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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357738 Radiation of wavelength $λ$ , is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted electron will be:

1

> v {(\frac{4}{3})}^{1 / 2}

2

< v {(\frac{4}{3})}^{1 / 2}

3

= v {(\frac{4}{3})}^{1 / 2}

4

= v {(\frac{3}{4})}^{1 / 2}

Explanation:

According to Einstein's photoelectric effectmaximum kinetic energy of a photoelectron,
$K E = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
Assume speed of faster photoelectron is $v^{'}$
when incident photon has wavelength $\frac{3 λ}{4}$ .
$∴ \frac{1}{2} m v^{' 2} = \frac{4 h c}{3 λ} - ϕ$
or, $\frac{1}{2} m v^{' 2} = \frac{4}{3} (\frac{1}{2} m v^{2} + ϕ) - ϕ$
$\begin{aligned} v^{'} = \sqrt{\frac{4}{3} v^{2} + \frac{2 ϕ}{3 m}} \\ ∴ v^{'} > \sqrt{\frac{4}{3}} v \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357739 According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal versus the frequency, of the incident radiation gives a straight line whose slope

1 Is the same for all metals and independent of the intensity of the radiation

2 Depends on the intensity of the radiation

3 Depends both on the intensity of the radiation and the metal used

4 Depends on the nature of the metals used

Explanation:

$h v = h v_{o} + K E_{max} \Rightarrow K E_{max} = h v - h v_{o}$
On comparing this equation with $y = m x + c$ we get $m = h =$ Universal constant.
So the slope does not depend on anything.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357740 In the figure shown electromagnetic radiations of wavelength $200 n m$ are incident on the metallic plate $A$ . The photo electrons are accelerated by a potential difference $10 V$ . These electrons strike another metal plate $B$ from which electromagnetic radiations are emitted. The minimum wavelength of the emitted photons is $100 n m$ . If the work function of the metal ' $A$ ' is found to be $1.9 N e V$ . The value of $N$ is ____ .
supporting img
Use $h v = 12400 e V$ $A^{\circ}$ , use $R c h = 13.6 e V .$

1

2.5 e V

2

5.2 e V

3

1.8 e V

4

3.8 e V

Explanation:

Energy of incident photons $= \frac{12400}{2000} e V$ .
Maximum kinetic energy of ejected electrons $= \frac{12400}{2000} - ϕ$
Maximum kinetic energy of electrons striking plate
$B = \frac{12400}{2000} - ϕ + 10$
Minimum wavelength of photons emitted from
$B = \frac{12400}{1000}$
$∴ \frac{12400}{2000} - ϕ + 10 = \frac{12400}{1000} ∴ ϕ = 3.8 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357741 Maximum velocity of the photoelectrons emitted by a metal surface $1.2 \times 10^{6} m s^{- 1}$ . Assuming the specific charge of the electron to be $1.8 \times 10^{11} C k g^{- 1},$ the value of the stopping potential in volt will be

1 2

2 3

3 4

4 6

Explanation:

Specific charge of electron,
$e / m = 1.8 \times 10^{11} C k g^{- 1}$
Maximum kinetic energy of photoelectrons
$\frac{1}{2} m v_{{max}^{2}} = e V_{s}$
where $V_{s}$ is the stopping potential
$\Rightarrow V_{s} = \frac{v_{{max}^{2}}}{2 (e / m)} = \frac{{(1.2 \times 10^{6})}^{2}}{2 \times 1.8 \times 10^{11}} = 4 V$

KCET - 2006

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357742 Work-function for caesium metal is $2.14 e V .$ Let a beam of light of frequency $6 \times 10^{14} H z$ is incident over the metal surface. Now, match the following columns and choose the correct option
Column I
Column II
A
maximum $K E$ of emitted photoelectrons (in $e V$ )
P
334 $k m s^{- 1}$
B
Minimum $K E$ of emitted photoelectrons (in $e V$
Q
345 $m V$
C
Stopping potential of material ( in $m V$ ) is
R
$0.345 e V$
D
Maximum speed of the emitted photoelectrons ( in $k m s^{- 1}$ )
S
0

