$
$(KE)_{max} = \frac{0.662 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV$
$(KE)_{max} = 0.41 eV$

VITEEE-2009 - AP EAMCET -2009

Dual nature of radiation and Matter

142105 The variation of photocurrent with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is shown in the graph, then:

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:

C Form the given graph, the magnitude of stopping potential,
$V_{0_{1}} < V_{0_{2}} < V_{0_{3}}$
Stopping potential is directly proportional to the incident frequency i.e. smaller the incident frequency the smaller the stopping potential.
Therefore, $v_{1} < v_{2} < v_{3}$

Karnataka CET-2016

Dual nature of radiation and Matter

142099 In photoelectric effect, initially when energy of electrons emitted is $E_{0}$ , de-Broglie wavelength associated with them is $λ_{0}$ . Now, energy is doubled then associated de- Broglie wavelength $λ^{'}$ is

1

λ^{'} = \frac{λ_{o}}{\sqrt{2}}

2

λ^{'} = \sqrt{2 λ_{0}}

3

λ^{'} = λ_{o}

4

λ^{'} = \frac{λ_{o}}{2}

Explanation:

A Let, $E$ be the energy and $p$ is momentum of electron,
$E = \frac{p^{2}}{2 m}$
And momentum $(p) = \sqrt{2 mE}$
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
Putting the value of $p$ from equation (i) in equation (ii), we get -
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ^{'}}{λ_{o}} = \sqrt{\frac{E}{E^{'}}}$
$λ^{'} = λ_{o} \sqrt{\frac{E}{2 E}}$
$λ^{'} = \frac{λ_{o}}{\sqrt{2}}$

VITEEE-2016

Dual nature of radiation and Matter

142101 When the momentum of a photon is changed by an amount $p^{'}$ then the corresponding change in the de-Broglie wavelength is found to be $0.20 %$ Then, the original momentum of the photon was

1

300 p^{'}

2

500 p^{'}

3

400 p^{'}

4

100 p^{'}

Explanation:

B Given that,
$\frac{Δ λ}{λ} = 0.2 % = \frac{0.2}{100} = \frac{1}{500}$
We know that,
According to the de-Broglie wavelength -
$λ = \frac{h}{p}$
$λ p = h$
$\frac{Δ p}{p} + \frac{Δ λ}{λ} = 0$
$| \frac{Δ p}{p} | = | \frac{- Δ λ}{λ} |$
$\frac{p^{'}}{p} = \frac{1}{500}$
$p = 500 p^{'}$

VITEEE-2015

Dual nature of radiation and Matter

142103 The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

1 double the earlier value

2 unchanged

3 more than doubled

4 less than doubled

Explanation:

C Given that, $v^{'} = 2 v$
We know that,
$Misplaced &$
$Misplaced &$
$= 2 h v - 2 ϕ_{o} + ϕ_{o}$
$= 2 (h v - ϕ_{o}) + ϕ_{o}$
From equation (i) and equation (ii), we get -
$K^{'} = 2 K + ϕ_{o}$
The kinetic energy of the photoelectrons will be more than double when incident frequency is double.

VITEEE-2011

Dual nature of radiation and Matter

142104 The work function of a certain metal is $3.31 \times 10^{- 19} J$ . Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \AA$ is (Given, $h = 6.62 \times 10^{- 34} J$ $s, c = 3 \times 10^{8} {ms}^{- 1}, e = 1.6 \times 10^{- 19} C$ )

1

2.48 eV

2

0.41 eV

3

2.07 eV

4

0.82 eV

Explanation:

B Given that ,
$ϕ_{o} = 3.31 \times 10^{- 19} J$
$λ = 5000 \AA = 5 \times 10^{- 7} m$
$h = 6.62 \times 10^{- 34} J - S$
$c = 3 \times 10^{8} m / s$
$e = 1.6 \times 10^{- 19} C$
According to the Einstein's photoelectric equation -
$(KE)_{max} = E - ϕ_{o}$
$(KE)_{max} = \frac{hc}{λ} - ϕ_{o}$
$(KE)_{max} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{5 \times 10^{- 7}} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 3.972 \times 10^{- 19} - 3.31 \times 10^{- 19}$
$(KE)_{max} = 0.662 \times 10^{- 19} J$
$(KE)_{max} = \frac{0.662 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV$
$(KE)_{max} = 0.41 eV$

VITEEE-2009 - AP EAMCET -2009

Dual nature of radiation and Matter

142105 The variation of photocurrent with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is shown in the graph, then:

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:

C Form the given graph, the magnitude of stopping potential,
$V_{0_{1}} < V_{0_{2}} < V_{0_{3}}$
Stopping potential is directly proportional to the incident frequency i.e. smaller the incident frequency the smaller the stopping potential.
Therefore, $v_{1} < v_{2} < v_{3}$

Karnataka CET-2016