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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357734 The work functions for metals $A, B$ and $C$ are respectively, $1.92 e V, 2 e V$ and $5 e V$ . According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength $4100$ $A^{\circ}$ is/are

1 Only

A

2 Only

C

3

A

and

B

4 None of these

Explanation:

Energy incident by the radiation of wavelength $4000 \overset{^{\circ}}{A}$ is
$E = \frac{h c}{λ} = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{4100 \times 10^{- 10}}$
$= 4.8 \times 10^{- 19} J$
$= \frac{4.8 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 3 e V$
Now, we have $W_{A} = 1.92 e V$ ,
$W_{B} = 2 e V$ and $W_{C} = 5 e V$
Since, $W_{A} < E$ and $W_{B} < E$
Hence, $A$ and $B$ will emit photoelectrons.
Metal $C$ will emit electrons. So, correct option is (2)

AIIMS - 2011

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357735 Light of frequency $4 v_{0}$ is incident on the metal of the threshold frequency $v_{0}$ . The maximum kinetic energy of the emitted photoelectrons is

1

2 h v_{0}

2

\frac{1}{2} h v_{0}

3

\frac{3}{2} h v_{0}

4

3 h v_{0}

Explanation:

$E = h v_{0} + K_{max} \Rightarrow K_{max} = 3 h v_{0}$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357736 When a metal surface is illuminated by light of wavelength $400 n m$ and $200 n m$ , the maximum velocities of the photoelectrons ejected are $v$ and 2 $v$ respectively. The work function of the metal is ( $h =$ planck's constant, $c =$ velocity of light in air)

1

2.5 h c \times 10^{6} J

2

h c \times 10^{6} J

3

1.6 h c \times 10^{6} J

4

0.5 h c \times 10^{6} J

Explanation:

By using $\frac{h c}{λ} = ϕ + \frac{1}{2} m v^{2}$
$\Rightarrow \frac{h c}{400 \times 10^{- 9}} = ϕ + \frac{1}{2} m v^{2} (1)$
and $\frac{h c}{200 \times 10^{- 9}} = ϕ + \frac{1}{2} m (2 v)^{2} (2)$
On solving (1) and (2)
$\frac{1}{2} m v^{2} = \frac{h c}{3} [\frac{1}{200 \times 10^{- 9}} - \frac{1}{400 \times 10^{- 9}}] (3)$
From equations (1) and (3), $ϕ = 1.6 h c \times 10^{6} J$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357737 A copper ball of radius $1 c m$ and work function $4.47 e V$ is irradiated with ultraviolet radiation of wavelength $2500 \overset{o}{A}$ . The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is:

1

5.5 \times 10^{- 13} C

2

7.5 \times 10^{- 13} C

3

4.5 \times 10^{- 12} C

4

2.5 \times 10^{- 11} C

Explanation:

When the sphere acquires a maximum charge at which the emitted photoelectrons can't escape from the sphere due to electrostatic attraction. The electron $K . E$ converts into $P . E$ . The $K . E$ of the photoelectron is
$K . E = h v - ϕ = P . E$
$\frac{h c}{λ} - 4.47 e V = \frac{1}{4 π ε_{0}} \frac{(1.6 \times 10^{- 19}) q}{10^{- 2} m}$
$\Rightarrow q = 5.5 \times 10^{- 13} C$

JEE - 2013

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357734 The work functions for metals $A, B$ and $C$ are respectively, $1.92 e V, 2 e V$ and $5 e V$ . According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength $4100$ $A^{\circ}$ is/are

1 Only

A

2 Only

C

3

A

and

B

4 None of these

Explanation:

Energy incident by the radiation of wavelength $4000 \overset{^{\circ}}{A}$ is
$E = \frac{h c}{λ} = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{4100 \times 10^{- 10}}$
$= 4.8 \times 10^{- 19} J$
$= \frac{4.8 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 3 e V$
Now, we have $W_{A} = 1.92 e V$ ,
$W_{B} = 2 e V$ and $W_{C} = 5 e V$
Since, $W_{A} < E$ and $W_{B} < E$
Hence, $A$ and $B$ will emit photoelectrons.
Metal $C$ will emit electrons. So, correct option is (2)

AIIMS - 2011

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357735 Light of frequency $4 v_{0}$ is incident on the metal of the threshold frequency $v_{0}$ . The maximum kinetic energy of the emitted photoelectrons is

