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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357764 Light of wavelength 5000 $A^{\circ}$ falls on a sensitive plate with photoelectric work function of 1.9 $e V$ . The kinetic energy of the photoelectron emitted will be

1 0.58

e V

2 2.48

e V

3 1.24

e V

4 1.16

e V

Explanation:

Kinetic energy of the emitted photoelectron is given by photoelectric equation. $K E = E_{i} - W_{o}$
Here, $E_{i} =$ energy of incident radiation $= \frac{h c}{λ}$
$W_{o} =$ work function
$K E = \frac{h c}{λ} - W_{o}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5000 \times 10^{- 10}}$
$- 1.9 \times 1.6 \times 10^{- 19} J$
$= \frac{1.98 \times 10^{- 25}}{5 \times 10^{- 7}} - 3.04 \times 10^{- 19} J$
$= 3.96 \times 10^{- 19} - 3.04 \times 10^{- 19} J$
$= \frac{0.92 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 0.575 e V$
$\approx 0.58 e V$ .
So correct option is (1)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357765 Ultraviolet radiations of 6.2 $e V$ fall on an aluminium surface (work function 4.2 $e V$ ) The kinetic energy in joules of the fastest electron emitted is approximately

1

3.2 \times 10^{- 21} J

2

3.2 \times 10^{- 19} J

3

3.2 \times 10^{- 17} J

4

3.2 \times 10^{- 15} J

Explanation:

Kinetic energy of the fastest electron emitted will be
$K E_{max} = E_{i} - W_{0}$
$\Rightarrow K E_{max} = 6.2 e V - 4.2 e V = 2 e V$
$\Rightarrow K E_{m a x} = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19} J$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357766 When a certain metal surface is illuminated with a light of wavelength $λ$ , the stopping potential is $V$ , when the same surface is illuminated by light of wavelength $2 λ$ , the stoping potential is $(\frac{V}{3})$ . The threshold wavelength for the surface is

1

\frac{8 λ}{3}

2

\frac{4 λ}{3}

3

4 λ

4

6 λ

Explanation:

Given a metal, wavelength of light used $= λ$
Stopping potential $= V$
If $λ_{0}$ be the threshold wavelength, then maximum kinetic energy of emitted electrons
$K_{max} = h c (\frac{1}{λ} - \frac{1}{λ_{0}}) (1)$
Again, wavelength of used light $λ^{'} = 2 λ$
Stopping potential $V^{'} = \frac{V}{3}$
then $K E_{max} = h c (\frac{1}{λ^{'}} - \frac{1}{λ_{0}})$
$\Rightarrow e V^{'} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})$
$\Rightarrow \frac{e V}{3} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}}) (2)$
From eq.(1) and (2), we have
$\Rightarrow \frac{e V}{\frac{e V}{3}} = \frac{h c (\frac{1}{λ} - \frac{1}{λ_{0}})}{h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})}$
$3 (\frac{1}{2 λ} - \frac{1}{λ_{0}}) = \frac{1}{λ} - \frac{1}{λ_{0}}$
$\Rightarrow λ_{0} = 4 λ$
So threshold wavelength is 4 times of wavelength of light.

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357767 The graph of stopping potential $v_{s}$ against frequency $v_{0}$ of incident radiation is plotted for two different metals $P$ and $Q$ as shown in the graph. $ϕ_{p}$ and $ϕ_{Q}$ are work - functions of $P$ and $Q$ respectively, then
supporting img

1

ϕ_{P} > ϕ_{Q}

2

ϕ_{P} < ϕ_{Q}

3

ϕ_{P} = ϕ_{Q}

4

v_{0}^{'} < v_{0}

Explanation:

The work - function of a surface is
$ϕ = h v_{0}$
where, $h =$ planck's constant
and $v_{0} =$ threshold frequency
From graph it is clear that
$\begin{aligned} {(v_{0})}_{P} < {(v_{0})}_{Q} \\ ∴ ϕ_{P} < ϕ_{Q} \end{aligned}$

