QBManager

Dual nature of radiation and Matter

142146 Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ . If the velocities of the photoelectrons of mass $m$ coming out are respectively $v_{1}$ and $v_{2}$ , then

1

v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})

2

v_{1} + v_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{1 / 2}

3

v_{1}^{2} v_{2}^{2} = \frac{2 h}{m} (f_{1} + f 2)

4

v_{1} - v_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{1 / 2}

Explanation:

A We know that,
${KE}_{max} = h v - h v_{o}$
$\frac{1}{2} {mv}_{1}^{2} = h v_{1} - h v_{o}$
$\frac{1}{2} {mv}_{2}^{2} = h v_{2} - h v_{o}$
Subtract equation (ii) by equation (i), we get -
$h (v_{1} - v_{2}) = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2})$
$h (f_{1} - f_{2}) = \frac{m}{2} (v_{1}^{2} - v_{2}^{2})$
$v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

Assam CEE-2016

Dual nature of radiation and Matter

142147 The graph $\frac{1}{λ}$ and stopping potential (V) of three metals having work function $ϕ_{1}, ϕ_{2}$ and $ϕ_{3}$ in an experiment of photoelectric effect is plotted as shown in the figure. Which one of the following statement is /are correct? [Here $λ$ is the wavelength of the incident rayl

(I) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 1 : 2 : 4$
(II) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 4 : 2 : 1$
(III) $Tan θ \propto \frac{hc}{e}$ where $h =$ Planck's constant, $c =$ speed of light
(IV) The violet colour-light can eject photoelectrons from metals 2 and 3

1 (i), (iii)

2 (i), (iv)

3 (ii), (iii)

4 (i), (ii) and (iv)

Explanation:

A From the graph,
For metal 1, $\frac{ϕ_{1}}{hc} = 0.001$
For metal 2, $\frac{ϕ_{2}}{hc} = 0.002$
For metal 3, $\frac{ϕ_{3}}{hc} = 0.004$
Ratio of work function,
$ϕ_{1} : ϕ_{2} : ϕ_{3} = 0.001 : 0.002 : 0.004$
$= 1 : 2 : 4$
Therefore option (i) is correct.
According to the Einstein photoelectric equation -
$eV = \frac{hc}{λ} - ϕ_{o}$
$V = [\frac{hc}{e}] [\frac{1}{λ}] - \frac{ϕ_{o}}{e}$
$∵$ It is an equation of straight line $∴$ Slope, $\tan θ = \frac{hc}{e}$
Therefore, option (iii) is correct.
For metal 2: $\frac{1}{λ_{2}} = 0.002 nm \Rightarrow λ_{2} = 500 nm$
For metal 3: $\frac{1}{λ_{3}} = 0.004 nm \Rightarrow λ_{3} = 250 nm$
So, metal 2 can eject violet colour light, as the wavelength of violet colour is around $400 nm$ and metal 2 shows the nearby value.
But the value of metal 3 does not lie in violet wavelength range.

JIPMER-2016

Dual nature of radiation and Matter

142148 Photoelectric emission is observed from a metallic surface for frequency $v_{1}$ and $v_{2}$ of the incident light $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1 : n$ , then the threshold frequency of the metallic surface is

1

\frac{(v_{1} - v_{2})}{(n - 1)}

2

\frac{(n v_{1} - v_{2})}{(n - 1)}

3

\frac{(n v_{2} - v_{1})}{(n - 1)}

4

\frac{(v_{1} - v_{2})}{n}

Explanation:

B From Einstein photoelectric equation, $h v - h v_{0} = 0$
Case-I : $h (v_{1} - v_{0}) = n_{1}$
Case-II : $h (v_{2} - v_{0}) = n_{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1} - v_{0}}{v_{2} - v_{0}} = \frac{n_{1}}{n_{2}} = \frac{1}{n}$
$n (v_{1} - v_{o}) = v_{2} - v_{o} [∵ \frac{n_{1}}{n_{2}} = \frac{1}{n}]$
$Or v_{0} = \frac{n v_{1} - v_{2}}{n - 1}$

