QBManager

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357708 Radiation of wavelength $6561$ $A^{\circ}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{- 4} T$ . If the radius of the largest circular path followed by the electrons is $10 m m,$ the work function of the metal is close to

1

1.8 e V

2

1.1 e V

3

0.8 e V

4

1.6 e V

Explanation:

Let the work function of metal surface be $ϕ$ .
$∴ K E_{max} = \frac{h c}{λ} - ϕ$
Again, $R_{max} = \frac{\sqrt{2 m K E_{max}}}{q B} = \frac{\sqrt{2 m (\frac{h c}{λ} - ϕ)}}{q B}$
$∴ \frac{R_{max} q^{2} B^{2}}{2 m} = \frac{h c}{λ} - ϕ$
$∴ ϕ = \frac{h c}{λ} - \frac{R_{max}^{2} q^{2} B^{2}}{2 m} = 1.0899 e V \approx 1.1 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357709 A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_{0}$ . If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$ , then the radius $r$ is given by

1

\frac{\sqrt{2 e (E - W_{0})}}{m B}

2

\frac{\sqrt{2 e (W_{0} - E)}}{e B}

3

\sqrt{\frac{2 m W_{0}}{e B}}

4

\frac{\sqrt{2 m (E - W_{0})}}{e B}

Explanation:

We know that Einstein equation,
$E - W_{0} = \frac{1}{2} m v^{2}$
$\Rightarrow \sqrt{\frac{2 (E - W_{0})}{m}} = v$
and a charged particle placed in uniform magnetic field experience a force
$\begin{aligned} F = \frac{m v^{2}}{r} \\ \Rightarrow e v B = \frac{m v^{2}}{r} \\ \Rightarrow r = \frac{m v}{e B} = \frac{\sqrt{2 m (E - W_{0})}}{e B} \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357710 $U V$ light of $4.13 e V$ is incident on a photosensitive metal surface having work function $3.13 e V .$ The maximum kinetic energy of ejected photoelectrons will be

1

4.13 e V

2

1 e V

3

3.13 e V

4

7.26 e V

Explanation:

By Einstein's photoelectric equation,
$K E_{max} = E - ϕ = 4 \cdot 13 - 3.13$
$\Rightarrow K . E_{max} = 1 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357711 The stopping potential $V_{s}$ for photoelectric emission from a metal surface is plotted along $Y$ -axis and frequency $v$ of incident light along $X$ -axis. A straight line is obtained as shown. Planck's constant is given by
supporting img

1 Slope of the line

2 Product of slope on the line and charge on the electron

3 Product of intercept along

Y

-axis and mass of the electron

4 Product of slope and mass of electron

Explanation:

$K_{max} = h v - ϕ$
$\Rightarrow e V_{S} = h v - ϕ \Rightarrow V_{S} = \frac{h}{e} v - \frac{ϕ}{e}$
Comparing this equation with $y = m x + c$ , we get slope $m = \frac{h}{e} \Rightarrow h = m \times e$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357712 The threshold frequency for a certain photosensitive metal is $ν_{0}$ . When it is illuminated by light of frequency $2 ν_{0}$ , the maximum velocity of photoelectrons is $v_{0}$ . What will be the maximum velocity of the electrons when the same metal is illuminated by light of frequency $5 ν_{0}$

1

\sqrt{2} V_{0}

2

2 V_{0}

3

2 \sqrt{2} V_{0}

4

4 V_{0}

Explanation:

Maximum velocity of photo electron is
$V_{m a x} = \sqrt{\frac{2 h (υ - υ_{0})}{m}}$
$V_{o} = \sqrt{\frac{2 h (2 υ_{0} - υ_{0})}{m}} (g i v e n, υ = 2 υ_{0})$
$V_{0} = \sqrt{\frac{2 h υ_{0}}{m}} (1)$
For the incident frequency $5 υ_{0}$
$h (5 υ_{0}) - ϕ = \frac{1}{2} m V^{2}$
$h (5 υ_{0}) - h υ_{0} = \frac{1}{2} m V^{2}$
$V = 2 \sqrt{\frac{2 h β υ_{0}}{m}} = 2 υ_{0}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357708 Radiation of wavelength $6561$ $A^{\circ}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{- 4} T$ . If the radius of the largest circular path followed by the electrons is $10 m m,$ the work function of the metal is close to

