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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357760 In a photoelectric experiment, the maximum velocity of photoelectrons emitted

1 Depends on intensity of incident radiation

2 Does not depend on cathode material

3 Depends on frequency of incident radiation

4 Does not depend on wavelength of incident radiation

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357761 When a piece of metal is illuminated by a monochromatic light of wavelength $λ$ , then stopping potential is $3 V_{s}$ . When same surface is illuminated by light of wavelength $2 λ$ , then stopping potential becomes $V_{s}$ . The value of threshold wavelength for photoelectric emission will be

1

(4 λ) / 3

2

6 λ

3

4 λ

4

8 λ

Explanation:

From photoelectricity equation,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e V_{s}$
where $λ =$ wavelength of incident light,
$λ_{0} =$ threshold wavelength
$V_{s} =$ stopping potential
For the given problem,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e \cdot 3 V_{s} (1) and$
$\frac{h c}{2 λ} = \frac{h c}{λ_{0}} + e \cdot V_{s} (2)$
Eliminating $V_{S}$ from these two equations,
$\frac{h c}{λ_{0}} = \frac{h c}{4 λ} \Rightarrow λ_{0} = 4 λ$

KCET - 2008

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357762 A metal surface is illuminated by light of two different wavelengths $248 n m$ and $310 n m$ . The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_{1}$ and $u_{2}$ respectively. If $u_{1} : u_{2} = 2 : 1$ and $h c = 1240 e V - n m$ , the work function of the metal is nearly

1

3.2 e V

2

3.7 e V

3

2.5 e V

4

2.8 e V

Explanation:

From Einstein's photoelectric equation, kinetic energy of photoelectrons,
$K = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
For $λ = 248 n m, v = u_{1}$
$∴ K_{1} = \frac{1}{2} m u_{1}^{2} = \frac{h c}{248} - ϕ (1)$
For $λ = 310 n m, v = u_{2}$
$∴ K_{2} = \frac{1}{2} m u_{2}^{2} = \frac{h c}{310} - ϕ (2)$
Divide eqn. (1) by eq. (2),
$\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{\frac{h c}{248} - ϕ}{\frac{h c}{310} - ϕ} = \frac{\frac{1240}{248} - ϕ}{\frac{1240}{310} - ϕ}$
$\Rightarrow {(\frac{2}{1})}^{2} = \frac{5 - ϕ}{4 - ϕ}$
$\Rightarrow ϕ = 3.7 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357763 When the light of frequency $2 ν_{0}$ (where $ν_{0}$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is
v
$_{1}$ . When the frequency of the incident radiation is increased to $5 ν_{0}$ the maximum velocity of electrons emitted from the same plate is
v
$_{2}$ . The ratio of
v
$_{1}$ to
v
$_{2}$ is

1

4 : 1

2

2 : 1

3

1 : 4

4

1 : 2

Explanation:

$E_{1} = h v - φ = h (2 υ_{0}) - υ h_{0} = υ h_{0}$
$E_{2} = 5 h υ_{0} - h_{0} υ = 4 h υ_{0}$
$\frac{E_{1}}{E_{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{4}$
$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

NEET - 2018

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357760 In a photoelectric experiment, the maximum velocity of photoelectrons emitted

1 Depends on intensity of incident radiation

2 Does not depend on cathode material

3 Depends on frequency of incident radiation

4 Does not depend on wavelength of incident radiation

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357761 When a piece of metal is illuminated by a monochromatic light of wavelength $λ$ , then stopping potential is $3 V_{s}$ . When same surface is illuminated by light of wavelength $2 λ$ , then stopping potential becomes $V_{s}$ . The value of threshold wavelength for photoelectric emission will be

1

(4 λ) / 3

2

6 λ

3

4 λ

4

8 λ

Explanation:

From photoelectricity equation,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e V_{s}$
where $λ =$ wavelength of incident light,
$λ_{0} =$ threshold wavelength
$V_{s} =$ stopping potential
For the given problem,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e \cdot 3 V_{s} (1) and$
$\frac{h c}{2 λ} = \frac{h c}{λ_{0}} + e \cdot V_{s} (2)$
Eliminating $V_{S}$ from these two equations,
$\frac{h c}{λ_{0}} = \frac{h c}{4 λ} \Rightarrow λ_{0} = 4 λ$

KCET - 2008

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357762 A metal surface is illuminated by light of two different wavelengths $248 n m$ and $310 n m$ . The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_{1}$ and $u_{2}$ respectively. If $u_{1} : u_{2} = 2 : 1$ and $h c = 1240 e V - n m$ , the work function of the metal is nearly

1

3.2 e V

2

3.7 e V

3

2.5 e V

4

2.8 e V

Explanation:

From Einstein's photoelectric equation, kinetic energy of photoelectrons,
$K = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
For $λ = 248 n m, v = u_{1}$
$∴ K_{1} = \frac{1}{2} m u_{1}^{2} = \frac{h c}{248} - ϕ (1)$
For $λ = 310 n m, v = u_{2}$
$∴ K_{2} = \frac{1}{2} m u_{2}^{2} = \frac{h c}{310} - ϕ (2)$
Divide eqn. (1) by eq. (2),
$\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{\frac{h c}{248} - ϕ}{\frac{h c}{310} - ϕ} = \frac{\frac{1240}{248} - ϕ}{\frac{1240}{310} - ϕ}$
$\Rightarrow {(\frac{2}{1})}^{2} = \frac{5 - ϕ}{4 - ϕ}$
$\Rightarrow ϕ = 3.7 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357763 When the light of frequency $2 ν_{0}$ (where $ν_{0}$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is
v
$_{1}$ . When the frequency of the incident radiation is increased to $5 ν_{0}$ the maximum velocity of electrons emitted from the same plate is
v
$_{2}$ . The ratio of
v
$_{1}$ to
v
$_{2}$ is

1

4 : 1

2

2 : 1

3

1 : 4

4

1 : 2

Explanation:

$E_{1} = h v - φ = h (2 υ_{0}) - υ h_{0} = υ h_{0}$
$E_{2} = 5 h υ_{0} - h_{0} υ = 4 h υ_{0}$
$\frac{E_{1}}{E_{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{4}$
$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

NEET - 2018

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357760 In a photoelectric experiment, the maximum velocity of photoelectrons emitted

1 Depends on intensity of incident radiation

2 Does not depend on cathode material

3 Depends on frequency of incident radiation

4 Does not depend on wavelength of incident radiation

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357761 When a piece of metal is illuminated by a monochromatic light of wavelength $λ$ , then stopping potential is $3 V_{s}$ . When same surface is illuminated by light of wavelength $2 λ$ , then stopping potential becomes $V_{s}$ . The value of threshold wavelength for photoelectric emission will be

1

(4 λ) / 3

2

6 λ

3

4 λ

4

8 λ

Explanation:

From photoelectricity equation,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e V_{s}$
where $λ =$ wavelength of incident light,
$λ_{0} =$ threshold wavelength
$V_{s} =$ stopping potential
For the given problem,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e \cdot 3 V_{s} (1) and$
$\frac{h c}{2 λ} = \frac{h c}{λ_{0}} + e \cdot V_{s} (2)$
Eliminating $V_{S}$ from these two equations,
$\frac{h c}{λ_{0}} = \frac{h c}{4 λ} \Rightarrow λ_{0} = 4 λ$

KCET - 2008

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357762 A metal surface is illuminated by light of two different wavelengths $248 n m$ and $310 n m$ . The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_{1}$ and $u_{2}$ respectively. If $u_{1} : u_{2} = 2 : 1$ and $h c = 1240 e V - n m$ , the work function of the metal is nearly

1

3.2 e V

2

3.7 e V

3

2.5 e V

4

2.8 e V

Explanation:

From Einstein's photoelectric equation, kinetic energy of photoelectrons,
$K = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
For $λ = 248 n m, v = u_{1}$
$∴ K_{1} = \frac{1}{2} m u_{1}^{2} = \frac{h c}{248} - ϕ (1)$
For $λ = 310 n m, v = u_{2}$
$∴ K_{2} = \frac{1}{2} m u_{2}^{2} = \frac{h c}{310} - ϕ (2)$
Divide eqn. (1) by eq. (2),
$\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{\frac{h c}{248} - ϕ}{\frac{h c}{310} - ϕ} = \frac{\frac{1240}{248} - ϕ}{\frac{1240}{310} - ϕ}$
$\Rightarrow {(\frac{2}{1})}^{2} = \frac{5 - ϕ}{4 - ϕ}$
$\Rightarrow ϕ = 3.7 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357763 When the light of frequency $2 ν_{0}$ (where $ν_{0}$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is
v
$_{1}$ . When the frequency of the incident radiation is increased to $5 ν_{0}$ the maximum velocity of electrons emitted from the same plate is
v
$_{2}$ . The ratio of
v
$_{1}$ to
v
$_{2}$ is

1

4 : 1

2

2 : 1

3

1 : 4

4

1 : 2

Explanation:

$E_{1} = h v - φ = h (2 υ_{0}) - υ h_{0} = υ h_{0}$
$E_{2} = 5 h υ_{0} - h_{0} υ = 4 h υ_{0}$
$\frac{E_{1}}{E_{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{4}$
$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

NEET - 2018

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357760 In a photoelectric experiment, the maximum velocity of photoelectrons emitted

1 Depends on intensity of incident radiation

2 Does not depend on cathode material

3 Depends on frequency of incident radiation

4 Does not depend on wavelength of incident radiation

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357761 When a piece of metal is illuminated by a monochromatic light of wavelength $λ$ , then stopping potential is $3 V_{s}$ . When same surface is illuminated by light of wavelength $2 λ$ , then stopping potential becomes $V_{s}$ . The value of threshold wavelength for photoelectric emission will be

1

(4 λ) / 3

2

6 λ

3

4 λ

4

8 λ

Explanation:

From photoelectricity equation,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e V_{s}$
where $λ =$ wavelength of incident light,
$λ_{0} =$ threshold wavelength
$V_{s} =$ stopping potential
For the given problem,
$\frac{h c}{λ} = \frac{h c}{λ_{0}} + e \cdot 3 V_{s} (1) and$
$\frac{h c}{2 λ} = \frac{h c}{λ_{0}} + e \cdot V_{s} (2)$
Eliminating $V_{S}$ from these two equations,
$\frac{h c}{λ_{0}} = \frac{h c}{4 λ} \Rightarrow λ_{0} = 4 λ$

KCET - 2008

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357762 A metal surface is illuminated by light of two different wavelengths $248 n m$ and $310 n m$ . The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_{1}$ and $u_{2}$ respectively. If $u_{1} : u_{2} = 2 : 1$ and $h c = 1240 e V - n m$ , the work function of the metal is nearly

1

3.2 e V

2

3.7 e V

3

2.5 e V

4

2.8 e V

Explanation:

From Einstein's photoelectric equation, kinetic energy of photoelectrons,
$K = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
For $λ = 248 n m, v = u_{1}$
$∴ K_{1} = \frac{1}{2} m u_{1}^{2} = \frac{h c}{248} - ϕ (1)$
For $λ = 310 n m, v = u_{2}$
$∴ K_{2} = \frac{1}{2} m u_{2}^{2} = \frac{h c}{310} - ϕ (2)$
Divide eqn. (1) by eq. (2),
$\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{\frac{h c}{248} - ϕ}{\frac{h c}{310} - ϕ} = \frac{\frac{1240}{248} - ϕ}{\frac{1240}{310} - ϕ}$
$\Rightarrow {(\frac{2}{1})}^{2} = \frac{5 - ϕ}{4 - ϕ}$
$\Rightarrow ϕ = 3.7 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357763 When the light of frequency $2 ν_{0}$ (where $ν_{0}$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is
v
$_{1}$ . When the frequency of the incident radiation is increased to $5 ν_{0}$ the maximum velocity of electrons emitted from the same plate is
v
$_{2}$ . The ratio of
v
$_{1}$ to
v
$_{2}$ is

1

4 : 1

2

2 : 1

3

1 : 4

4

1 : 2

Explanation:

$E_{1} = h v - φ = h (2 υ_{0}) - υ h_{0} = υ h_{0}$
$E_{2} = 5 h υ_{0} - h_{0} υ = 4 h υ_{0}$
$\frac{E_{1}}{E_{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{4}$
$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

NEET - 2018