EAMCET-2001

Dual nature of radiation and Matter

142150 In photoelectric effect, the work function of a metal is $3.5 eV$ . The emitted electrons can be stopped by applying a potential of $- 1.2 V$ . Then,

1 the energy of the incident photons is

4.7 eV

2 the energy of the incident photons is

2.3 eV

3 if higher frequency photons be used, the photoelectric current will rise

4 When the energy of photons is

3.5 eV

, the photoelectric current will be maximum

Explanation:

A Given that, $ϕ = 3.5 eV$
$V_{o} = - 1.2 V$
or ${eV}_{o} = - 1.2 eV$
Einstein photoelectric effect equation-
$(K.E)_{max} = h v - ϕ$
${eV}_{o} = h v - ϕ$
$1.2 = h v - 3.5$
$1.2 - h v + 3.5 = 0$
$h v = 4.7 eV$
Hence, the energy of the incident photons is $4.7 eV$ .

AIPMT - 1994

Dual nature of radiation and Matter

142146 Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ . If the velocities of the photoelectrons of mass $m$ coming out are respectively $v_{1}$ and $v_{2}$ , then

1

v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})

2

v_{1} + v_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{1 / 2}

3

v_{1}^{2} v_{2}^{2} = \frac{2 h}{m} (f_{1} + f 2)

4

v_{1} - v_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{1 / 2}

Explanation:

A We know that,
${KE}_{max} = h v - h v_{o}$
$\frac{1}{2} {mv}_{1}^{2} = h v_{1} - h v_{o}$
$\frac{1}{2} {mv}_{2}^{2} = h v_{2} - h v_{o}$
Subtract equation (ii) by equation (i), we get -
$h (v_{1} - v_{2}) = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2})$
$h (f_{1} - f_{2}) = \frac{m}{2} (v_{1}^{2} - v_{2}^{2})$
$v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

Assam CEE-2016

Dual nature of radiation and Matter

142147 The graph $\frac{1}{λ}$ and stopping potential (V) of three metals having work function $ϕ_{1}, ϕ_{2}$ and $ϕ_{3}$ in an experiment of photoelectric effect is plotted as shown in the figure. Which one of the following statement is /are correct? [Here $λ$ is the wavelength of the incident rayl

(I) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 1 : 2 : 4$
(II) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 4 : 2 : 1$
(III) $Tan θ \propto \frac{hc}{e}$ where $h =$ Planck's constant, $c =$ speed of light
(IV) The violet colour-light can eject photoelectrons from metals 2 and 3

1 (i), (iii)

2 (i), (iv)

3 (ii), (iii)

4 (i), (ii) and (iv)

Explanation:

A From the graph,
For metal 1, $\frac{ϕ_{1}}{hc} = 0.001$
For metal 2, $\frac{ϕ_{2}}{hc} = 0.002$
For metal 3, $\frac{ϕ_{3}}{hc} = 0.004$
Ratio of work function,
$ϕ_{1} : ϕ_{2} : ϕ_{3} = 0.001 : 0.002 : 0.004$
$= 1 : 2 : 4$
Therefore option (i) is correct.
According to the Einstein photoelectric equation -
$eV = \frac{hc}{λ} - ϕ_{o}$
$V = [\frac{hc}{e}] [\frac{1}{λ}] - \frac{ϕ_{o}}{e}$
$∵$ It is an equation of straight line $∴$ Slope, $\tan θ = \frac{hc}{e}$
Therefore, option (iii) is correct.
For metal 2: $\frac{1}{λ_{2}} = 0.002 nm \Rightarrow λ_{2} = 500 nm$
For metal 3: $\frac{1}{λ_{3}} = 0.004 nm \Rightarrow λ_{3} = 250 nm$
So, metal 2 can eject violet colour light, as the wavelength of violet colour is around $400 nm$ and metal 2 shows the nearby value.
But the value of metal 3 does not lie in violet wavelength range.

JIPMER-2016

Dual nature of radiation and Matter

142148 Photoelectric emission is observed from a metallic surface for frequency $v_{1}$ and $v_{2}$ of the incident light $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1 : n$ , then the threshold frequency of the metallic surface is

1

\frac{(v_{1} - v_{2})}{(n - 1)}

2

\frac{(n v_{1} - v_{2})}{(n - 1)}

3

\frac{(n v_{2} - v_{1})}{(n - 1)}

4

\frac{(v_{1} - v_{2})}{n}

Explanation:

B From Einstein photoelectric equation, $h v - h v_{0} = 0$
Case-I : $h (v_{1} - v_{0}) = n_{1}$
Case-II : $h (v_{2} - v_{0}) = n_{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1} - v_{0}}{v_{2} - v_{0}} = \frac{n_{1}}{n_{2}} = \frac{1}{n}$
$n (v_{1} - v_{o}) = v_{2} - v_{o} [∵ \frac{n_{1}}{n_{2}} = \frac{1}{n}]$
$Or v_{0} = \frac{n v_{1} - v_{2}}{n - 1}$

EAMCET-2001

Dual nature of radiation and Matter

142150 In photoelectric effect, the work function of a metal is $3.5 eV$ . The emitted electrons can be stopped by applying a potential of $- 1.2 V$ . Then,

1 the energy of the incident photons is

4.7 eV

2 the energy of the incident photons is

2.3 eV

3 if higher frequency photons be used, the photoelectric current will rise

4 When the energy of photons is

3.5 eV

, the photoelectric current will be maximum

Explanation:

A Given that, $ϕ = 3.5 eV$
$V_{o} = - 1.2 V$
or ${eV}_{o} = - 1.2 eV$
Einstein photoelectric effect equation-
$(K.E)_{max} = h v - ϕ$
${eV}_{o} = h v - ϕ$
$1.2 = h v - 3.5$
$1.2 - h v + 3.5 = 0$
$h v = 4.7 eV$
Hence, the energy of the incident photons is $4.7 eV$ .

AIPMT - 1994

Dual nature of radiation and Matter

142146 Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ . If the velocities of the photoelectrons of mass $m$ coming out are respectively $v_{1}$ and $v_{2}$ , then

1

v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})

2

v_{1} + v_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{1 / 2}

3

v_{1}^{2} v_{2}^{2} = \frac{2 h}{m} (f_{1} + f 2)

4

v_{1} - v_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{1 / 2}

Explanation:

A We know that,
${KE}_{max} = h v - h v_{o}$
$\frac{1}{2} {mv}_{1}^{2} = h v_{1} - h v_{o}$
$\frac{1}{2} {mv}_{2}^{2} = h v_{2} - h v_{o}$
Subtract equation (ii) by equation (i), we get -
$h (v_{1} - v_{2}) = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2})$
$h (f_{1} - f_{2}) = \frac{m}{2} (v_{1}^{2} - v_{2}^{2})$
$v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

Assam CEE-2016

Dual nature of radiation and Matter

142147 The graph $\frac{1}{λ}$ and stopping potential (V) of three metals having work function $ϕ_{1}, ϕ_{2}$ and $ϕ_{3}$ in an experiment of photoelectric effect is plotted as shown in the figure. Which one of the following statement is /are correct? [Here $λ$ is the wavelength of the incident rayl

(I) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 1 : 2 : 4$
(II) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 4 : 2 : 1$
(III) $Tan θ \propto \frac{hc}{e}$ where $h =$ Planck's constant, $c =$ speed of light
(IV) The violet colour-light can eject photoelectrons from metals 2 and 3

1 (i), (iii)

2 (i), (iv)

3 (ii), (iii)

4 (i), (ii) and (iv)

Explanation:

A From the graph,
For metal 1, $\frac{ϕ_{1}}{hc} = 0.001$
For metal 2, $\frac{ϕ_{2}}{hc} = 0.002$
For metal 3, $\frac{ϕ_{3}}{hc} = 0.004$
Ratio of work function,
$ϕ_{1} : ϕ_{2} : ϕ_{3} = 0.001 : 0.002 : 0.004$
$= 1 : 2 : 4$
Therefore option (i) is correct.
According to the Einstein photoelectric equation -
$eV = \frac{hc}{λ} - ϕ_{o}$
$V = [\frac{hc}{e}] [\frac{1}{λ}] - \frac{ϕ_{o}}{e}$
$∵$ It is an equation of straight line $∴$ Slope, $\tan θ = \frac{hc}{e}$
Therefore, option (iii) is correct.
For metal 2: $\frac{1}{λ_{2}} = 0.002 nm \Rightarrow λ_{2} = 500 nm$
For metal 3: $\frac{1}{λ_{3}} = 0.004 nm \Rightarrow λ_{3} = 250 nm$
So, metal 2 can eject violet colour light, as the wavelength of violet colour is around $400 nm$ and metal 2 shows the nearby value.
But the value of metal 3 does not lie in violet wavelength range.

JIPMER-2016

Dual nature of radiation and Matter

142148 Photoelectric emission is observed from a metallic surface for frequency $v_{1}$ and $v_{2}$ of the incident light $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1 : n$ , then the threshold frequency of the metallic surface is

1

\frac{(v_{1} - v_{2})}{(n - 1)}

2

\frac{(n v_{1} - v_{2})}{(n - 1)}

3

\frac{(n v_{2} - v_{1})}{(n - 1)}

4

\frac{(v_{1} - v_{2})}{n}

Explanation:

B From Einstein photoelectric equation, $h v - h v_{0} = 0$
Case-I : $h (v_{1} - v_{0}) = n_{1}$
Case-II : $h (v_{2} - v_{0}) = n_{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1} - v_{0}}{v_{2} - v_{0}} = \frac{n_{1}}{n_{2}} = \frac{1}{n}$
$n (v_{1} - v_{o}) = v_{2} - v_{o} [∵ \frac{n_{1}}{n_{2}} = \frac{1}{n}]$
$Or v_{0} = \frac{n v_{1} - v_{2}}{n - 1}$

EAMCET-2001

Dual nature of radiation and Matter

142150 In photoelectric effect, the work function of a metal is $3.5 eV$ . The emitted electrons can be stopped by applying a potential of $- 1.2 V$ . Then,

1 the energy of the incident photons is

4.7 eV

2 the energy of the incident photons is

2.3 eV

3 if higher frequency photons be used, the photoelectric current will rise

4 When the energy of photons is

3.5 eV

, the photoelectric current will be maximum

Explanation:

A Given that, $ϕ = 3.5 eV$
$V_{o} = - 1.2 V$
or ${eV}_{o} = - 1.2 eV$
Einstein photoelectric effect equation-
$(K.E)_{max} = h v - ϕ$
${eV}_{o} = h v - ϕ$
$1.2 = h v - 3.5$
$1.2 - h v + 3.5 = 0$
$h v = 4.7 eV$
Hence, the energy of the incident photons is $4.7 eV$ .

AIPMT - 1994

Dual nature of radiation and Matter

142146 Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ . If the velocities of the photoelectrons of mass $m$ coming out are respectively $v_{1}$ and $v_{2}$ , then

1

v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})

2

v_{1} + v_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{1 / 2}

3

v_{1}^{2} v_{2}^{2} = \frac{2 h}{m} (f_{1} + f 2)

4

v_{1} - v_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{1 / 2}

Explanation:

A We know that,
${KE}_{max} = h v - h v_{o}$
$\frac{1}{2} {mv}_{1}^{2} = h v_{1} - h v_{o}$
$\frac{1}{2} {mv}_{2}^{2} = h v_{2} - h v_{o}$
Subtract equation (ii) by equation (i), we get -
$h (v_{1} - v_{2}) = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2})$
$h (f_{1} - f_{2}) = \frac{m}{2} (v_{1}^{2} - v_{2}^{2})$
$v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

Assam CEE-2016

Dual nature of radiation and Matter

142147 The graph $\frac{1}{λ}$ and stopping potential (V) of three metals having work function $ϕ_{1}, ϕ_{2}$ and $ϕ_{3}$ in an experiment of photoelectric effect is plotted as shown in the figure. Which one of the following statement is /are correct? [Here $λ$ is the wavelength of the incident rayl

(I) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 1 : 2 : 4$
(II) Ratio of work functions $ϕ_{1} : ϕ_{2} : ϕ_{3} = 4 : 2 : 1$
(III) $Tan θ \propto \frac{hc}{e}$ where $h =$ Planck's constant, $c =$ speed of light
(IV) The violet colour-light can eject photoelectrons from metals 2 and 3

1 (i), (iii)

2 (i), (iv)

3 (ii), (iii)

4 (i), (ii) and (iv)

Explanation:

A From the graph,
For metal 1, $\frac{ϕ_{1}}{hc} = 0.001$
For metal 2, $\frac{ϕ_{2}}{hc} = 0.002$
For metal 3, $\frac{ϕ_{3}}{hc} = 0.004$
Ratio of work function,
$ϕ_{1} : ϕ_{2} : ϕ_{3} = 0.001 : 0.002 : 0.004$
$= 1 : 2 : 4$
Therefore option (i) is correct.
According to the Einstein photoelectric equation -
$eV = \frac{hc}{λ} - ϕ_{o}$
$V = [\frac{hc}{e}] [\frac{1}{λ}] - \frac{ϕ_{o}}{e}$
$∵$ It is an equation of straight line $∴$ Slope, $\tan θ = \frac{hc}{e}$
Therefore, option (iii) is correct.
For metal 2: $\frac{1}{λ_{2}} = 0.002 nm \Rightarrow λ_{2} = 500 nm$
For metal 3: $\frac{1}{λ_{3}} = 0.004 nm \Rightarrow λ_{3} = 250 nm$
So, metal 2 can eject violet colour light, as the wavelength of violet colour is around $400 nm$ and metal 2 shows the nearby value.
But the value of metal 3 does not lie in violet wavelength range.

JIPMER-2016

Dual nature of radiation and Matter

142148 Photoelectric emission is observed from a metallic surface for frequency $v_{1}$ and $v_{2}$ of the incident light $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1 : n$ , then the threshold frequency of the metallic surface is

1

\frac{(v_{1} - v_{2})}{(n - 1)}

2

\frac{(n v_{1} - v_{2})}{(n - 1)}

3

\frac{(n v_{2} - v_{1})}{(n - 1)}

4

\frac{(v_{1} - v_{2})}{n}

Explanation:

B From Einstein photoelectric equation, $h v - h v_{0} = 0$
Case-I : $h (v_{1} - v_{0}) = n_{1}$
Case-II : $h (v_{2} - v_{0}) = n_{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{v_{1} - v_{0}}{v_{2} - v_{0}} = \frac{n_{1}}{n_{2}} = \frac{1}{n}$
$n (v_{1} - v_{o}) = v_{2} - v_{o} [∵ \frac{n_{1}}{n_{2}} = \frac{1}{n}]$
$Or v_{0} = \frac{n v_{1} - v_{2}}{n - 1}$

EAMCET-2001

Dual nature of radiation and Matter

142150 In photoelectric effect, the work function of a metal is $3.5 eV$ . The emitted electrons can be stopped by applying a potential of $- 1.2 V$ . Then,

1 the energy of the incident photons is

4.7 eV

2 the energy of the incident photons is

2.3 eV

3 if higher frequency photons be used, the photoelectric current will rise

4 When the energy of photons is

3.5 eV

, the photoelectric current will be maximum

Explanation:

A Given that, $ϕ = 3.5 eV$
$V_{o} = - 1.2 V$
or ${eV}_{o} = - 1.2 eV$
Einstein photoelectric effect equation-
$(K.E)_{max} = h v - ϕ$
${eV}_{o} = h v - ϕ$
$1.2 = h v - 3.5$
$1.2 - h v + 3.5 = 0$
$h v = 4.7 eV$
Hence, the energy of the incident photons is $4.7 eV$ .

AIPMT - 1994

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