1

1.8 e V

2

1.1 e V

3

0.8 e V

4

1.6 e V

Explanation:

Let the work function of metal surface be $ϕ$ .
$∴ K E_{max} = \frac{h c}{λ} - ϕ$
Again, $R_{max} = \frac{\sqrt{2 m K E_{max}}}{q B} = \frac{\sqrt{2 m (\frac{h c}{λ} - ϕ)}}{q B}$
$∴ \frac{R_{max} q^{2} B^{2}}{2 m} = \frac{h c}{λ} - ϕ$
$∴ ϕ = \frac{h c}{λ} - \frac{R_{max}^{2} q^{2} B^{2}}{2 m} = 1.0899 e V \approx 1.1 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357709 A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_{0}$ . If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$ , then the radius $r$ is given by

1

\frac{\sqrt{2 e (E - W_{0})}}{m B}

2

\frac{\sqrt{2 e (W_{0} - E)}}{e B}

3

\sqrt{\frac{2 m W_{0}}{e B}}

4

\frac{\sqrt{2 m (E - W_{0})}}{e B}

Explanation:

We know that Einstein equation,
$E - W_{0} = \frac{1}{2} m v^{2}$
$\Rightarrow \sqrt{\frac{2 (E - W_{0})}{m}} = v$
and a charged particle placed in uniform magnetic field experience a force
$\begin{aligned} F = \frac{m v^{2}}{r} \\ \Rightarrow e v B = \frac{m v^{2}}{r} \\ \Rightarrow r = \frac{m v}{e B} = \frac{\sqrt{2 m (E - W_{0})}}{e B} \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357710 $U V$ light of $4.13 e V$ is incident on a photosensitive metal surface having work function $3.13 e V .$ The maximum kinetic energy of ejected photoelectrons will be

1

4.13 e V

2

1 e V

3

3.13 e V

4

7.26 e V

Explanation:

By Einstein's photoelectric equation,
$K E_{max} = E - ϕ = 4 \cdot 13 - 3.13$
$\Rightarrow K . E_{max} = 1 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357711 The stopping potential $V_{s}$ for photoelectric emission from a metal surface is plotted along $Y$ -axis and frequency $v$ of incident light along $X$ -axis. A straight line is obtained as shown. Planck's constant is given by
supporting img

1 Slope of the line

2 Product of slope on the line and charge on the electron

3 Product of intercept along

Y

-axis and mass of the electron

4 Product of slope and mass of electron

Explanation:

$K_{max} = h v - ϕ$
$\Rightarrow e V_{S} = h v - ϕ \Rightarrow V_{S} = \frac{h}{e} v - \frac{ϕ}{e}$
Comparing this equation with $y = m x + c$ , we get slope $m = \frac{h}{e} \Rightarrow h = m \times e$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357712 The threshold frequency for a certain photosensitive metal is $ν_{0}$ . When it is illuminated by light of frequency $2 ν_{0}$ , the maximum velocity of photoelectrons is $v_{0}$ . What will be the maximum velocity of the electrons when the same metal is illuminated by light of frequency $5 ν_{0}$

1

\sqrt{2} V_{0}

2

2 V_{0}

3

2 \sqrt{2} V_{0}

4

4 V_{0}

Explanation:

Maximum velocity of photo electron is
$V_{m a x} = \sqrt{\frac{2 h (υ - υ_{0})}{m}}$
$V_{o} = \sqrt{\frac{2 h (2 υ_{0} - υ_{0})}{m}} (g i v e n, υ = 2 υ_{0})$
$V_{0} = \sqrt{\frac{2 h υ_{0}}{m}} (1)$
For the incident frequency $5 υ_{0}$
$h (5 υ_{0}) - ϕ = \frac{1}{2} m V^{2}$
$h (5 υ_{0}) - h υ_{0} = \frac{1}{2} m V^{2}$
$V = 2 \sqrt{\frac{2 h β υ_{0}}{m}} = 2 υ_{0}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357708 Radiation of wavelength $6561$ $A^{\circ}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{- 4} T$ . If the radius of the largest circular path followed by the electrons is $10 m m,$ the work function of the metal is close to

