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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357828 Experimental study of photoelectric effect shows that
I.
Photocurrent $\propto$ intensity of light
II.
Saturation current $\propto$ intensity of light
III.
Photoemission occurs only at frequency greater than thresold frequency.

1 Only I is correct

2 I and II are correct

3 I and III are correct

4 I, II and III are correct

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357829 Match the quantities of Column I with their values in Column II and choose the correct option from the codes given below.
Column I
Column II
A
Maximum number of photons absorbed by an electron in a photo electric effect
P
$\frac{ϕ_{0}}{v_{0}}$
B
Number of photons each of energy $E$ emitted by a source of power $P$ per second
Q
$\frac{P}{E}$
C
Energy of a photon
R
$h v$
D
Planck's constant in terms of work function and threshold frequency
S
1

1

A - Q, B - S, C - P, D - Q

2

A - S, B - Q, C - R, D - P

3

A - Q, B - S, C - R, D - P

4

A - R, B - P, C - Q, D - S

Explanation:

A. An electron absorbs a single photon of light and if the energy carried by photon is more than work function of the metal, the electron is emitted.
B. From a source,
Total emitted power $= P ($ total energy per second)
Energy of each photon $= E$
So, number of photons emitted by the source $=$ $P / E$ .
C. Energy of a photons $= E = h v$
D. For the cut-off or threshold frequency, the energy $h v_{0}$ of the incident radiation must be equal to work function $ϕ_{0}$ , so that $h v_{0} = ϕ_{0}$
$\Rightarrow h = \frac{ϕ_{0}}{v_{0}}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357830 When the amplitude of the light wave incident on a photometal sheet is increased then

1 The photoelectric current increases

2 The photoelectric current remains unchanged

3 The stopping potential increases

4 The stopping potential decreases

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357831 Energy from the sun is received on earth at the rate of $= 2 c a l c m^{- 2} m i n^{- 1}$ . If average wavelength of solar light be taken as $5500 \overset{^{\circ}}{A}$ , then how many photons are received on the earth $(c m^{- 2} m i n^{- 1})$ ?
(Take, $h = 6.6 \times 10^{- 34} J - s, 1 c a l = 4.2 J$ )

1

1.5 \times 10^{13}

2

2.9 \times 10^{13}

3

2.3 \times 10^{19}

4

1.75 \times 10^{19}

Explanation:

Energy received from the sun
$= 2 c a l c m^{- 2} m i n^{- 1}$
$= 8.4 J - c m^{- 2} m i n^{- 1}$
Energy of one photon received from sun,
$E = \frac{h c}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5500 \times 10^{- 10}}$
$= 3.6 \times 10^{- 19} J$
$∴$ Number of photons reaching the earth per $c m^{2}$ per minute will be
$n = \frac{Energy received from sun}{Energy of one photon}$
$n = \frac{8.4}{3.6 \times 10^{- 19}} = 2.3 \times 10^{19}$

AIIMS - 2007

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357832 A laser used to weld detached retains emits light with a wavelength $652 n m$ in pulses that are of $20 m s$ duration. The average power during each pulse is $0.6 W$ . The energy in each pulse and in a single photon are

1

7.5 \times 10^{15} e V, 2.7 e V

2

6.5 \times 10^{16} e V, 2.9 e V

3

6.5 \times 10^{16} e V, 2.7 e V

4

7.5 \times 10^{16} e V, 1.9 e V

Explanation:

$E = p t; E ($ in each photon $) = \frac{h c}{λ}$
$λ = 652 n m$
$E = P t \Rightarrow E = (0.6) 20 \times 10^{- 3} = 12 \times 10^{- 3} J$
$\Rightarrow E = \frac{12 \times 10^{- 3}}{1.6 \times 10^{- 19}} = \frac{12}{6} \times \frac{10^{- 3}}{10^{- 20}}$
$= \frac{3}{4} \times 10^{17} = 0.75 \times 10^{17}$
$= 7.5 \times 10^{16} e V$
Energy of photon $= \frac{h c}{λ} = \frac{1240}{652} = 1.9 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357828 Experimental study of photoelectric effect shows that
I.
Photocurrent $\propto$ intensity of light
II.
Saturation current $\propto$ intensity of light
III.
Photoemission occurs only at frequency greater than thresold frequency.

