QBManager

Dual nature of radiation and Matter

141988 The value of Plank's constant, if the slope of the graph of stopping potential vs frequency of incident light is $4 \times 10^{- 15} Vs$ is

1

6.0 \times 120^{- 34} Js

2

6.2 \times 10^{- 34} Js

3

6.4 \times 10^{- 34} Js

4

6.6 \times 10^{- 34} Js

Explanation:

C According to the Einstein photoelectric equation -
${eV}_{o} = h v - ϕ$
$V_{o} = \frac{h v}{e} - \frac{ϕ}{e}$
$∵ slope = \frac{h}{e} = 4 \times 10^{- 15}$
$h = 4 \times 10^{- 15} \times 1.6 \times 10^{- 19}$
$h = 6.4 \times 10^{- 34} Js$

Shift-I

Dual nature of radiation and Matter

141989 Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 nm$ . If the stopping potential for the ejected electrons is $2.8 eV$ , then the work function of the metal is (Take he $= 1240 eV . nm$ )

1

5.2 eV

2

4.4 eV

3

3.8 eV

4

2.2 eV

Explanation:

D Given that, $λ = 248 nm, {eV}_{o} = 2.8 eV$ , hc $=$ $1240 eV, nm$
Einstein's equation of photoelectric effect-
$K.E. = h v - ϕ [∵ v = \frac{c}{λ}]$
$KE = \frac{hc}{λ} - ϕ_{o}$
${eV}_{o} = \frac{1240}{248} - ϕ_{o}$
$2.8 = 5 - ϕ_{o}$
$ϕ_{o} = 2.2 eV$

Shift-I

Dual nature of radiation and Matter

141991 In a photo electric experiment, the wavelength of the light incident on the metal is changed from $200 nm$ to $400 nm$ . The decrease in the stopping potential is close to
[Use hc $= 1240 eV$ -nm where $h =$ Planck's constant and $c$ is velocity of light]

1

3.1 V

2

2.8 V

3

4.2 V

4

1.2 V

Explanation:

A Given that,
$λ_{1} = 200 nm, λ_{2} = 400 nm, hc = 1240 eV nm$
According to the Einstein's photoelectric equation -
$\frac{hc}{λ_{1}} = ϕ_{o} + {eV}_{1}$
$\frac{hc}{λ_{2}} = ϕ_{o} + {eV}_{2}$
Subtracting equation (ii) in equation (i), we get -
$hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = e (V_{1} - V_{2})$
$V_{1} - V_{2} = \frac{hc}{e} [\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}}]$
$= \frac{1240 eV}{e} [\frac{400 - 200}{200 \times 400}]$
$= 1240 \times \frac{1}{400}$
$= 3.1 V$

Shift-II

Dual nature of radiation and Matter

141992 When monochromatic light falls on a photosensitive metal, an electron is emitted with maximum velocity $1.6 \times 10^{6} m / s$ . Find the stopping potential.
[charge of electron $= 1.6 \times 10^{- 19} C$ , mass of electron $= 9 \times 10^{- 31} kg]$

1

7.2 V

2

14.4 V

3

21.6 V

4

28.8 V

Explanation:

A Given that,
$V_{max} = 1.6 \times 10^{6} m / s$
$e = 1.6 \times 10^{- 19} C$
$m_{e} = 9 \times 10^{- 31} kg$
We know that,
Stopping potential-
$\frac{1}{2} {mv}_{max}^{2} = {eV}_{o}$
$V_{o} = \frac{{mv}_{max}^{2}}{2 e}$
$V_{o} = \frac{9 \times 10^{- 31} \times {(1.6 \times 10^{6})}^{2}}{2 \times 1.6 \times 10^{- 19}}$
$V_{o} = 7.2 V$

Shift-II

Dual nature of radiation and Matter

141993 In a photoelectric effect, the maximum energy of the photoelectron is attained with exposure of $2000 A$ light. If the maximum kinetic energy of the photoelectron is $3 eV$ , the threshold wavelength will be
(use hc $= 1240 ev . nm$ )