1

1.8 e V

2

1.1 e V

3

0.8 e V

4

1.6 e V

Explanation:

Let the work function of metal surface be $ϕ$ .
$∴ K E_{max} = \frac{h c}{λ} - ϕ$
Again, $R_{max} = \frac{\sqrt{2 m K E_{max}}}{q B} = \frac{\sqrt{2 m (\frac{h c}{λ} - ϕ)}}{q B}$
$∴ \frac{R_{max} q^{2} B^{2}}{2 m} = \frac{h c}{λ} - ϕ$
$∴ ϕ = \frac{h c}{λ} - \frac{R_{max}^{2} q^{2} B^{2}}{2 m} = 1.0899 e V \approx 1.1 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357709 A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_{0}$ . If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$ , then the radius $r$ is given by

1

\frac{\sqrt{2 e (E - W_{0})}}{m B}

2

\frac{\sqrt{2 e (W_{0} - E)}}{e B}

3

\sqrt{\frac{2 m W_{0}}{e B}}

4

\frac{\sqrt{2 m (E - W_{0})}}{e B}

Explanation:

We know that Einstein equation,
$E - W_{0} = \frac{1}{2} m v^{2}$
$\Rightarrow \sqrt{\frac{2 (E - W_{0})}{m}} = v$
and a charged particle placed in uniform magnetic field experience a force
$\begin{aligned} F = \frac{m v^{2}}{r} \\ \Rightarrow e v B = \frac{m v^{2}}{r} \\ \Rightarrow r = \frac{m v}{e B} = \frac{\sqrt{2 m (E - W_{0})}}{e B} \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357710 $U V$ light of $4.13 e V$ is incident on a photosensitive metal surface having work function $3.13 e V .$ The maximum kinetic energy of ejected photoelectrons will be

1

4.13 e V

2

1 e V

3

3.13 e V

4

7.26 e V

Explanation:

By Einstein's photoelectric equation,
$K E_{max} = E - ϕ = 4 \cdot 13 - 3.13$
$\Rightarrow K . E_{max} = 1 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357711 The stopping potential $V_{s}$ for photoelectric emission from a metal surface is plotted along $Y$ -axis and frequency $v$ of incident light along $X$ -axis. A straight line is obtained as shown. Planck's constant is given by
supporting img

1 Slope of the line

2 Product of slope on the line and charge on the electron

3 Product of intercept along

Y

-axis and mass of the electron

4 Product of slope and mass of electron

Explanation:

$K_{max} = h v - ϕ$
$\Rightarrow e V_{S} = h v - ϕ \Rightarrow V_{S} = \frac{h}{e} v - \frac{ϕ}{e}$
Comparing this equation with $y = m x + c$ , we get slope $m = \frac{h}{e} \Rightarrow h = m \times e$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357712 The threshold frequency for a certain photosensitive metal is $ν_{0}$ . When it is illuminated by light of frequency $2 ν_{0}$ , the maximum velocity of photoelectrons is $v_{0}$ . What will be the maximum velocity of the electrons when the same metal is illuminated by light of frequency $5 ν_{0}$

1

\sqrt{2} V_{0}

2

2 V_{0}

3

2 \sqrt{2} V_{0}

4

4 V_{0}

Explanation:

Maximum velocity of photo electron is
$V_{m a x} = \sqrt{\frac{2 h (υ - υ_{0})}{m}}$
$V_{o} = \sqrt{\frac{2 h (2 υ_{0} - υ_{0})}{m}} (g i v e n, υ = 2 υ_{0})$
$V_{0} = \sqrt{\frac{2 h υ_{0}}{m}} (1)$
For the incident frequency $5 υ_{0}$
$h (5 υ_{0}) - ϕ = \frac{1}{2} m V^{2}$
$h (5 υ_{0}) - h υ_{0} = \frac{1}{2} m V^{2}$
$V = 2 \sqrt{\frac{2 h β υ_{0}}{m}} = 2 υ_{0}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357708 Radiation of wavelength $6561$ $A^{\circ}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{- 4} T$ . If the radius of the largest circular path followed by the electrons is $10 m m,$ the work function of the metal is close to