1

2 h v_{0}

2

\frac{1}{2} h v_{0}

3

\frac{3}{2} h v_{0}

4

3 h v_{0}

Explanation:

$E = h v_{0} + K_{max} \Rightarrow K_{max} = 3 h v_{0}$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357736 When a metal surface is illuminated by light of wavelength $400 n m$ and $200 n m$ , the maximum velocities of the photoelectrons ejected are $v$ and 2 $v$ respectively. The work function of the metal is ( $h =$ planck's constant, $c =$ velocity of light in air)

1

2.5 h c \times 10^{6} J

2

h c \times 10^{6} J

3

1.6 h c \times 10^{6} J

4

0.5 h c \times 10^{6} J

Explanation:

By using $\frac{h c}{λ} = ϕ + \frac{1}{2} m v^{2}$
$\Rightarrow \frac{h c}{400 \times 10^{- 9}} = ϕ + \frac{1}{2} m v^{2} (1)$
and $\frac{h c}{200 \times 10^{- 9}} = ϕ + \frac{1}{2} m (2 v)^{2} (2)$
On solving (1) and (2)
$\frac{1}{2} m v^{2} = \frac{h c}{3} [\frac{1}{200 \times 10^{- 9}} - \frac{1}{400 \times 10^{- 9}}] (3)$
From equations (1) and (3), $ϕ = 1.6 h c \times 10^{6} J$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357737 A copper ball of radius $1 c m$ and work function $4.47 e V$ is irradiated with ultraviolet radiation of wavelength $2500 \overset{o}{A}$ . The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is:

1

5.5 \times 10^{- 13} C

2

7.5 \times 10^{- 13} C

3

4.5 \times 10^{- 12} C

4

2.5 \times 10^{- 11} C

Explanation:

When the sphere acquires a maximum charge at which the emitted photoelectrons can't escape from the sphere due to electrostatic attraction. The electron $K . E$ converts into $P . E$ . The $K . E$ of the photoelectron is
$K . E = h v - ϕ = P . E$
$\frac{h c}{λ} - 4.47 e V = \frac{1}{4 π ε_{0}} \frac{(1.6 \times 10^{- 19}) q}{10^{- 2} m}$
$\Rightarrow q = 5.5 \times 10^{- 13} C$

JEE - 2013

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357734 The work functions for metals $A, B$ and $C$ are respectively, $1.92 e V, 2 e V$ and $5 e V$ . According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength $4100$ $A^{\circ}$ is/are

1 Only

A

2 Only

C

3

A

and

B

4 None of these

Explanation:

Energy incident by the radiation of wavelength $4000 \overset{^{\circ}}{A}$ is
$E = \frac{h c}{λ} = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{4100 \times 10^{- 10}}$
$= 4.8 \times 10^{- 19} J$
$= \frac{4.8 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 3 e V$
Now, we have $W_{A} = 1.92 e V$ ,
$W_{B} = 2 e V$ and $W_{C} = 5 e V$
Since, $W_{A} < E$ and $W_{B} < E$
Hence, $A$ and $B$ will emit photoelectrons.
Metal $C$ will emit electrons. So, correct option is (2)

AIIMS - 2011

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357735 Light of frequency $4 v_{0}$ is incident on the metal of the threshold frequency $v_{0}$ . The maximum kinetic energy of the emitted photoelectrons is

1

2 h v_{0}

2

\frac{1}{2} h v_{0}

3

\frac{3}{2} h v_{0}

4

3 h v_{0}

Explanation:

$E = h v_{0} + K_{max} \Rightarrow K_{max} = 3 h v_{0}$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357736 When a metal surface is illuminated by light of wavelength $400 n m$ and $200 n m$ , the maximum velocities of the photoelectrons ejected are $v$ and 2 $v$ respectively. The work function of the metal is ( $h =$ planck's constant, $c =$ velocity of light in air)

1

2.5 h c \times 10^{6} J

2

h c \times 10^{6} J

3

1.6 h c \times 10^{6} J

4

0.5 h c \times 10^{6} J

Explanation:

By using $\frac{h c}{λ} = ϕ + \frac{1}{2} m v^{2}$
$\Rightarrow \frac{h c}{400 \times 10^{- 9}} = ϕ + \frac{1}{2} m v^{2} (1)$
and $\frac{h c}{200 \times 10^{- 9}} = ϕ + \frac{1}{2} m (2 v)^{2} (2)$
On solving (1) and (2)
$\frac{1}{2} m v^{2} = \frac{h c}{3} [\frac{1}{200 \times 10^{- 9}} - \frac{1}{400 \times 10^{- 9}}] (3)$
From equations (1) and (3), $ϕ = 1.6 h c \times 10^{6} J$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357737 A copper ball of radius $1 c m$ and work function $4.47 e V$ is irradiated with ultraviolet radiation of wavelength $2500 \overset{o}{A}$ . The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is:

1

5.5 \times 10^{- 13} C

2

7.5 \times 10^{- 13} C

3

4.5 \times 10^{- 12} C

4

2.5 \times 10^{- 11} C

Explanation:

When the sphere acquires a maximum charge at which the emitted photoelectrons can't escape from the sphere due to electrostatic attraction. The electron $K . E$ converts into $P . E$ . The $K . E$ of the photoelectron is
$K . E = h v - ϕ = P . E$
$\frac{h c}{λ} - 4.47 e V = \frac{1}{4 π ε_{0}} \frac{(1.6 \times 10^{- 19}) q}{10^{- 2} m}$
$\Rightarrow q = 5.5 \times 10^{- 13} C$

JEE - 2013

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357734 The work functions for metals $A, B$ and $C$ are respectively, $1.92 e V, 2 e V$ and $5 e V$ . According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength $4100$ $A^{\circ}$ is/are

1 Only

A

2 Only

C

3

A

and

B

4 None of these

Explanation:

Energy incident by the radiation of wavelength $4000 \overset{^{\circ}}{A}$ is
$E = \frac{h c}{λ} = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{4100 \times 10^{- 10}}$
$= 4.8 \times 10^{- 19} J$
$= \frac{4.8 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 3 e V$
Now, we have $W_{A} = 1.92 e V$ ,
$W_{B} = 2 e V$ and $W_{C} = 5 e V$
Since, $W_{A} < E$ and $W_{B} < E$
Hence, $A$ and $B$ will emit photoelectrons.
Metal $C$ will emit electrons. So, correct option is (2)

AIIMS - 2011

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357735 Light of frequency $4 v_{0}$ is incident on the metal of the threshold frequency $v_{0}$ . The maximum kinetic energy of the emitted photoelectrons is

1

2 h v_{0}

2

\frac{1}{2} h v_{0}

3

\frac{3}{2} h v_{0}

4

3 h v_{0}

Explanation:

$E = h v_{0} + K_{max} \Rightarrow K_{max} = 3 h v_{0}$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357736 When a metal surface is illuminated by light of wavelength $400 n m$ and $200 n m$ , the maximum velocities of the photoelectrons ejected are $v$ and 2 $v$ respectively. The work function of the metal is ( $h =$ planck's constant, $c =$ velocity of light in air)

1

2.5 h c \times 10^{6} J

2

h c \times 10^{6} J

3

1.6 h c \times 10^{6} J

4

0.5 h c \times 10^{6} J

Explanation:

By using $\frac{h c}{λ} = ϕ + \frac{1}{2} m v^{2}$
$\Rightarrow \frac{h c}{400 \times 10^{- 9}} = ϕ + \frac{1}{2} m v^{2} (1)$
and $\frac{h c}{200 \times 10^{- 9}} = ϕ + \frac{1}{2} m (2 v)^{2} (2)$
On solving (1) and (2)
$\frac{1}{2} m v^{2} = \frac{h c}{3} [\frac{1}{200 \times 10^{- 9}} - \frac{1}{400 \times 10^{- 9}}] (3)$
From equations (1) and (3), $ϕ = 1.6 h c \times 10^{6} J$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357737 A copper ball of radius $1 c m$ and work function $4.47 e V$ is irradiated with ultraviolet radiation of wavelength $2500 \overset{o}{A}$ . The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is:

1

5.5 \times 10^{- 13} C

2

7.5 \times 10^{- 13} C

3

4.5 \times 10^{- 12} C

4

2.5 \times 10^{- 11} C

Explanation:

When the sphere acquires a maximum charge at which the emitted photoelectrons can't escape from the sphere due to electrostatic attraction. The electron $K . E$ converts into $P . E$ . The $K . E$ of the photoelectron is
$K . E = h v - ϕ = P . E$
$\frac{h c}{λ} - 4.47 e V = \frac{1}{4 π ε_{0}} \frac{(1.6 \times 10^{- 19}) q}{10^{- 2} m}$
$\Rightarrow q = 5.5 \times 10^{- 13} C$

JEE - 2013

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