1 Only I is correct

2 I and II are correct

3 I and III are correct

4 I, II and III are correct

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357829 Match the quantities of Column I with their values in Column II and choose the correct option from the codes given below.
Column I
Column II
A
Maximum number of photons absorbed by an electron in a photo electric effect
P
$\frac{ϕ_{0}}{v_{0}}$
B
Number of photons each of energy $E$ emitted by a source of power $P$ per second
Q
$\frac{P}{E}$
C
Energy of a photon
R
$h v$
D
Planck's constant in terms of work function and threshold frequency
S
1

1

A - Q, B - S, C - P, D - Q

2

A - S, B - Q, C - R, D - P

3

A - Q, B - S, C - R, D - P

4

A - R, B - P, C - Q, D - S

Explanation:

A. An electron absorbs a single photon of light and if the energy carried by photon is more than work function of the metal, the electron is emitted.
B. From a source,
Total emitted power $= P ($ total energy per second)
Energy of each photon $= E$
So, number of photons emitted by the source $=$ $P / E$ .
C. Energy of a photons $= E = h v$
D. For the cut-off or threshold frequency, the energy $h v_{0}$ of the incident radiation must be equal to work function $ϕ_{0}$ , so that $h v_{0} = ϕ_{0}$
$\Rightarrow h = \frac{ϕ_{0}}{v_{0}}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357830 When the amplitude of the light wave incident on a photometal sheet is increased then

1 The photoelectric current increases

2 The photoelectric current remains unchanged

3 The stopping potential increases

4 The stopping potential decreases

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357831 Energy from the sun is received on earth at the rate of $= 2 c a l c m^{- 2} m i n^{- 1}$ . If average wavelength of solar light be taken as $5500 \overset{^{\circ}}{A}$ , then how many photons are received on the earth $(c m^{- 2} m i n^{- 1})$ ?
(Take, $h = 6.6 \times 10^{- 34} J - s, 1 c a l = 4.2 J$ )

1

1.5 \times 10^{13}

2

2.9 \times 10^{13}

3

2.3 \times 10^{19}

4

1.75 \times 10^{19}

Explanation:

Energy received from the sun
$= 2 c a l c m^{- 2} m i n^{- 1}$
$= 8.4 J - c m^{- 2} m i n^{- 1}$
Energy of one photon received from sun,
$E = \frac{h c}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5500 \times 10^{- 10}}$
$= 3.6 \times 10^{- 19} J$
$∴$ Number of photons reaching the earth per $c m^{2}$ per minute will be
$n = \frac{Energy received from sun}{Energy of one photon}$
$n = \frac{8.4}{3.6 \times 10^{- 19}} = 2.3 \times 10^{19}$

AIIMS - 2007

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357832 A laser used to weld detached retains emits light with a wavelength $652 n m$ in pulses that are of $20 m s$ duration. The average power during each pulse is $0.6 W$ . The energy in each pulse and in a single photon are

1

7.5 \times 10^{15} e V, 2.7 e V

2

6.5 \times 10^{16} e V, 2.9 e V

3

6.5 \times 10^{16} e V, 2.7 e V

4

7.5 \times 10^{16} e V, 1.9 e V

Explanation:

$E = p t; E ($ in each photon $) = \frac{h c}{λ}$
$λ = 652 n m$
$E = P t \Rightarrow E = (0.6) 20 \times 10^{- 3} = 12 \times 10^{- 3} J$
$\Rightarrow E = \frac{12 \times 10^{- 3}}{1.6 \times 10^{- 19}} = \frac{12}{6} \times \frac{10^{- 3}}{10^{- 20}}$
$= \frac{3}{4} \times 10^{17} = 0.75 \times 10^{17}$
$= 7.5 \times 10^{16} e V$
Energy of photon $= \frac{h c}{λ} = \frac{1240}{652} = 1.9 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357828 Experimental study of photoelectric effect shows that
I.
Photocurrent $\propto$ intensity of light
II.
Saturation current $\propto$ intensity of light
III.
Photoemission occurs only at frequency greater than thresold frequency.