1

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

2

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

3

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

4

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

Explanation:

Maximum $K . E$ is given by
$K_{m a x} = h f - ϕ_{0}$
$= \frac{6.62 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} - 2.14$
$= 2.485 - 2.140 = 0.345 e V$
$K_{m i n} = 0, (s o B \to 4)$
Stopping potential, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.345 e V}{e}$
$\Rightarrow V_{0} = 0.345 V = 345 m V$
Maximum speed of emitted electrons,
$K_{m a x} = \frac{1}{2} m v_{m a x}^{2}$
$o r v_{m a x} = \sqrt{\frac{2 K_{m a x}}{m}} = \sqrt{\frac{2 \times 0.345 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$
$= 3.34 \times 10^{5} m s^{- 1} = 334 k m s^{- 1}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357738 Radiation of wavelength $λ$ , is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted electron will be:

1

> v {(\frac{4}{3})}^{1 / 2}

2

< v {(\frac{4}{3})}^{1 / 2}

3

= v {(\frac{4}{3})}^{1 / 2}

4

= v {(\frac{3}{4})}^{1 / 2}

Explanation:

According to Einstein's photoelectric effectmaximum kinetic energy of a photoelectron,
$K E = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
Assume speed of faster photoelectron is $v^{'}$
when incident photon has wavelength $\frac{3 λ}{4}$ .
$∴ \frac{1}{2} m v^{' 2} = \frac{4 h c}{3 λ} - ϕ$
or, $\frac{1}{2} m v^{' 2} = \frac{4}{3} (\frac{1}{2} m v^{2} + ϕ) - ϕ$
$\begin{aligned} v^{'} = \sqrt{\frac{4}{3} v^{2} + \frac{2 ϕ}{3 m}} \\ ∴ v^{'} > \sqrt{\frac{4}{3}} v \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357739 According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal versus the frequency, of the incident radiation gives a straight line whose slope

1 Is the same for all metals and independent of the intensity of the radiation

2 Depends on the intensity of the radiation

3 Depends both on the intensity of the radiation and the metal used

4 Depends on the nature of the metals used

Explanation:

$h v = h v_{o} + K E_{max} \Rightarrow K E_{max} = h v - h v_{o}$
On comparing this equation with $y = m x + c$ we get $m = h =$ Universal constant.
So the slope does not depend on anything.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357740 In the figure shown electromagnetic radiations of wavelength $200 n m$ are incident on the metallic plate $A$ . The photo electrons are accelerated by a potential difference $10 V$ . These electrons strike another metal plate $B$ from which electromagnetic radiations are emitted. The minimum wavelength of the emitted photons is $100 n m$ . If the work function of the metal ' $A$ ' is found to be $1.9 N e V$ . The value of $N$ is ____ .
supporting img
Use $h v = 12400 e V$ $A^{\circ}$ , use $R c h = 13.6 e V .$

1

2.5 e V

2

5.2 e V

3

1.8 e V

4

3.8 e V

Explanation:

Energy of incident photons $= \frac{12400}{2000} e V$ .
Maximum kinetic energy of ejected electrons $= \frac{12400}{2000} - ϕ$
Maximum kinetic energy of electrons striking plate
$B = \frac{12400}{2000} - ϕ + 10$
Minimum wavelength of photons emitted from
$B = \frac{12400}{1000}$
$∴ \frac{12400}{2000} - ϕ + 10 = \frac{12400}{1000} ∴ ϕ = 3.8 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357741 Maximum velocity of the photoelectrons emitted by a metal surface $1.2 \times 10^{6} m s^{- 1}$ . Assuming the specific charge of the electron to be $1.8 \times 10^{11} C k g^{- 1},$ the value of the stopping potential in volt will be

1 2

2 3

3 4

4 6

Explanation:

Specific charge of electron,
$e / m = 1.8 \times 10^{11} C k g^{- 1}$
Maximum kinetic energy of photoelectrons
$\frac{1}{2} m v_{{max}^{2}} = e V_{s}$
where $V_{s}$ is the stopping potential
$\Rightarrow V_{s} = \frac{v_{{max}^{2}}}{2 (e / m)} = \frac{{(1.2 \times 10^{6})}^{2}}{2 \times 1.8 \times 10^{11}} = 4 V$

KCET - 2006

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357742 Work-function for caesium metal is $2.14 e V .$ Let a beam of light of frequency $6 \times 10^{14} H z$ is incident over the metal surface. Now, match the following columns and choose the correct option
Column I
Column II
A
maximum $K E$ of emitted photoelectrons (in $e V$ )
P
334 $k m s^{- 1}$
B
Minimum $K E$ of emitted photoelectrons (in $e V$
Q
345 $m V$
C
Stopping potential of material ( in $m V$ ) is
R
$0.345 e V$
D
Maximum speed of the emitted photoelectrons ( in $k m s^{- 1}$ )
S
0

1

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

2

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

3

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

4

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

Explanation:

Maximum $K . E$ is given by
$K_{m a x} = h f - ϕ_{0}$
$= \frac{6.62 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} - 2.14$
$= 2.485 - 2.140 = 0.345 e V$
$K_{m i n} = 0, (s o B \to 4)$
Stopping potential, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.345 e V}{e}$
$\Rightarrow V_{0} = 0.345 V = 345 m V$
Maximum speed of emitted electrons,
$K_{m a x} = \frac{1}{2} m v_{m a x}^{2}$
$o r v_{m a x} = \sqrt{\frac{2 K_{m a x}}{m}} = \sqrt{\frac{2 \times 0.345 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$
$= 3.34 \times 10^{5} m s^{- 1} = 334 k m s^{- 1}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357738 Radiation of wavelength $λ$ , is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted electron will be:

1

> v {(\frac{4}{3})}^{1 / 2}

2

< v {(\frac{4}{3})}^{1 / 2}

3

= v {(\frac{4}{3})}^{1 / 2}

4

= v {(\frac{3}{4})}^{1 / 2}

Explanation:

According to Einstein's photoelectric effectmaximum kinetic energy of a photoelectron,
$K E = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
Assume speed of faster photoelectron is $v^{'}$
when incident photon has wavelength $\frac{3 λ}{4}$ .
$∴ \frac{1}{2} m v^{' 2} = \frac{4 h c}{3 λ} - ϕ$
or, $\frac{1}{2} m v^{' 2} = \frac{4}{3} (\frac{1}{2} m v^{2} + ϕ) - ϕ$
$\begin{aligned} v^{'} = \sqrt{\frac{4}{3} v^{2} + \frac{2 ϕ}{3 m}} \\ ∴ v^{'} > \sqrt{\frac{4}{3}} v \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357739 According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal versus the frequency, of the incident radiation gives a straight line whose slope

1 Is the same for all metals and independent of the intensity of the radiation

2 Depends on the intensity of the radiation

3 Depends both on the intensity of the radiation and the metal used

4 Depends on the nature of the metals used

Explanation:

$h v = h v_{o} + K E_{max} \Rightarrow K E_{max} = h v - h v_{o}$
On comparing this equation with $y = m x + c$ we get $m = h =$ Universal constant.
So the slope does not depend on anything.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357740 In the figure shown electromagnetic radiations of wavelength $200 n m$ are incident on the metallic plate $A$ . The photo electrons are accelerated by a potential difference $10 V$ . These electrons strike another metal plate $B$ from which electromagnetic radiations are emitted. The minimum wavelength of the emitted photons is $100 n m$ . If the work function of the metal ' $A$ ' is found to be $1.9 N e V$ . The value of $N$ is ____ .
supporting img
Use $h v = 12400 e V$ $A^{\circ}$ , use $R c h = 13.6 e V .$

1

2.5 e V

2

5.2 e V

3

1.8 e V

4

3.8 e V

Explanation:

Energy of incident photons $= \frac{12400}{2000} e V$ .
Maximum kinetic energy of ejected electrons $= \frac{12400}{2000} - ϕ$
Maximum kinetic energy of electrons striking plate
$B = \frac{12400}{2000} - ϕ + 10$
Minimum wavelength of photons emitted from
$B = \frac{12400}{1000}$
$∴ \frac{12400}{2000} - ϕ + 10 = \frac{12400}{1000} ∴ ϕ = 3.8 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357741 Maximum velocity of the photoelectrons emitted by a metal surface $1.2 \times 10^{6} m s^{- 1}$ . Assuming the specific charge of the electron to be $1.8 \times 10^{11} C k g^{- 1},$ the value of the stopping potential in volt will be

1 2

2 3

3 4

4 6

Explanation:

Specific charge of electron,
$e / m = 1.8 \times 10^{11} C k g^{- 1}$
Maximum kinetic energy of photoelectrons
$\frac{1}{2} m v_{{max}^{2}} = e V_{s}$
where $V_{s}$ is the stopping potential
$\Rightarrow V_{s} = \frac{v_{{max}^{2}}}{2 (e / m)} = \frac{{(1.2 \times 10^{6})}^{2}}{2 \times 1.8 \times 10^{11}} = 4 V$

KCET - 2006

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357742 Work-function for caesium metal is $2.14 e V .$ Let a beam of light of frequency $6 \times 10^{14} H z$ is incident over the metal surface. Now, match the following columns and choose the correct option
Column I
Column II
A
maximum $K E$ of emitted photoelectrons (in $e V$ )
P
334 $k m s^{- 1}$
B
Minimum $K E$ of emitted photoelectrons (in $e V$
Q
345 $m V$
C
Stopping potential of material ( in $m V$ ) is
R
$0.345 e V$
D
Maximum speed of the emitted photoelectrons ( in $k m s^{- 1}$ )
S
0

1

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

2

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

3

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

4

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

Explanation:

Maximum $K . E$ is given by
$K_{m a x} = h f - ϕ_{0}$
$= \frac{6.62 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} - 2.14$
$= 2.485 - 2.140 = 0.345 e V$
$K_{m i n} = 0, (s o B \to 4)$
Stopping potential, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.345 e V}{e}$
$\Rightarrow V_{0} = 0.345 V = 345 m V$
Maximum speed of emitted electrons,
$K_{m a x} = \frac{1}{2} m v_{m a x}^{2}$
$o r v_{m a x} = \sqrt{\frac{2 K_{m a x}}{m}} = \sqrt{\frac{2 \times 0.345 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$
$= 3.34 \times 10^{5} m s^{- 1} = 334 k m s^{- 1}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357738 Radiation of wavelength $λ$ , is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted electron will be:

1

> v {(\frac{4}{3})}^{1 / 2}

2

< v {(\frac{4}{3})}^{1 / 2}

3

= v {(\frac{4}{3})}^{1 / 2}

4

= v {(\frac{3}{4})}^{1 / 2}

Explanation:

According to Einstein's photoelectric effectmaximum kinetic energy of a photoelectron,
$K E = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
Assume speed of faster photoelectron is $v^{'}$
when incident photon has wavelength $\frac{3 λ}{4}$ .
$∴ \frac{1}{2} m v^{' 2} = \frac{4 h c}{3 λ} - ϕ$
or, $\frac{1}{2} m v^{' 2} = \frac{4}{3} (\frac{1}{2} m v^{2} + ϕ) - ϕ$
$\begin{aligned} v^{'} = \sqrt{\frac{4}{3} v^{2} + \frac{2 ϕ}{3 m}} \\ ∴ v^{'} > \sqrt{\frac{4}{3}} v \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357739 According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal versus the frequency, of the incident radiation gives a straight line whose slope

1 Is the same for all metals and independent of the intensity of the radiation

2 Depends on the intensity of the radiation

3 Depends both on the intensity of the radiation and the metal used

4 Depends on the nature of the metals used

Explanation:

$h v = h v_{o} + K E_{max} \Rightarrow K E_{max} = h v - h v_{o}$
On comparing this equation with $y = m x + c$ we get $m = h =$ Universal constant.
So the slope does not depend on anything.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357740 In the figure shown electromagnetic radiations of wavelength $200 n m$ are incident on the metallic plate $A$ . The photo electrons are accelerated by a potential difference $10 V$ . These electrons strike another metal plate $B$ from which electromagnetic radiations are emitted. The minimum wavelength of the emitted photons is $100 n m$ . If the work function of the metal ' $A$ ' is found to be $1.9 N e V$ . The value of $N$ is ____ .
supporting img
Use $h v = 12400 e V$ $A^{\circ}$ , use $R c h = 13.6 e V .$

1

2.5 e V

2

5.2 e V

3

1.8 e V

4

3.8 e V

Explanation:

Energy of incident photons $= \frac{12400}{2000} e V$ .
Maximum kinetic energy of ejected electrons $= \frac{12400}{2000} - ϕ$
Maximum kinetic energy of electrons striking plate
$B = \frac{12400}{2000} - ϕ + 10$
Minimum wavelength of photons emitted from
$B = \frac{12400}{1000}$
$∴ \frac{12400}{2000} - ϕ + 10 = \frac{12400}{1000} ∴ ϕ = 3.8 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357741 Maximum velocity of the photoelectrons emitted by a metal surface $1.2 \times 10^{6} m s^{- 1}$ . Assuming the specific charge of the electron to be $1.8 \times 10^{11} C k g^{- 1},$ the value of the stopping potential in volt will be

1 2

2 3

3 4

4 6

Explanation:

Specific charge of electron,
$e / m = 1.8 \times 10^{11} C k g^{- 1}$
Maximum kinetic energy of photoelectrons
$\frac{1}{2} m v_{{max}^{2}} = e V_{s}$
where $V_{s}$ is the stopping potential
$\Rightarrow V_{s} = \frac{v_{{max}^{2}}}{2 (e / m)} = \frac{{(1.2 \times 10^{6})}^{2}}{2 \times 1.8 \times 10^{11}} = 4 V$

KCET - 2006

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357742 Work-function for caesium metal is $2.14 e V .$ Let a beam of light of frequency $6 \times 10^{14} H z$ is incident over the metal surface. Now, match the following columns and choose the correct option
Column I
Column II
A
maximum $K E$ of emitted photoelectrons (in $e V$ )
P
334 $k m s^{- 1}$
B
Minimum $K E$ of emitted photoelectrons (in $e V$
Q
345 $m V$
C
Stopping potential of material ( in $m V$ ) is
R
$0.345 e V$
D
Maximum speed of the emitted photoelectrons ( in $k m s^{- 1}$ )
S
0

1

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

2

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

3

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

4

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

Explanation:

Maximum $K . E$ is given by
$K_{m a x} = h f - ϕ_{0}$
$= \frac{6.62 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} - 2.14$
$= 2.485 - 2.140 = 0.345 e V$
$K_{m i n} = 0, (s o B \to 4)$
Stopping potential, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.345 e V}{e}$
$\Rightarrow V_{0} = 0.345 V = 345 m V$
Maximum speed of emitted electrons,
$K_{m a x} = \frac{1}{2} m v_{m a x}^{2}$
$o r v_{m a x} = \sqrt{\frac{2 K_{m a x}}{m}} = \sqrt{\frac{2 \times 0.345 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$
$= 3.34 \times 10^{5} m s^{- 1} = 334 k m s^{- 1}$
So option (2) is correct.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357738 Radiation of wavelength $λ$ , is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted electron will be:

1

> v {(\frac{4}{3})}^{1 / 2}

2

< v {(\frac{4}{3})}^{1 / 2}

3

= v {(\frac{4}{3})}^{1 / 2}

4

= v {(\frac{3}{4})}^{1 / 2}

Explanation:

According to Einstein's photoelectric effectmaximum kinetic energy of a photoelectron,
$K E = \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
Assume speed of faster photoelectron is $v^{'}$
when incident photon has wavelength $\frac{3 λ}{4}$ .
$∴ \frac{1}{2} m v^{' 2} = \frac{4 h c}{3 λ} - ϕ$
or, $\frac{1}{2} m v^{' 2} = \frac{4}{3} (\frac{1}{2} m v^{2} + ϕ) - ϕ$
$\begin{aligned} v^{'} = \sqrt{\frac{4}{3} v^{2} + \frac{2 ϕ}{3 m}} \\ ∴ v^{'} > \sqrt{\frac{4}{3}} v \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357739 According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal versus the frequency, of the incident radiation gives a straight line whose slope

1 Is the same for all metals and independent of the intensity of the radiation

2 Depends on the intensity of the radiation

3 Depends both on the intensity of the radiation and the metal used

4 Depends on the nature of the metals used

Explanation:

$h v = h v_{o} + K E_{max} \Rightarrow K E_{max} = h v - h v_{o}$
On comparing this equation with $y = m x + c$ we get $m = h =$ Universal constant.
So the slope does not depend on anything.