1

387.5 nm

2

402.5 nm

3

339.5 nm

4

445.5 nm

Explanation:

A Given that, $(KE)_{max} = 3 eV$
$hc = 1240 eV$
$λ = 2000 \AA = 200 nm$
Maximum kinetic energy of photoelectron is-
$(KE)_{max} = \frac{hc}{λ} - \frac{hc}{λ_{o}}$
$3 eV = \frac{1240}{200} - \frac{hc}{λ_{o}}$
$\frac{hc}{λ_{o}} = 3.2 eV$
$λ_{o} = \frac{hc}{3.2}$
$= \frac{1240}{3.2} = 387.5 nm$

Shift-I

Dual nature of radiation and Matter

141988 The value of Plank's constant, if the slope of the graph of stopping potential vs frequency of incident light is $4 \times 10^{- 15} Vs$ is

1

6.0 \times 120^{- 34} Js

2

6.2 \times 10^{- 34} Js

3

6.4 \times 10^{- 34} Js

4

6.6 \times 10^{- 34} Js

Explanation:

C According to the Einstein photoelectric equation -
${eV}_{o} = h v - ϕ$
$V_{o} = \frac{h v}{e} - \frac{ϕ}{e}$
$∵ slope = \frac{h}{e} = 4 \times 10^{- 15}$
$h = 4 \times 10^{- 15} \times 1.6 \times 10^{- 19}$
$h = 6.4 \times 10^{- 34} Js$

Shift-I

Dual nature of radiation and Matter

141989 Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 nm$ . If the stopping potential for the ejected electrons is $2.8 eV$ , then the work function of the metal is (Take he $= 1240 eV . nm$ )

1

5.2 eV

2

4.4 eV

3

3.8 eV

4

2.2 eV

Explanation:

D Given that, $λ = 248 nm, {eV}_{o} = 2.8 eV$ , hc $=$ $1240 eV, nm$
Einstein's equation of photoelectric effect-
$K.E. = h v - ϕ [∵ v = \frac{c}{λ}]$
$KE = \frac{hc}{λ} - ϕ_{o}$
${eV}_{o} = \frac{1240}{248} - ϕ_{o}$
$2.8 = 5 - ϕ_{o}$
$ϕ_{o} = 2.2 eV$

Shift-I

Dual nature of radiation and Matter

141991 In a photo electric experiment, the wavelength of the light incident on the metal is changed from $200 nm$ to $400 nm$ . The decrease in the stopping potential is close to
[Use hc $= 1240 eV$ -nm where $h =$ Planck's constant and $c$ is velocity of light]

1

3.1 V

2

2.8 V

3

4.2 V

4

1.2 V

Explanation:

A Given that,
$λ_{1} = 200 nm, λ_{2} = 400 nm, hc = 1240 eV nm$
According to the Einstein's photoelectric equation -
$\frac{hc}{λ_{1}} = ϕ_{o} + {eV}_{1}$
$\frac{hc}{λ_{2}} = ϕ_{o} + {eV}_{2}$
Subtracting equation (ii) in equation (i), we get -
$hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = e (V_{1} - V_{2})$
$V_{1} - V_{2} = \frac{hc}{e} [\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}}]$
$= \frac{1240 eV}{e} [\frac{400 - 200}{200 \times 400}]$
$= 1240 \times \frac{1}{400}$
$= 3.1 V$

Shift-II

Dual nature of radiation and Matter

141992 When monochromatic light falls on a photosensitive metal, an electron is emitted with maximum velocity $1.6 \times 10^{6} m / s$ . Find the stopping potential.
[charge of electron $= 1.6 \times 10^{- 19} C$ , mass of electron $= 9 \times 10^{- 31} kg]$

1

7.2 V

2

14.4 V

3

21.6 V

4

28.8 V

Explanation:

A Given that,
$V_{max} = 1.6 \times 10^{6} m / s$
$e = 1.6 \times 10^{- 19} C$
$m_{e} = 9 \times 10^{- 31} kg$
We know that,
Stopping potential-
$\frac{1}{2} {mv}_{max}^{2} = {eV}_{o}$
$V_{o} = \frac{{mv}_{max}^{2}}{2 e}$
$V_{o} = \frac{9 \times 10^{- 31} \times {(1.6 \times 10^{6})}^{2}}{2 \times 1.6 \times 10^{- 19}}$
$V_{o} = 7.2 V$

Shift-II

Dual nature of radiation and Matter

141993 In a photoelectric effect, the maximum energy of the photoelectron is attained with exposure of $2000 A$ light. If the maximum kinetic energy of the photoelectron is $3 eV$ , the threshold wavelength will be
(use hc $= 1240 ev . nm$ )

1

387.5 nm

2

402.5 nm

3

339.5 nm

4

445.5 nm

Explanation:

A Given that, $(KE)_{max} = 3 eV$
$hc = 1240 eV$
$λ = 2000 \AA = 200 nm$
Maximum kinetic energy of photoelectron is-
$(KE)_{max} = \frac{hc}{λ} - \frac{hc}{λ_{o}}$
$3 eV = \frac{1240}{200} - \frac{hc}{λ_{o}}$
$\frac{hc}{λ_{o}} = 3.2 eV$
$λ_{o} = \frac{hc}{3.2}$
$= \frac{1240}{3.2} = 387.5 nm$

Shift-I

Dual nature of radiation and Matter

141988 The value of Plank's constant, if the slope of the graph of stopping potential vs frequency of incident light is $4 \times 10^{- 15} Vs$ is

1

6.0 \times 120^{- 34} Js

2

6.2 \times 10^{- 34} Js

3

6.4 \times 10^{- 34} Js

4

6.6 \times 10^{- 34} Js

Explanation:

C According to the Einstein photoelectric equation -
${eV}_{o} = h v - ϕ$
$V_{o} = \frac{h v}{e} - \frac{ϕ}{e}$
$∵ slope = \frac{h}{e} = 4 \times 10^{- 15}$
$h = 4 \times 10^{- 15} \times 1.6 \times 10^{- 19}$
$h = 6.4 \times 10^{- 34} Js$

Shift-I

Dual nature of radiation and Matter

141989 Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 nm$ . If the stopping potential for the ejected electrons is $2.8 eV$ , then the work function of the metal is (Take he $= 1240 eV . nm$ )

1

5.2 eV

2

4.4 eV

3

3.8 eV

4

2.2 eV

Explanation:

D Given that, $λ = 248 nm, {eV}_{o} = 2.8 eV$ , hc $=$ $1240 eV, nm$
Einstein's equation of photoelectric effect-
$K.E. = h v - ϕ [∵ v = \frac{c}{λ}]$
$KE = \frac{hc}{λ} - ϕ_{o}$
${eV}_{o} = \frac{1240}{248} - ϕ_{o}$
$2.8 = 5 - ϕ_{o}$
$ϕ_{o} = 2.2 eV$

Shift-I

Dual nature of radiation and Matter

141991 In a photo electric experiment, the wavelength of the light incident on the metal is changed from $200 nm$ to $400 nm$ . The decrease in the stopping potential is close to
[Use hc $= 1240 eV$ -nm where $h =$ Planck's constant and $c$ is velocity of light]

1

3.1 V

2

2.8 V

3

4.2 V

4

1.2 V

Explanation:

A Given that,
$λ_{1} = 200 nm, λ_{2} = 400 nm, hc = 1240 eV nm$
According to the Einstein's photoelectric equation -
$\frac{hc}{λ_{1}} = ϕ_{o} + {eV}_{1}$
$\frac{hc}{λ_{2}} = ϕ_{o} + {eV}_{2}$
Subtracting equation (ii) in equation (i), we get -
$hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = e (V_{1} - V_{2})$
$V_{1} - V_{2} = \frac{hc}{e} [\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}}]$
$= \frac{1240 eV}{e} [\frac{400 - 200}{200 \times 400}]$
$= 1240 \times \frac{1}{400}$
$= 3.1 V$

Shift-II

Dual nature of radiation and Matter

141992 When monochromatic light falls on a photosensitive metal, an electron is emitted with maximum velocity $1.6 \times 10^{6} m / s$ . Find the stopping potential.
[charge of electron $= 1.6 \times 10^{- 19} C$ , mass of electron $= 9 \times 10^{- 31} kg]$

1

7.2 V

2

14.4 V

3

21.6 V

4

28.8 V

Explanation:

A Given that,
$V_{max} = 1.6 \times 10^{6} m / s$
$e = 1.6 \times 10^{- 19} C$
$m_{e} = 9 \times 10^{- 31} kg$
We know that,
Stopping potential-
$\frac{1}{2} {mv}_{max}^{2} = {eV}_{o}$
$V_{o} = \frac{{mv}_{max}^{2}}{2 e}$
$V_{o} = \frac{9 \times 10^{- 31} \times {(1.6 \times 10^{6})}^{2}}{2 \times 1.6 \times 10^{- 19}}$
$V_{o} = 7.2 V$

Shift-II

Dual nature of radiation and Matter

141993 In a photoelectric effect, the maximum energy of the photoelectron is attained with exposure of $2000 A$ light. If the maximum kinetic energy of the photoelectron is $3 eV$ , the threshold wavelength will be
(use hc $= 1240 ev . nm$ )

1

387.5 nm

2

402.5 nm

3

339.5 nm

4

445.5 nm

Explanation:

A Given that, $(KE)_{max} = 3 eV$
$hc = 1240 eV$
$λ = 2000 \AA = 200 nm$
Maximum kinetic energy of photoelectron is-
$(KE)_{max} = \frac{hc}{λ} - \frac{hc}{λ_{o}}$
$3 eV = \frac{1240}{200} - \frac{hc}{λ_{o}}$
$\frac{hc}{λ_{o}} = 3.2 eV$
$λ_{o} = \frac{hc}{3.2}$
$= \frac{1240}{3.2} = 387.5 nm$

Shift-I

Dual nature of radiation and Matter

141988 The value of Plank's constant, if the slope of the graph of stopping potential vs frequency of incident light is $4 \times 10^{- 15} Vs$ is

1

6.0 \times 120^{- 34} Js

2

6.2 \times 10^{- 34} Js

3

6.4 \times 10^{- 34} Js

4

6.6 \times 10^{- 34} Js

Explanation:

C According to the Einstein photoelectric equation -
${eV}_{o} = h v - ϕ$
$V_{o} = \frac{h v}{e} - \frac{ϕ}{e}$
$∵ slope = \frac{h}{e} = 4 \times 10^{- 15}$
$h = 4 \times 10^{- 15} \times 1.6 \times 10^{- 19}$
$h = 6.4 \times 10^{- 34} Js$

Shift-I

Dual nature of radiation and Matter

141989 Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 nm$ . If the stopping potential for the ejected electrons is $2.8 eV$ , then the work function of the metal is (Take he $= 1240 eV . nm$ )

1

5.2 eV

2

4.4 eV

3

3.8 eV

4

2.2 eV

Explanation:

D Given that, $λ = 248 nm, {eV}_{o} = 2.8 eV$ , hc $=$ $1240 eV, nm$
Einstein's equation of photoelectric effect-
$K.E. = h v - ϕ [∵ v = \frac{c}{λ}]$
$KE = \frac{hc}{λ} - ϕ_{o}$
${eV}_{o} = \frac{1240}{248} - ϕ_{o}$
$2.8 = 5 - ϕ_{o}$
$ϕ_{o} = 2.2 eV$