1 Only I is correct

2 I and II are correct

3 I and III are correct

4 I, II and III are correct

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357829 Match the quantities of Column I with their values in Column II and choose the correct option from the codes given below.
Column I
Column II
A
Maximum number of photons absorbed by an electron in a photo electric effect
P
$\frac{ϕ_{0}}{v_{0}}$
B
Number of photons each of energy $E$ emitted by a source of power $P$ per second
Q
$\frac{P}{E}$
C
Energy of a photon
R
$h v$
D
Planck's constant in terms of work function and threshold frequency
S
1

1

A - Q, B - S, C - P, D - Q

2

A - S, B - Q, C - R, D - P

3

A - Q, B - S, C - R, D - P

4

A - R, B - P, C - Q, D - S

Explanation:

A. An electron absorbs a single photon of light and if the energy carried by photon is more than work function of the metal, the electron is emitted.
B. From a source,
Total emitted power $= P ($ total energy per second)
Energy of each photon $= E$
So, number of photons emitted by the source $=$ $P / E$ .
C. Energy of a photons $= E = h v$
D. For the cut-off or threshold frequency, the energy $h v_{0}$ of the incident radiation must be equal to work function $ϕ_{0}$ , so that $h v_{0} = ϕ_{0}$
$\Rightarrow h = \frac{ϕ_{0}}{v_{0}}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357830 When the amplitude of the light wave incident on a photometal sheet is increased then

1 The photoelectric current increases

2 The photoelectric current remains unchanged

3 The stopping potential increases

4 The stopping potential decreases

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357831 Energy from the sun is received on earth at the rate of $= 2 c a l c m^{- 2} m i n^{- 1}$ . If average wavelength of solar light be taken as $5500 \overset{^{\circ}}{A}$ , then how many photons are received on the earth $(c m^{- 2} m i n^{- 1})$ ?
(Take, $h = 6.6 \times 10^{- 34} J - s, 1 c a l = 4.2 J$ )

1

1.5 \times 10^{13}

2

2.9 \times 10^{13}

3

2.3 \times 10^{19}

4

1.75 \times 10^{19}

Explanation:

Energy received from the sun
$= 2 c a l c m^{- 2} m i n^{- 1}$
$= 8.4 J - c m^{- 2} m i n^{- 1}$
Energy of one photon received from sun,
$E = \frac{h c}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5500 \times 10^{- 10}}$
$= 3.6 \times 10^{- 19} J$
$∴$ Number of photons reaching the earth per $c m^{2}$ per minute will be
$n = \frac{Energy received from sun}{Energy of one photon}$
$n = \frac{8.4}{3.6 \times 10^{- 19}} = 2.3 \times 10^{19}$

AIIMS - 2007

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357832 A laser used to weld detached retains emits light with a wavelength $652 n m$ in pulses that are of $20 m s$ duration. The average power during each pulse is $0.6 W$ . The energy in each pulse and in a single photon are

1

7.5 \times 10^{15} e V, 2.7 e V

2

6.5 \times 10^{16} e V, 2.9 e V

3

6.5 \times 10^{16} e V, 2.7 e V

4

7.5 \times 10^{16} e V, 1.9 e V

Explanation:

$E = p t; E ($ in each photon $) = \frac{h c}{λ}$
$λ = 652 n m$
$E = P t \Rightarrow E = (0.6) 20 \times 10^{- 3} = 12 \times 10^{- 3} J$
$\Rightarrow E = \frac{12 \times 10^{- 3}}{1.6 \times 10^{- 19}} = \frac{12}{6} \times \frac{10^{- 3}}{10^{- 20}}$
$= \frac{3}{4} \times 10^{17} = 0.75 \times 10^{17}$
$= 7.5 \times 10^{16} e V$
Energy of photon $= \frac{h c}{λ} = \frac{1240}{652} = 1.9 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357828 Experimental study of photoelectric effect shows that
I.
Photocurrent $\propto$ intensity of light
II.
Saturation current $\propto$ intensity of light
III.
Photoemission occurs only at frequency greater than thresold frequency.