MHTCET - 2020

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357764 Light of wavelength 5000 $A^{\circ}$ falls on a sensitive plate with photoelectric work function of 1.9 $e V$ . The kinetic energy of the photoelectron emitted will be

1 0.58

e V

2 2.48

e V

3 1.24

e V

4 1.16

e V

Explanation:

Kinetic energy of the emitted photoelectron is given by photoelectric equation. $K E = E_{i} - W_{o}$
Here, $E_{i} =$ energy of incident radiation $= \frac{h c}{λ}$
$W_{o} =$ work function
$K E = \frac{h c}{λ} - W_{o}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5000 \times 10^{- 10}}$
$- 1.9 \times 1.6 \times 10^{- 19} J$
$= \frac{1.98 \times 10^{- 25}}{5 \times 10^{- 7}} - 3.04 \times 10^{- 19} J$
$= 3.96 \times 10^{- 19} - 3.04 \times 10^{- 19} J$
$= \frac{0.92 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 0.575 e V$
$\approx 0.58 e V$ .
So correct option is (1)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357765 Ultraviolet radiations of 6.2 $e V$ fall on an aluminium surface (work function 4.2 $e V$ ) The kinetic energy in joules of the fastest electron emitted is approximately

1

3.2 \times 10^{- 21} J

2

3.2 \times 10^{- 19} J

3

3.2 \times 10^{- 17} J

4

3.2 \times 10^{- 15} J

Explanation:

Kinetic energy of the fastest electron emitted will be
$K E_{max} = E_{i} - W_{0}$
$\Rightarrow K E_{max} = 6.2 e V - 4.2 e V = 2 e V$
$\Rightarrow K E_{m a x} = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19} J$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357766 When a certain metal surface is illuminated with a light of wavelength $λ$ , the stopping potential is $V$ , when the same surface is illuminated by light of wavelength $2 λ$ , the stoping potential is $(\frac{V}{3})$ . The threshold wavelength for the surface is

1

\frac{8 λ}{3}

2

\frac{4 λ}{3}

3

4 λ

4

6 λ

Explanation:

Given a metal, wavelength of light used $= λ$
Stopping potential $= V$
If $λ_{0}$ be the threshold wavelength, then maximum kinetic energy of emitted electrons
$K_{max} = h c (\frac{1}{λ} - \frac{1}{λ_{0}}) (1)$
Again, wavelength of used light $λ^{'} = 2 λ$
Stopping potential $V^{'} = \frac{V}{3}$
then $K E_{max} = h c (\frac{1}{λ^{'}} - \frac{1}{λ_{0}})$
$\Rightarrow e V^{'} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})$
$\Rightarrow \frac{e V}{3} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}}) (2)$
From eq.(1) and (2), we have
$\Rightarrow \frac{e V}{\frac{e V}{3}} = \frac{h c (\frac{1}{λ} - \frac{1}{λ_{0}})}{h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})}$
$3 (\frac{1}{2 λ} - \frac{1}{λ_{0}}) = \frac{1}{λ} - \frac{1}{λ_{0}}$
$\Rightarrow λ_{0} = 4 λ$
So threshold wavelength is 4 times of wavelength of light.

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357767 The graph of stopping potential $v_{s}$ against frequency $v_{0}$ of incident radiation is plotted for two different metals $P$ and $Q$ as shown in the graph. $ϕ_{p}$ and $ϕ_{Q}$ are work - functions of $P$ and $Q$ respectively, then
supporting img

1

ϕ_{P} > ϕ_{Q}

2

ϕ_{P} < ϕ_{Q}

3

ϕ_{P} = ϕ_{Q}

4

v_{0}^{'} < v_{0}

Explanation:

The work - function of a surface is
$ϕ = h v_{0}$
where, $h =$ planck's constant
and $v_{0} =$ threshold frequency
From graph it is clear that
$\begin{aligned} {(v_{0})}_{P} < {(v_{0})}_{Q} \\ ∴ ϕ_{P} < ϕ_{Q} \end{aligned}$