Shift-I

Dual nature of radiation and Matter

141991 In a photo electric experiment, the wavelength of the light incident on the metal is changed from $200 nm$ to $400 nm$ . The decrease in the stopping potential is close to
[Use hc $= 1240 eV$ -nm where $h =$ Planck's constant and $c$ is velocity of light]

1

3.1 V

2

2.8 V

3

4.2 V

4

1.2 V

Explanation:

A Given that,
$λ_{1} = 200 nm, λ_{2} = 400 nm, hc = 1240 eV nm$
According to the Einstein's photoelectric equation -
$\frac{hc}{λ_{1}} = ϕ_{o} + {eV}_{1}$
$\frac{hc}{λ_{2}} = ϕ_{o} + {eV}_{2}$
Subtracting equation (ii) in equation (i), we get -
$hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = e (V_{1} - V_{2})$
$V_{1} - V_{2} = \frac{hc}{e} [\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}}]$
$= \frac{1240 eV}{e} [\frac{400 - 200}{200 \times 400}]$
$= 1240 \times \frac{1}{400}$
$= 3.1 V$

Shift-II

Dual nature of radiation and Matter

141992 When monochromatic light falls on a photosensitive metal, an electron is emitted with maximum velocity $1.6 \times 10^{6} m / s$ . Find the stopping potential.
[charge of electron $= 1.6 \times 10^{- 19} C$ , mass of electron $= 9 \times 10^{- 31} kg]$

1

7.2 V

2

14.4 V

3

21.6 V

4

28.8 V

Explanation:

A Given that,
$V_{max} = 1.6 \times 10^{6} m / s$
$e = 1.6 \times 10^{- 19} C$
$m_{e} = 9 \times 10^{- 31} kg$
We know that,
Stopping potential-
$\frac{1}{2} {mv}_{max}^{2} = {eV}_{o}$
$V_{o} = \frac{{mv}_{max}^{2}}{2 e}$
$V_{o} = \frac{9 \times 10^{- 31} \times {(1.6 \times 10^{6})}^{2}}{2 \times 1.6 \times 10^{- 19}}$
$V_{o} = 7.2 V$

Shift-II

Dual nature of radiation and Matter

141993 In a photoelectric effect, the maximum energy of the photoelectron is attained with exposure of $2000 A$ light. If the maximum kinetic energy of the photoelectron is $3 eV$ , the threshold wavelength will be
(use hc $= 1240 ev . nm$ )

1

387.5 nm

2

402.5 nm

3

339.5 nm

4

445.5 nm

Explanation:

A Given that, $(KE)_{max} = 3 eV$
$hc = 1240 eV$
$λ = 2000 \AA = 200 nm$
Maximum kinetic energy of photoelectron is-
$(KE)_{max} = \frac{hc}{λ} - \frac{hc}{λ_{o}}$
$3 eV = \frac{1240}{200} - \frac{hc}{λ_{o}}$
$\frac{hc}{λ_{o}} = 3.2 eV$
$λ_{o} = \frac{hc}{3.2}$
$= \frac{1240}{3.2} = 387.5 nm$

Shift-I

Dual nature of radiation and Matter

141988 The value of Plank's constant, if the slope of the graph of stopping potential vs frequency of incident light is $4 \times 10^{- 15} Vs$ is

1

6.0 \times 120^{- 34} Js

2

6.2 \times 10^{- 34} Js

3

6.4 \times 10^{- 34} Js

4

6.6 \times 10^{- 34} Js

Explanation:

C According to the Einstein photoelectric equation -
${eV}_{o} = h v - ϕ$
$V_{o} = \frac{h v}{e} - \frac{ϕ}{e}$
$∵ slope = \frac{h}{e} = 4 \times 10^{- 15}$
$h = 4 \times 10^{- 15} \times 1.6 \times 10^{- 19}$
$h = 6.4 \times 10^{- 34} Js$