1 Only I is correct

2 I and II are correct

3 I and III are correct

4 I, II and III are correct

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357829 Match the quantities of Column I with their values in Column II and choose the correct option from the codes given below.
Column I
Column II
A
Maximum number of photons absorbed by an electron in a photo electric effect
P
$\frac{ϕ_{0}}{v_{0}}$
B
Number of photons each of energy $E$ emitted by a source of power $P$ per second
Q
$\frac{P}{E}$
C
Energy of a photon
R
$h v$
D
Planck's constant in terms of work function and threshold frequency
S
1

1

A - Q, B - S, C - P, D - Q

2

A - S, B - Q, C - R, D - P

3

A - Q, B - S, C - R, D - P

4

A - R, B - P, C - Q, D - S

Explanation:

A. An electron absorbs a single photon of light and if the energy carried by photon is more than work function of the metal, the electron is emitted.
B. From a source,
Total emitted power $= P ($ total energy per second)
Energy of each photon $= E$
So, number of photons emitted by the source $=$ $P / E$ .
C. Energy of a photons $= E = h v$
D. For the cut-off or threshold frequency, the energy $h v_{0}$ of the incident radiation must be equal to work function $ϕ_{0}$ , so that $h v_{0} = ϕ_{0}$
$\Rightarrow h = \frac{ϕ_{0}}{v_{0}}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357830 When the amplitude of the light wave incident on a photometal sheet is increased then

1 The photoelectric current increases

2 The photoelectric current remains unchanged

3 The stopping potential increases

4 The stopping potential decreases

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357831 Energy from the sun is received on earth at the rate of $= 2 c a l c m^{- 2} m i n^{- 1}$ . If average wavelength of solar light be taken as $5500 \overset{^{\circ}}{A}$ , then how many photons are received on the earth $(c m^{- 2} m i n^{- 1})$ ?
(Take, $h = 6.6 \times 10^{- 34} J - s, 1 c a l = 4.2 J$ )

1

1.5 \times 10^{13}

2

2.9 \times 10^{13}

3

2.3 \times 10^{19}

4

1.75 \times 10^{19}

Explanation:

Energy received from the sun
$= 2 c a l c m^{- 2} m i n^{- 1}$
$= 8.4 J - c m^{- 2} m i n^{- 1}$
Energy of one photon received from sun,
$E = \frac{h c}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5500 \times 10^{- 10}}$
$= 3.6 \times 10^{- 19} J$
$∴$ Number of photons reaching the earth per $c m^{2}$ per minute will be
$n = \frac{Energy received from sun}{Energy of one photon}$
$n = \frac{8.4}{3.6 \times 10^{- 19}} = 2.3 \times 10^{19}$

AIIMS - 2007

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357832 A laser used to weld detached retains emits light with a wavelength $652 n m$ in pulses that are of $20 m s$ duration. The average power during each pulse is $0.6 W$ . The energy in each pulse and in a single photon are

1

7.5 \times 10^{15} e V, 2.7 e V

2

6.5 \times 10^{16} e V, 2.9 e V

3

6.5 \times 10^{16} e V, 2.7 e V

4

7.5 \times 10^{16} e V, 1.9 e V

Explanation:

$E = p t; E ($ in each photon $) = \frac{h c}{λ}$
$λ = 652 n m$
$E = P t \Rightarrow E = (0.6) 20 \times 10^{- 3} = 12 \times 10^{- 3} J$
$\Rightarrow E = \frac{12 \times 10^{- 3}}{1.6 \times 10^{- 19}} = \frac{12}{6} \times \frac{10^{- 3}}{10^{- 20}}$
$= \frac{3}{4} \times 10^{17} = 0.75 \times 10^{17}$
$= 7.5 \times 10^{16} e V$
Energy of photon $= \frac{h c}{λ} = \frac{1240}{652} = 1.9 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357828 Experimental study of photoelectric effect shows that
I.
Photocurrent $\propto$ intensity of light
II.
Saturation current $\propto$ intensity of light
III.
Photoemission occurs only at frequency greater than thresold frequency.

1 Only I is correct

2 I and II are correct

3 I and III are correct

4 I, II and III are correct

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357829 Match the quantities of Column I with their values in Column II and choose the correct option from the codes given below.
Column I
Column II
A
Maximum number of photons absorbed by an electron in a photo electric effect
P
$\frac{ϕ_{0}}{v_{0}}$
B
Number of photons each of energy $E$ emitted by a source of power $P$ per second
Q
$\frac{P}{E}$
C
Energy of a photon
R
$h v$
D
Planck's constant in terms of work function and threshold frequency
S
1

1

A - Q, B - S, C - P, D - Q

2

A - S, B - Q, C - R, D - P

3

A - Q, B - S, C - R, D - P

4

A - R, B - P, C - Q, D - S

Explanation:

A. An electron absorbs a single photon of light and if the energy carried by photon is more than work function of the metal, the electron is emitted.
B. From a source,
Total emitted power $= P ($ total energy per second)
Energy of each photon $= E$
So, number of photons emitted by the source $=$ $P / E$ .
C. Energy of a photons $= E = h v$
D. For the cut-off or threshold frequency, the energy $h v_{0}$ of the incident radiation must be equal to work function $ϕ_{0}$ , so that $h v_{0} = ϕ_{0}$
$\Rightarrow h = \frac{ϕ_{0}}{v_{0}}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357830 When the amplitude of the light wave incident on a photometal sheet is increased then

1 The photoelectric current increases

2 The photoelectric current remains unchanged

3 The stopping potential increases

4 The stopping potential decreases

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357831 Energy from the sun is received on earth at the rate of $= 2 c a l c m^{- 2} m i n^{- 1}$ . If average wavelength of solar light be taken as $5500 \overset{^{\circ}}{A}$ , then how many photons are received on the earth $(c m^{- 2} m i n^{- 1})$ ?
(Take, $h = 6.6 \times 10^{- 34} J - s, 1 c a l = 4.2 J$ )

1

1.5 \times 10^{13}

2

2.9 \times 10^{13}

3

2.3 \times 10^{19}

4

1.75 \times 10^{19}

Explanation:

Energy received from the sun
$= 2 c a l c m^{- 2} m i n^{- 1}$
$= 8.4 J - c m^{- 2} m i n^{- 1}$
Energy of one photon received from sun,
$E = \frac{h c}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5500 \times 10^{- 10}}$
$= 3.6 \times 10^{- 19} J$
$∴$ Number of photons reaching the earth per $c m^{2}$ per minute will be
$n = \frac{Energy received from sun}{Energy of one photon}$
$n = \frac{8.4}{3.6 \times 10^{- 19}} = 2.3 \times 10^{19}$

AIIMS - 2007

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357832 A laser used to weld detached retains emits light with a wavelength $652 n m$ in pulses that are of $20 m s$ duration. The average power during each pulse is $0.6 W$ . The energy in each pulse and in a single photon are

1

7.5 \times 10^{15} e V, 2.7 e V

2

6.5 \times 10^{16} e V, 2.9 e V

3

6.5 \times 10^{16} e V, 2.7 e V

4

7.5 \times 10^{16} e V, 1.9 e V

Explanation:

$E = p t; E ($ in each photon $) = \frac{h c}{λ}$
$λ = 652 n m$
$E = P t \Rightarrow E = (0.6) 20 \times 10^{- 3} = 12 \times 10^{- 3} J$
$\Rightarrow E = \frac{12 \times 10^{- 3}}{1.6 \times 10^{- 19}} = \frac{12}{6} \times \frac{10^{- 3}}{10^{- 20}}$
$= \frac{3}{4} \times 10^{17} = 0.75 \times 10^{17}$
$= 7.5 \times 10^{16} e V$
Energy of photon $= \frac{h c}{λ} = \frac{1240}{652} = 1.9 e V$

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