QBManager

Dual nature of radiation and Matter

142274 The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$ . Then the work function of the metal must be

1

\frac{h v}{2}

2 hv

3

2 hv

4 none of the above

Explanation:

A We know that,
$hv = ϕ + V_{s}$
Now, the frequency is increased to $\frac{3 v}{2}$
$\frac{3 hv}{2} = ϕ + 2 V_{s}$
From equation (i) and (ii) we get -
$ϕ = \frac{hv}{2}$
Work function $= \frac{hv}{2}$

AIIMS-2009

Dual nature of radiation and Matter

142275 A photon of energy $4 eV$ is incident on a metal surface whose work function is $2 eV$ . The minimum reverse potential to be applied for stopping the emission of electrons is

1

2 V

2

4 V

3

6 V

4

8 V

Explanation:

A Given,
$ϕ = 2 eV, h v = 4 eV$
According to Einstein's photoelectric equation-
$E_{K} = \frac{1}{2} {mv}_{max}^{2} = h v - ϕ$
$\frac{1}{2} {mv}_{max}^{2} = 4 eV - 2 eV$
Stopping potential-
${eV}_{0} = 2 eV$
$V_{0} = 2 V$

AIIMS-2004

Dual nature of radiation and Matter

142277 Two identical metal plates show photoelectric effect by a light of wavelength $λ_{1}$ on plate 1 and $λ_{2}$ on plate 2 (where $λ_{1} = 2 λ_{2}$ ). The maximum kinetic energy will be-

1

2 K_{2} = K_{1}

2

K_{1} < K_{2} / 2

3

K_{1} > K_{2} / 2

4

2 K_{1} = K_{2}

Explanation:

B Maximum kinetic energy,
$K = \frac{hc}{λ} - ϕ_{o}$
Condition I-
$∵ λ_{1} = 2 λ_{2}$
$K_{1} = \frac{hc}{2 λ_{2}} - ϕ_{o}$
Condition II-
$∵ λ = λ_{2}$
$K_{2} = \frac{hc}{λ_{2}} - ϕ_{o}$
Dividing both side of the above equation by 2 ,
$\frac{K_{2}}{2} = \frac{hc}{2 λ_{2}} - \frac{ϕ_{o}}{2}$
From equation (i) and (ii), we get-
$\frac{K_{2}}{2} - K_{1} = \frac{ϕ_{o}}{2}$
$\frac{K_{2}}{2} = K_{1} + \frac{ϕ_{o}}{2}$
So, from equation (iii) we can say,
$\frac{K_{2}}{2} > K_{1}$

BCECE-2016

Dual nature of radiation and Matter

142278 Light with an energy flux of $18 W / {cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 {cm}^{2}$ , then the total momentum delivered on the surface during a span of 30 min is-

1

2.16 \times 10^{- 3} kg - m / s

2

1.52 \times 10^{- 5} kg - m / s

3

8.31 \times 10^{- 8} kg - m / s

4

18.2 \times 10^{- 6} kg - m / s

Explanation:

A Given, $I = 18 W / {cm}^{2}, A = 20 {cm}^{2}, Δ t = 30$ $min = 30 \times 60 = 1800 \sec$
Total energy falling on the surface
$U = IA Δ t$
$U = 18 \times 20 \times 1800 J$
$U = 6.48 \times 10^{5} J$
Momentum, $p = \frac{U}{c}$
$p = \frac{6.48 \times 10^{5}}{3 \times 10^{8}}$
$p = 2.16 \times 10^{- 3} {kgms}^{- 1}$

BCECE-2014

Dual nature of radiation and Matter

142279 No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 nm$ . The work function of the metal is approximately equal to-

1

3 \times 10^{- 16} J

2

3 \times 10^{- 19} J

3

3 \times 10^{- 20} J

4

3 \times 10^{- 22} J

Explanation:

B Given that,
$λ_{o} = 6000 \overset{\circ}{A} = 6 \times 10^{- 7} m, h = 6.6 \times 10^{- 34} Js$
Since, we know that work function is given as-
$ϕ_{o} = \frac{hc}{λ_{o}}$
$ϕ_{o} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6 \times 10^{- 7}}$
$ϕ_{o} = 3 \times 10^{- 19} J$

BCECE-2009

Dual nature of radiation and Matter

142274 The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$ . Then the work function of the metal must be

1

\frac{h v}{2}

2 hv

3

2 hv

4 none of the above

Explanation:

A We know that,
$hv = ϕ + V_{s}$
Now, the frequency is increased to $\frac{3 v}{2}$
$\frac{3 hv}{2} = ϕ + 2 V_{s}$
From equation (i) and (ii) we get -
$ϕ = \frac{hv}{2}$
Work function $= \frac{hv}{2}$

AIIMS-2009

Dual nature of radiation and Matter

142275 A photon of energy $4 eV$ is incident on a metal surface whose work function is $2 eV$ . The minimum reverse potential to be applied for stopping the emission of electrons is

1

2 V

2

4 V

3

6 V

4

8 V

Explanation:

A Given,
$ϕ = 2 eV, h v = 4 eV$
According to Einstein's photoelectric equation-
$E_{K} = \frac{1}{2} {mv}_{max}^{2} = h v - ϕ$
$\frac{1}{2} {mv}_{max}^{2} = 4 eV - 2 eV$
Stopping potential-
${eV}_{0} = 2 eV$
$V_{0} = 2 V$

AIIMS-2004

Dual nature of radiation and Matter

142277 Two identical metal plates show photoelectric effect by a light of wavelength $λ_{1}$ on plate 1 and $λ_{2}$ on plate 2 (where $λ_{1} = 2 λ_{2}$ ). The maximum kinetic energy will be-

1

2 K_{2} = K_{1}

2

K_{1} < K_{2} / 2

3

K_{1} > K_{2} / 2

4

2 K_{1} = K_{2}

Explanation:

B Maximum kinetic energy,
$K = \frac{hc}{λ} - ϕ_{o}$
Condition I-
$∵ λ_{1} = 2 λ_{2}$
$K_{1} = \frac{hc}{2 λ_{2}} - ϕ_{o}$
Condition II-
$∵ λ = λ_{2}$
$K_{2} = \frac{hc}{λ_{2}} - ϕ_{o}$
Dividing both side of the above equation by 2 ,
$\frac{K_{2}}{2} = \frac{hc}{2 λ_{2}} - \frac{ϕ_{o}}{2}$
From equation (i) and (ii), we get-
$\frac{K_{2}}{2} - K_{1} = \frac{ϕ_{o}}{2}$
$\frac{K_{2}}{2} = K_{1} + \frac{ϕ_{o}}{2}$
So, from equation (iii) we can say,
$\frac{K_{2}}{2} > K_{1}$

BCECE-2016

Dual nature of radiation and Matter

142278 Light with an energy flux of $18 W / {cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 {cm}^{2}$ , then the total momentum delivered on the surface during a span of 30 min is-

1

2.16 \times 10^{- 3} kg - m / s

2

1.52 \times 10^{- 5} kg - m / s

3

8.31 \times 10^{- 8} kg - m / s

4

18.2 \times 10^{- 6} kg - m / s

Explanation:

A Given, $I = 18 W / {cm}^{2}, A = 20 {cm}^{2}, Δ t = 30$ $min = 30 \times 60 = 1800 \sec$
Total energy falling on the surface
$U = IA Δ t$
$U = 18 \times 20 \times 1800 J$
$U = 6.48 \times 10^{5} J$
Momentum, $p = \frac{U}{c}$
$p = \frac{6.48 \times 10^{5}}{3 \times 10^{8}}$
$p = 2.16 \times 10^{- 3} {kgms}^{- 1}$

BCECE-2014

Dual nature of radiation and Matter

142279 No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 nm$ . The work function of the metal is approximately equal to-

1

3 \times 10^{- 16} J

2

3 \times 10^{- 19} J

3

3 \times 10^{- 20} J

4

3 \times 10^{- 22} J

Explanation:

B Given that,
$λ_{o} = 6000 \overset{\circ}{A} = 6 \times 10^{- 7} m, h = 6.6 \times 10^{- 34} Js$
Since, we know that work function is given as-
$ϕ_{o} = \frac{hc}{λ_{o}}$
$ϕ_{o} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6 \times 10^{- 7}}$
$ϕ_{o} = 3 \times 10^{- 19} J$

BCECE-2009

Dual nature of radiation and Matter

142274 The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$ . Then the work function of the metal must be

1

\frac{h v}{2}

2 hv

3

2 hv

4 none of the above

Explanation:

A We know that,
$hv = ϕ + V_{s}$
Now, the frequency is increased to $\frac{3 v}{2}$
$\frac{3 hv}{2} = ϕ + 2 V_{s}$
From equation (i) and (ii) we get -
$ϕ = \frac{hv}{2}$
Work function $= \frac{hv}{2}$

AIIMS-2009

Dual nature of radiation and Matter

142275 A photon of energy $4 eV$ is incident on a metal surface whose work function is $2 eV$ . The minimum reverse potential to be applied for stopping the emission of electrons is

1

2 V

2

4 V

3

6 V

4

8 V

Explanation:

A Given,
$ϕ = 2 eV, h v = 4 eV$
According to Einstein's photoelectric equation-
$E_{K} = \frac{1}{2} {mv}_{max}^{2} = h v - ϕ$
$\frac{1}{2} {mv}_{max}^{2} = 4 eV - 2 eV$
Stopping potential-
${eV}_{0} = 2 eV$
$V_{0} = 2 V$

AIIMS-2004

Dual nature of radiation and Matter

142277 Two identical metal plates show photoelectric effect by a light of wavelength $λ_{1}$ on plate 1 and $λ_{2}$ on plate 2 (where $λ_{1} = 2 λ_{2}$ ). The maximum kinetic energy will be-

1

2 K_{2} = K_{1}

2

K_{1} < K_{2} / 2

3

K_{1} > K_{2} / 2

4

2 K_{1} = K_{2}

Explanation:

B Maximum kinetic energy,
$K = \frac{hc}{λ} - ϕ_{o}$
Condition I-
$∵ λ_{1} = 2 λ_{2}$
$K_{1} = \frac{hc}{2 λ_{2}} - ϕ_{o}$
Condition II-
$∵ λ = λ_{2}$
$K_{2} = \frac{hc}{λ_{2}} - ϕ_{o}$
Dividing both side of the above equation by 2 ,
$\frac{K_{2}}{2} = \frac{hc}{2 λ_{2}} - \frac{ϕ_{o}}{2}$
From equation (i) and (ii), we get-
$\frac{K_{2}}{2} - K_{1} = \frac{ϕ_{o}}{2}$
$\frac{K_{2}}{2} = K_{1} + \frac{ϕ_{o}}{2}$
So, from equation (iii) we can say,
$\frac{K_{2}}{2} > K_{1}$

BCECE-2016

Dual nature of radiation and Matter

142278 Light with an energy flux of $18 W / {cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 {cm}^{2}$ , then the total momentum delivered on the surface during a span of 30 min is-

1

2.16 \times 10^{- 3} kg - m / s

2

1.52 \times 10^{- 5} kg - m / s

3

8.31 \times 10^{- 8} kg - m / s

4

18.2 \times 10^{- 6} kg - m / s

Explanation:

A Given, $I = 18 W / {cm}^{2}, A = 20 {cm}^{2}, Δ t = 30$ $min = 30 \times 60 = 1800 \sec$
Total energy falling on the surface
$U = IA Δ t$
$U = 18 \times 20 \times 1800 J$
$U = 6.48 \times 10^{5} J$
Momentum, $p = \frac{U}{c}$
$p = \frac{6.48 \times 10^{5}}{3 \times 10^{8}}$
$p = 2.16 \times 10^{- 3} {kgms}^{- 1}$

BCECE-2014

Dual nature of radiation and Matter

142279 No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 nm$ . The work function of the metal is approximately equal to-

1

3 \times 10^{- 16} J

2

3 \times 10^{- 19} J

3

3 \times 10^{- 20} J

4

3 \times 10^{- 22} J

Explanation:

B Given that,
$λ_{o} = 6000 \overset{\circ}{A} = 6 \times 10^{- 7} m, h = 6.6 \times 10^{- 34} Js$
Since, we know that work function is given as-
$ϕ_{o} = \frac{hc}{λ_{o}}$
$ϕ_{o} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6 \times 10^{- 7}}$
$ϕ_{o} = 3 \times 10^{- 19} J$

BCECE-2009

Dual nature of radiation and Matter

142274 The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$ . Then the work function of the metal must be

1

\frac{h v}{2}

2 hv

3

2 hv

4 none of the above

Explanation:

A We know that,
$hv = ϕ + V_{s}$
Now, the frequency is increased to $\frac{3 v}{2}$
$\frac{3 hv}{2} = ϕ + 2 V_{s}$
From equation (i) and (ii) we get -
$ϕ = \frac{hv}{2}$
Work function $= \frac{hv}{2}$

AIIMS-2009

Dual nature of radiation and Matter

142275 A photon of energy $4 eV$ is incident on a metal surface whose work function is $2 eV$ . The minimum reverse potential to be applied for stopping the emission of electrons is

1

2 V

2

4 V

3

6 V

4

8 V

Explanation:

A Given,
$ϕ = 2 eV, h v = 4 eV$
According to Einstein's photoelectric equation-
$E_{K} = \frac{1}{2} {mv}_{max}^{2} = h v - ϕ$
$\frac{1}{2} {mv}_{max}^{2} = 4 eV - 2 eV$
Stopping potential-
${eV}_{0} = 2 eV$
$V_{0} = 2 V$

AIIMS-2004

Dual nature of radiation and Matter

142277 Two identical metal plates show photoelectric effect by a light of wavelength $λ_{1}$ on plate 1 and $λ_{2}$ on plate 2 (where $λ_{1} = 2 λ_{2}$ ). The maximum kinetic energy will be-

1

2 K_{2} = K_{1}

2

K_{1} < K_{2} / 2

3

K_{1} > K_{2} / 2

4

2 K_{1} = K_{2}

Explanation:

B Maximum kinetic energy,
$K = \frac{hc}{λ} - ϕ_{o}$
Condition I-
$∵ λ_{1} = 2 λ_{2}$
$K_{1} = \frac{hc}{2 λ_{2}} - ϕ_{o}$
Condition II-
$∵ λ = λ_{2}$
$K_{2} = \frac{hc}{λ_{2}} - ϕ_{o}$
Dividing both side of the above equation by 2 ,
$\frac{K_{2}}{2} = \frac{hc}{2 λ_{2}} - \frac{ϕ_{o}}{2}$
From equation (i) and (ii), we get-
$\frac{K_{2}}{2} - K_{1} = \frac{ϕ_{o}}{2}$
$\frac{K_{2}}{2} = K_{1} + \frac{ϕ_{o}}{2}$
So, from equation (iii) we can say,
$\frac{K_{2}}{2} > K_{1}$

BCECE-2016

Dual nature of radiation and Matter

142278 Light with an energy flux of $18 W / {cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 {cm}^{2}$ , then the total momentum delivered on the surface during a span of 30 min is-

1

2.16 \times 10^{- 3} kg - m / s

2

1.52 \times 10^{- 5} kg - m / s

3

8.31 \times 10^{- 8} kg - m / s

4

18.2 \times 10^{- 6} kg - m / s

Explanation:

A Given, $I = 18 W / {cm}^{2}, A = 20 {cm}^{2}, Δ t = 30$ $min = 30 \times 60 = 1800 \sec$
Total energy falling on the surface
$U = IA Δ t$
$U = 18 \times 20 \times 1800 J$
$U = 6.48 \times 10^{5} J$
Momentum, $p = \frac{U}{c}$
$p = \frac{6.48 \times 10^{5}}{3 \times 10^{8}}$
$p = 2.16 \times 10^{- 3} {kgms}^{- 1}$

BCECE-2014

Dual nature of radiation and Matter

142279 No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 nm$ . The work function of the metal is approximately equal to-

1

3 \times 10^{- 16} J

2

3 \times 10^{- 19} J

3

3 \times 10^{- 20} J

4

3 \times 10^{- 22} J

Explanation:

B Given that,
$λ_{o} = 6000 \overset{\circ}{A} = 6 \times 10^{- 7} m, h = 6.6 \times 10^{- 34} Js$
Since, we know that work function is given as-
$ϕ_{o} = \frac{hc}{λ_{o}}$
$ϕ_{o} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6 \times 10^{- 7}}$
$ϕ_{o} = 3 \times 10^{- 19} J$

BCECE-2009

Dual nature of radiation and Matter

142274 The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$ . Then the work function of the metal must be

1

\frac{h v}{2}

2 hv

3

2 hv

4 none of the above

Explanation:

A We know that,
$hv = ϕ + V_{s}$
Now, the frequency is increased to $\frac{3 v}{2}$
$\frac{3 hv}{2} = ϕ + 2 V_{s}$
From equation (i) and (ii) we get -
$ϕ = \frac{hv}{2}$
Work function $= \frac{hv}{2}$

AIIMS-2009

Dual nature of radiation and Matter

142275 A photon of energy $4 eV$ is incident on a metal surface whose work function is $2 eV$ . The minimum reverse potential to be applied for stopping the emission of electrons is

1

2 V

2

4 V

3

6 V

4

8 V

Explanation:

A Given,
$ϕ = 2 eV, h v = 4 eV$
According to Einstein's photoelectric equation-
$E_{K} = \frac{1}{2} {mv}_{max}^{2} = h v - ϕ$
$\frac{1}{2} {mv}_{max}^{2} = 4 eV - 2 eV$
Stopping potential-
${eV}_{0} = 2 eV$
$V_{0} = 2 V$

AIIMS-2004

Dual nature of radiation and Matter

142277 Two identical metal plates show photoelectric effect by a light of wavelength $λ_{1}$ on plate 1 and $λ_{2}$ on plate 2 (where $λ_{1} = 2 λ_{2}$ ). The maximum kinetic energy will be-

1

2 K_{2} = K_{1}

2

K_{1} < K_{2} / 2

3

K_{1} > K_{2} / 2

4

2 K_{1} = K_{2}

Explanation:

B Maximum kinetic energy,
$K = \frac{hc}{λ} - ϕ_{o}$
Condition I-
$∵ λ_{1} = 2 λ_{2}$
$K_{1} = \frac{hc}{2 λ_{2}} - ϕ_{o}$
Condition II-
$∵ λ = λ_{2}$
$K_{2} = \frac{hc}{λ_{2}} - ϕ_{o}$
Dividing both side of the above equation by 2 ,
$\frac{K_{2}}{2} = \frac{hc}{2 λ_{2}} - \frac{ϕ_{o}}{2}$
From equation (i) and (ii), we get-
$\frac{K_{2}}{2} - K_{1} = \frac{ϕ_{o}}{2}$
$\frac{K_{2}}{2} = K_{1} + \frac{ϕ_{o}}{2}$
So, from equation (iii) we can say,
$\frac{K_{2}}{2} > K_{1}$

BCECE-2016

Dual nature of radiation and Matter

142278 Light with an energy flux of $18 W / {cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 {cm}^{2}$ , then the total momentum delivered on the surface during a span of 30 min is-

1

2.16 \times 10^{- 3} kg - m / s

2

1.52 \times 10^{- 5} kg - m / s

3

8.31 \times 10^{- 8} kg - m / s

4

18.2 \times 10^{- 6} kg - m / s

Explanation:

A Given, $I = 18 W / {cm}^{2}, A = 20 {cm}^{2}, Δ t = 30$ $min = 30 \times 60 = 1800 \sec$
Total energy falling on the surface
$U = IA Δ t$
$U = 18 \times 20 \times 1800 J$
$U = 6.48 \times 10^{5} J$
Momentum, $p = \frac{U}{c}$
$p = \frac{6.48 \times 10^{5}}{3 \times 10^{8}}$
$p = 2.16 \times 10^{- 3} {kgms}^{- 1}$

BCECE-2014

Dual nature of radiation and Matter

142279 No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 nm$ . The work function of the metal is approximately equal to-

1

3 \times 10^{- 16} J

2

3 \times 10^{- 19} J

3

3 \times 10^{- 20} J

4

3 \times 10^{- 22} J

Explanation:

B Given that,
$λ_{o} = 6000 \overset{\circ}{A} = 6 \times 10^{- 7} m, h = 6.6 \times 10^{- 34} Js$
Since, we know that work function is given as-
$ϕ_{o} = \frac{hc}{λ_{o}}$
$ϕ_{o} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6 \times 10^{- 7}}$
$ϕ_{o} = 3 \times 10^{- 19} J$

BCECE-2009