Shift-I

Dual nature of radiation and Matter

141989 Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 nm$ . If the stopping potential for the ejected electrons is $2.8 eV$ , then the work function of the metal is (Take he $= 1240 eV . nm$ )

1

5.2 eV

2

4.4 eV

3

3.8 eV

4

2.2 eV

Explanation:

D Given that, $λ = 248 nm, {eV}_{o} = 2.8 eV$ , hc $=$ $1240 eV, nm$
Einstein's equation of photoelectric effect-
$K.E. = h v - ϕ [∵ v = \frac{c}{λ}]$
$KE = \frac{hc}{λ} - ϕ_{o}$
${eV}_{o} = \frac{1240}{248} - ϕ_{o}$
$2.8 = 5 - ϕ_{o}$
$ϕ_{o} = 2.2 eV$

Shift-I

Dual nature of radiation and Matter

141991 In a photo electric experiment, the wavelength of the light incident on the metal is changed from $200 nm$ to $400 nm$ . The decrease in the stopping potential is close to
[Use hc $= 1240 eV$ -nm where $h =$ Planck's constant and $c$ is velocity of light]

1

3.1 V

2

2.8 V

3

4.2 V

4

1.2 V

Explanation:

A Given that,
$λ_{1} = 200 nm, λ_{2} = 400 nm, hc = 1240 eV nm$
According to the Einstein's photoelectric equation -
$\frac{hc}{λ_{1}} = ϕ_{o} + {eV}_{1}$
$\frac{hc}{λ_{2}} = ϕ_{o} + {eV}_{2}$
Subtracting equation (ii) in equation (i), we get -
$hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = e (V_{1} - V_{2})$
$V_{1} - V_{2} = \frac{hc}{e} [\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}}]$
$= \frac{1240 eV}{e} [\frac{400 - 200}{200 \times 400}]$
$= 1240 \times \frac{1}{400}$
$= 3.1 V$

Shift-II

Dual nature of radiation and Matter

141992 When monochromatic light falls on a photosensitive metal, an electron is emitted with maximum velocity $1.6 \times 10^{6} m / s$ . Find the stopping potential.
[charge of electron $= 1.6 \times 10^{- 19} C$ , mass of electron $= 9 \times 10^{- 31} kg]$

1

7.2 V

2

14.4 V

3

21.6 V

4

28.8 V

Explanation:

A Given that,
$V_{max} = 1.6 \times 10^{6} m / s$
$e = 1.6 \times 10^{- 19} C$
$m_{e} = 9 \times 10^{- 31} kg$
We know that,
Stopping potential-
$\frac{1}{2} {mv}_{max}^{2} = {eV}_{o}$
$V_{o} = \frac{{mv}_{max}^{2}}{2 e}$
$V_{o} = \frac{9 \times 10^{- 31} \times {(1.6 \times 10^{6})}^{2}}{2 \times 1.6 \times 10^{- 19}}$
$V_{o} = 7.2 V$

Shift-II

Dual nature of radiation and Matter

141993 In a photoelectric effect, the maximum energy of the photoelectron is attained with exposure of $2000 A$ light. If the maximum kinetic energy of the photoelectron is $3 eV$ , the threshold wavelength will be
(use hc $= 1240 ev . nm$ )

1

387.5 nm

2

402.5 nm

3

339.5 nm

4

445.5 nm

Explanation:

A Given that, $(KE)_{max} = 3 eV$
$hc = 1240 eV$
$λ = 2000 \AA = 200 nm$
Maximum kinetic energy of photoelectron is-
$(KE)_{max} = \frac{hc}{λ} - \frac{hc}{λ_{o}}$
$3 eV = \frac{1240}{200} - \frac{hc}{λ_{o}}$
$\frac{hc}{λ_{o}} = 3.2 eV$
$λ_{o} = \frac{hc}{3.2}$
$= \frac{1240}{3.2} = 387.5 nm$

Shift-I