MHTCET - 2020

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357764 Light of wavelength 5000 $A^{\circ}$ falls on a sensitive plate with photoelectric work function of 1.9 $e V$ . The kinetic energy of the photoelectron emitted will be

1 0.58

e V

2 2.48

e V

3 1.24

e V

4 1.16

e V

Explanation:

Kinetic energy of the emitted photoelectron is given by photoelectric equation. $K E = E_{i} - W_{o}$
Here, $E_{i} =$ energy of incident radiation $= \frac{h c}{λ}$
$W_{o} =$ work function
$K E = \frac{h c}{λ} - W_{o}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5000 \times 10^{- 10}}$
$- 1.9 \times 1.6 \times 10^{- 19} J$
$= \frac{1.98 \times 10^{- 25}}{5 \times 10^{- 7}} - 3.04 \times 10^{- 19} J$
$= 3.96 \times 10^{- 19} - 3.04 \times 10^{- 19} J$
$= \frac{0.92 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 0.575 e V$
$\approx 0.58 e V$ .
So correct option is (1)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357765 Ultraviolet radiations of 6.2 $e V$ fall on an aluminium surface (work function 4.2 $e V$ ) The kinetic energy in joules of the fastest electron emitted is approximately

1

3.2 \times 10^{- 21} J

2

3.2 \times 10^{- 19} J

3

3.2 \times 10^{- 17} J

4

3.2 \times 10^{- 15} J

Explanation:

Kinetic energy of the fastest electron emitted will be
$K E_{max} = E_{i} - W_{0}$
$\Rightarrow K E_{max} = 6.2 e V - 4.2 e V = 2 e V$
$\Rightarrow K E_{m a x} = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19} J$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357766 When a certain metal surface is illuminated with a light of wavelength $λ$ , the stopping potential is $V$ , when the same surface is illuminated by light of wavelength $2 λ$ , the stoping potential is $(\frac{V}{3})$ . The threshold wavelength for the surface is

1

\frac{8 λ}{3}

2

\frac{4 λ}{3}

3

4 λ

4

6 λ

Explanation:

Given a metal, wavelength of light used $= λ$
Stopping potential $= V$
If $λ_{0}$ be the threshold wavelength, then maximum kinetic energy of emitted electrons
$K_{max} = h c (\frac{1}{λ} - \frac{1}{λ_{0}}) (1)$
Again, wavelength of used light $λ^{'} = 2 λ$
Stopping potential $V^{'} = \frac{V}{3}$
then $K E_{max} = h c (\frac{1}{λ^{'}} - \frac{1}{λ_{0}})$
$\Rightarrow e V^{'} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})$
$\Rightarrow \frac{e V}{3} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}}) (2)$
From eq.(1) and (2), we have
$\Rightarrow \frac{e V}{\frac{e V}{3}} = \frac{h c (\frac{1}{λ} - \frac{1}{λ_{0}})}{h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})}$
$3 (\frac{1}{2 λ} - \frac{1}{λ_{0}}) = \frac{1}{λ} - \frac{1}{λ_{0}}$
$\Rightarrow λ_{0} = 4 λ$
So threshold wavelength is 4 times of wavelength of light.

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357767 The graph of stopping potential $v_{s}$ against frequency $v_{0}$ of incident radiation is plotted for two different metals $P$ and $Q$ as shown in the graph. $ϕ_{p}$ and $ϕ_{Q}$ are work - functions of $P$ and $Q$ respectively, then
supporting img

1

ϕ_{P} > ϕ_{Q}

2

ϕ_{P} < ϕ_{Q}

3

ϕ_{P} = ϕ_{Q}

4

v_{0}^{'} < v_{0}

Explanation:

The work - function of a surface is
$ϕ = h v_{0}$
where, $h =$ planck's constant
and $v_{0} =$ threshold frequency
From graph it is clear that
$\begin{aligned} {(v_{0})}_{P} < {(v_{0})}_{Q} \\ ∴ ϕ_{P} < ϕ_{Q} \end{aligned}$

MHTCET - 2020

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357764 Light of wavelength 5000 $A^{\circ}$ falls on a sensitive plate with photoelectric work function of 1.9 $e V$ . The kinetic energy of the photoelectron emitted will be

1 0.58

e V

2 2.48

e V

3 1.24

e V

4 1.16

e V

Explanation:

Kinetic energy of the emitted photoelectron is given by photoelectric equation. $K E = E_{i} - W_{o}$
Here, $E_{i} =$ energy of incident radiation $= \frac{h c}{λ}$
$W_{o} =$ work function
$K E = \frac{h c}{λ} - W_{o}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5000 \times 10^{- 10}}$
$- 1.9 \times 1.6 \times 10^{- 19} J$
$= \frac{1.98 \times 10^{- 25}}{5 \times 10^{- 7}} - 3.04 \times 10^{- 19} J$
$= 3.96 \times 10^{- 19} - 3.04 \times 10^{- 19} J$
$= \frac{0.92 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V = 0.575 e V$
$\approx 0.58 e V$ .
So correct option is (1)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357765 Ultraviolet radiations of 6.2 $e V$ fall on an aluminium surface (work function 4.2 $e V$ ) The kinetic energy in joules of the fastest electron emitted is approximately

1

3.2 \times 10^{- 21} J

2

3.2 \times 10^{- 19} J

3

3.2 \times 10^{- 17} J

4

3.2 \times 10^{- 15} J

Explanation:

Kinetic energy of the fastest electron emitted will be
$K E_{max} = E_{i} - W_{0}$
$\Rightarrow K E_{max} = 6.2 e V - 4.2 e V = 2 e V$
$\Rightarrow K E_{m a x} = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19} J$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357766 When a certain metal surface is illuminated with a light of wavelength $λ$ , the stopping potential is $V$ , when the same surface is illuminated by light of wavelength $2 λ$ , the stoping potential is $(\frac{V}{3})$ . The threshold wavelength for the surface is

1

\frac{8 λ}{3}

2

\frac{4 λ}{3}

3

4 λ

4

6 λ

Explanation:

Given a metal, wavelength of light used $= λ$
Stopping potential $= V$
If $λ_{0}$ be the threshold wavelength, then maximum kinetic energy of emitted electrons
$K_{max} = h c (\frac{1}{λ} - \frac{1}{λ_{0}}) (1)$
Again, wavelength of used light $λ^{'} = 2 λ$
Stopping potential $V^{'} = \frac{V}{3}$
then $K E_{max} = h c (\frac{1}{λ^{'}} - \frac{1}{λ_{0}})$
$\Rightarrow e V^{'} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})$
$\Rightarrow \frac{e V}{3} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}}) (2)$
From eq.(1) and (2), we have
$\Rightarrow \frac{e V}{\frac{e V}{3}} = \frac{h c (\frac{1}{λ} - \frac{1}{λ_{0}})}{h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})}$
$3 (\frac{1}{2 λ} - \frac{1}{λ_{0}}) = \frac{1}{λ} - \frac{1}{λ_{0}}$
$\Rightarrow λ_{0} = 4 λ$
So threshold wavelength is 4 times of wavelength of light.

MHTCET - 2019

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357767 The graph of stopping potential $v_{s}$ against frequency $v_{0}$ of incident radiation is plotted for two different metals $P$ and $Q$ as shown in the graph. $ϕ_{p}$ and $ϕ_{Q}$ are work - functions of $P$ and $Q$ respectively, then
supporting img

1

ϕ_{P} > ϕ_{Q}

2

ϕ_{P} < ϕ_{Q}

3

ϕ_{P} = ϕ_{Q}

4

v_{0}^{'} < v_{0}

Explanation:

The work - function of a surface is
$ϕ = h v_{0}$
where, $h =$ planck's constant
and $v_{0} =$ threshold frequency
From graph it is clear that
$\begin{aligned} {(v_{0})}_{P} < {(v_{0})}_{Q} \\ ∴ ϕ_{P} < ϕ_{Q} \end{aligned}$

MHTCET - 2020