1

1.8 e V

2

1.1 e V

3

0.8 e V

4

1.6 e V

Explanation:

Let the work function of metal surface be $ϕ$ .
$∴ K E_{max} = \frac{h c}{λ} - ϕ$
Again, $R_{max} = \frac{\sqrt{2 m K E_{max}}}{q B} = \frac{\sqrt{2 m (\frac{h c}{λ} - ϕ)}}{q B}$
$∴ \frac{R_{max} q^{2} B^{2}}{2 m} = \frac{h c}{λ} - ϕ$
$∴ ϕ = \frac{h c}{λ} - \frac{R_{max}^{2} q^{2} B^{2}}{2 m} = 1.0899 e V \approx 1.1 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357709 A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_{0}$ . If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$ , then the radius $r$ is given by

1

\frac{\sqrt{2 e (E - W_{0})}}{m B}

2

\frac{\sqrt{2 e (W_{0} - E)}}{e B}

3

\sqrt{\frac{2 m W_{0}}{e B}}

4

\frac{\sqrt{2 m (E - W_{0})}}{e B}

Explanation:

We know that Einstein equation,
$E - W_{0} = \frac{1}{2} m v^{2}$
$\Rightarrow \sqrt{\frac{2 (E - W_{0})}{m}} = v$
and a charged particle placed in uniform magnetic field experience a force
$\begin{aligned} F = \frac{m v^{2}}{r} \\ \Rightarrow e v B = \frac{m v^{2}}{r} \\ \Rightarrow r = \frac{m v}{e B} = \frac{\sqrt{2 m (E - W_{0})}}{e B} \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357710 $U V$ light of $4.13 e V$ is incident on a photosensitive metal surface having work function $3.13 e V .$ The maximum kinetic energy of ejected photoelectrons will be

1

4.13 e V

2

1 e V

3

3.13 e V

4

7.26 e V

Explanation:

By Einstein's photoelectric equation,
$K E_{max} = E - ϕ = 4 \cdot 13 - 3.13$
$\Rightarrow K . E_{max} = 1 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357711 The stopping potential $V_{s}$ for photoelectric emission from a metal surface is plotted along $Y$ -axis and frequency $v$ of incident light along $X$ -axis. A straight line is obtained as shown. Planck's constant is given by
supporting img

1 Slope of the line

2 Product of slope on the line and charge on the electron

3 Product of intercept along

Y

-axis and mass of the electron

4 Product of slope and mass of electron

Explanation:

$K_{max} = h v - ϕ$
$\Rightarrow e V_{S} = h v - ϕ \Rightarrow V_{S} = \frac{h}{e} v - \frac{ϕ}{e}$
Comparing this equation with $y = m x + c$ , we get slope $m = \frac{h}{e} \Rightarrow h = m \times e$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357712 The threshold frequency for a certain photosensitive metal is $ν_{0}$ . When it is illuminated by light of frequency $2 ν_{0}$ , the maximum velocity of photoelectrons is $v_{0}$ . What will be the maximum velocity of the electrons when the same metal is illuminated by light of frequency $5 ν_{0}$

1

\sqrt{2} V_{0}

2

2 V_{0}

3

2 \sqrt{2} V_{0}

4

4 V_{0}

Explanation:

Maximum velocity of photo electron is
$V_{m a x} = \sqrt{\frac{2 h (υ - υ_{0})}{m}}$
$V_{o} = \sqrt{\frac{2 h (2 υ_{0} - υ_{0})}{m}} (g i v e n, υ = 2 υ_{0})$
$V_{0} = \sqrt{\frac{2 h υ_{0}}{m}} (1)$
For the incident frequency $5 υ_{0}$
$h (5 υ_{0}) - ϕ = \frac{1}{2} m V^{2}$
$h (5 υ_{0}) - h υ_{0} = \frac{1}{2} m V^{2}$
$V = 2 \sqrt{\frac{2 h β υ_{0}}{m}} = 2 υ_{0}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357708 Radiation of wavelength $6561$ $A^{\circ}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{- 4} T$ . If the radius of the largest circular path followed by the electrons is $10 m m,$ the work function of the metal is close to