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357740 In the figure shown electromagnetic radiations of wavelength $200 n m$ are incident on the metallic plate $A$ . The photo electrons are accelerated by a potential difference $10 V$ . These electrons strike another metal plate $B$ from which electromagnetic radiations are emitted. The minimum wavelength of the emitted photons is $100 n m$ . If the work function of the metal ' $A$ ' is found to be $1.9 N e V$ . The value of $N$ is ____ .
supporting img
Use $h v = 12400 e V$ $A^{\circ}$ , use $R c h = 13.6 e V .$

1

2.5 e V

2

5.2 e V

3

1.8 e V

4

3.8 e V

Explanation:

Energy of incident photons $= \frac{12400}{2000} e V$ .
Maximum kinetic energy of ejected electrons $= \frac{12400}{2000} - ϕ$
Maximum kinetic energy of electrons striking plate
$B = \frac{12400}{2000} - ϕ + 10$
Minimum wavelength of photons emitted from
$B = \frac{12400}{1000}$
$∴ \frac{12400}{2000} - ϕ + 10 = \frac{12400}{1000} ∴ ϕ = 3.8 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357741 Maximum velocity of the photoelectrons emitted by a metal surface $1.2 \times 10^{6} m s^{- 1}$ . Assuming the specific charge of the electron to be $1.8 \times 10^{11} C k g^{- 1},$ the value of the stopping potential in volt will be

1 2

2 3

3 4

4 6

Explanation:

Specific charge of electron,
$e / m = 1.8 \times 10^{11} C k g^{- 1}$
Maximum kinetic energy of photoelectrons
$\frac{1}{2} m v_{{max}^{2}} = e V_{s}$
where $V_{s}$ is the stopping potential
$\Rightarrow V_{s} = \frac{v_{{max}^{2}}}{2 (e / m)} = \frac{{(1.2 \times 10^{6})}^{2}}{2 \times 1.8 \times 10^{11}} = 4 V$

KCET - 2006

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357742 Work-function for caesium metal is $2.14 e V .$ Let a beam of light of frequency $6 \times 10^{14} H z$ is incident over the metal surface. Now, match the following columns and choose the correct option
Column I
Column II
A
maximum $K E$ of emitted photoelectrons (in $e V$ )
P
334 $k m s^{- 1}$
B
Minimum $K E$ of emitted photoelectrons (in $e V$
Q
345 $m V$
C
Stopping potential of material ( in $m V$ ) is
R
$0.345 e V$
D
Maximum speed of the emitted photoelectrons ( in $k m s^{- 1}$ )
S
0

1

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

2

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

3

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

4

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

Explanation:

Maximum $K . E$ is given by
$K_{m a x} = h f - ϕ_{0}$
$= \frac{6.62 \times 10^{- 34} \times 6 \times 10^{14}}{1.6 \times 10^{- 19}} - 2.14$
$= 2.485 - 2.140 = 0.345 e V$
$K_{m i n} = 0, (s o B \to 4)$
Stopping potential, $V_{0} = \frac{K_{m a x}}{e} = \frac{0.345 e V}{e}$
$\Rightarrow V_{0} = 0.345 V = 345 m V$
Maximum speed of emitted electrons,
$K_{m a x} = \frac{1}{2} m v_{m a x}^{2}$
$o r v_{m a x} = \sqrt{\frac{2 K_{m a x}}{m}} = \sqrt{\frac{2 \times 0.345 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$
$= 3.34 \times 10^{5} m s^{- 1} = 334 k m s^{- 1}$
So option (2) is correct.

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Question Bank Series

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