1

1.8 e V

2

1.1 e V

3

0.8 e V

4

1.6 e V

Explanation:

Let the work function of metal surface be $ϕ$ .
$∴ K E_{max} = \frac{h c}{λ} - ϕ$
Again, $R_{max} = \frac{\sqrt{2 m K E_{max}}}{q B} = \frac{\sqrt{2 m (\frac{h c}{λ} - ϕ)}}{q B}$
$∴ \frac{R_{max} q^{2} B^{2}}{2 m} = \frac{h c}{λ} - ϕ$
$∴ ϕ = \frac{h c}{λ} - \frac{R_{max}^{2} q^{2} B^{2}}{2 m} = 1.0899 e V \approx 1.1 e V$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357709 A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_{0}$ . If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$ , then the radius $r$ is given by

1

\frac{\sqrt{2 e (E - W_{0})}}{m B}

2

\frac{\sqrt{2 e (W_{0} - E)}}{e B}

3

\sqrt{\frac{2 m W_{0}}{e B}}

4

\frac{\sqrt{2 m (E - W_{0})}}{e B}

Explanation:

We know that Einstein equation,
$E - W_{0} = \frac{1}{2} m v^{2}$
$\Rightarrow \sqrt{\frac{2 (E - W_{0})}{m}} = v$
and a charged particle placed in uniform magnetic field experience a force
$\begin{aligned} F = \frac{m v^{2}}{r} \\ \Rightarrow e v B = \frac{m v^{2}}{r} \\ \Rightarrow r = \frac{m v}{e B} = \frac{\sqrt{2 m (E - W_{0})}}{e B} \end{aligned}$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357710 $U V$ light of $4.13 e V$ is incident on a photosensitive metal surface having work function $3.13 e V .$ The maximum kinetic energy of ejected photoelectrons will be

1

4.13 e V

2

1 e V

3

3.13 e V

4

7.26 e V

Explanation:

By Einstein's photoelectric equation,
$K E_{max} = E - ϕ = 4 \cdot 13 - 3.13$
$\Rightarrow K . E_{max} = 1 e V$

JEE - 2024

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357711 The stopping potential $V_{s}$ for photoelectric emission from a metal surface is plotted along $Y$ -axis and frequency $v$ of incident light along $X$ -axis. A straight line is obtained as shown. Planck's constant is given by
supporting img

1 Slope of the line

2 Product of slope on the line and charge on the electron

3 Product of intercept along

Y

-axis and mass of the electron

4 Product of slope and mass of electron

Explanation:

$K_{max} = h v - ϕ$
$\Rightarrow e V_{S} = h v - ϕ \Rightarrow V_{S} = \frac{h}{e} v - \frac{ϕ}{e}$
Comparing this equation with $y = m x + c$ , we get slope $m = \frac{h}{e} \Rightarrow h = m \times e$ .

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357712 The threshold frequency for a certain photosensitive metal is $ν_{0}$ . When it is illuminated by light of frequency $2 ν_{0}$ , the maximum velocity of photoelectrons is $v_{0}$ . What will be the maximum velocity of the electrons when the same metal is illuminated by light of frequency $5 ν_{0}$

1

\sqrt{2} V_{0}

2

2 V_{0}

3

2 \sqrt{2} V_{0}

4

4 V_{0}

Explanation:

Maximum velocity of photo electron is
$V_{m a x} = \sqrt{\frac{2 h (υ - υ_{0})}{m}}$
$V_{o} = \sqrt{\frac{2 h (2 υ_{0} - υ_{0})}{m}} (g i v e n, υ = 2 υ_{0})$
$V_{0} = \sqrt{\frac{2 h υ_{0}}{m}} (1)$
For the incident frequency $5 υ_{0}$
$h (5 υ_{0}) - ϕ = \frac{1}{2} m V^{2}$
$h (5 υ_{0}) - h υ_{0} = \frac{1}{2} m V^{2}$
$V = 2 \sqrt{\frac{2 h β υ_{0}}{m}} = 2 υ_{0}$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: