QBManager

Dual nature of radiation and Matter

142325 Light of wavelength $500 nm$ is incident on a metal with work function $2.28 eV$ . The deBroglie wavelength of the emitted electron is

1

< 2.8 \times 10^{- 16} m

2

< 2.8 \times 10^{- 9} m

3

\geq 2.8 \times 10^{- 9} m

4

\leq 2.8 \times 10^{- 12} m

Explanation:

C Given that,
$λ = 500 nm = 500 \times 10^{- 9} m$
$ϕ = 2.28 eV$
According to photoelectric equation-
$hv = ϕ_{o} + K_{max}$
$\frac{hc}{λ} = ϕ_{o} + K_{max}$
$\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9}} = 2.28 + K_{max}$
$K_{max} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 2.28$
$K_{max} = 2.48 eV - 2.28 eV$
$K_{max} = 0.2 eV$
According to the de-Broglie wavelength of electric-
$λ = \frac{h}{\sqrt{2 mK}}$
$λ = \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$λ = 2.8 \times 10^{- 9} m$
$λ \geq 2.8 \times 10^{- 9} m$

AIPMT - 2015

Dual nature of radiation and Matter

142327 The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency $v$ of the incident radiation is doubled. The work function of this metal is

1

\frac{h v}{4}

2

\frac{h v}{3}

3

\frac{h v}{2}

4

\frac{2 h v}{3}

Explanation:

D Given that,
$v_{initial} = 2 v$
${(v_{max})}_{initial} = 2 v_{max}$
Photoelectric equation-
${KE}_{max} = h v - ϕ_{0}$
$\frac{1}{2} {mv}_{max}^{2} = h v - ϕ_{0}$
$\frac{1}{2} m {(2 v_{max})}^{2} = 2 h v - ϕ_{0}$
$4 \times \frac{1}{2} {mv}_{max}^{2} = 2 h v - ϕ_{0}$
$4 (h v - ϕ_{0}) = 2 h v - ϕ_{0}$
$4 h v - 4 ϕ_{0} = 2 h v - ϕ_{0}$
$4 hv - 2 hv = 4 ϕ_{0} - ϕ_{0}$
$2 h v = 3 ϕ_{0}$
$ϕ_{0} = \frac{2}{3} hv$

AP EAMCET(Medical)-2007

Dual nature of radiation and Matter

142328 Light rays of wavelengths $6000 \AA$ and photon intensity $39.6 watt / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surfaced emit photoelectron, then the number of electrons emitted per second per unit area from the surface will be:
(Planck constant $= 6.64 \times 10^{- 34} J - s$ , velocity of light $= 3 \times 10^{- 8} m / s$ )

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{16}

Explanation:

C Given,
Wavelength of light $(λ) = 6000 \AA$
Intensity $= 39.6 W / m^{2}$
The number of photons incident per sec per unit area-
$\frac{n}{At} = \frac{I λ}{hc} = \frac{39.6 \times 6000 \times 10^{- 10}}{6.64 \times 10^{- 34} \times 3 \times 10^{8}}$
$⊔ 12 \times 10^{19}$
The number of electrons emitted $= 1 %$ of incident photon
$\frac{Ne}{At} = \frac{1}{100} \times (\frac{n}{At})$
$= \frac{1}{100} \times 12 \times 10^{19}$
$\frac{Ne}{At} = 12 \times 10^{17}$

AP EAMCET(Medical)-2004

Dual nature of radiation and Matter

142329 The work functions of the metals $A$ and $B$ are in the ratio $1 : 2$ . If the light of frequencies $f$ and $2 f$ are incident on metal surfaces of $A$ and $B$ respectively. The ratio of maximum kinetic energy of photo electrons emitted is [ $f$ is greater than threshold frequency of $A$ . $2 f$ is greater than threshold frequency of $B$ ]:

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

B Given, $\frac{ϕ_{A}}{ϕ_{B}} = \frac{1}{2} \Rightarrow ϕ_{B} = 2 ϕ_{A}$
We know that, $E_{max} = h v - ϕ$
${(E_{A})}_{max} = hf - (ϕ)_{A}$
${(E_{B})}_{max} = h \cdot 2 f - (ϕ)_{B}$
$= h \cdot 2 f - 2 ϕ_{A}$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{h \cdot f - ϕ_{A}}{h \cdot 2 f - 2 ϕ_{A}}$
$= \frac{1}{2} (\frac{hf - ϕ_{A}}{hf - ϕ_{A}})$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{1}{2}$
So, ${(E_{A})}_{max} : {(E_{B})}_{max} = 1 : 2$

AP EAMCET(Medical)-2000

Dual nature of radiation and Matter

142331 Photons of energy $2.0 eV$ and wavelength $λ$ fall on a metal plate and release photoctrons with a maximum velocity v. By decreasing $λ$ by $25 %$ the maximum velocity of photoelectrons is doubled. The work function of the material of the metal plate in $eV$ is

1 2.22

2 1.985

3

2 / 35

4 1.80

Explanation:

D Given that,
Case-I
$\frac{hc}{λ} = 2 eV$
$\frac{hc}{λ} = ϕ + \frac{1}{2} ({mv}^{2})$
$\frac{hc}{λ} - ϕ = \frac{1}{2} {mv}^{2}$
Case-II
$\frac{hc}{(\frac{3 λ}{4})} = ϕ + \frac{1}{2} m (2 v)^{2}$
$\frac{hc}{(\frac{3 λ}{4})} - ϕ = \frac{1}{2} m (2 v)^{2}$
From equation (i) and (ii), we get-
$\frac{4 hc}{λ} (1 - \frac{1}{3}) = 3 ϕ$
$\frac{4 hc}{λ} (\frac{2}{3}) = 3 ϕ$
$\frac{4 \times 2 \times 2}{3} = 3 ϕ [\frac{hc}{λ} = 2 eV]$
$ϕ = \frac{16}{9}$

EAMCET-2000

Dual nature of radiation and Matter

142325 Light of wavelength $500 nm$ is incident on a metal with work function $2.28 eV$ . The deBroglie wavelength of the emitted electron is

1

< 2.8 \times 10^{- 16} m

2

< 2.8 \times 10^{- 9} m

3

\geq 2.8 \times 10^{- 9} m

4

\leq 2.8 \times 10^{- 12} m

Explanation:

C Given that,
$λ = 500 nm = 500 \times 10^{- 9} m$
$ϕ = 2.28 eV$
According to photoelectric equation-
$hv = ϕ_{o} + K_{max}$
$\frac{hc}{λ} = ϕ_{o} + K_{max}$
$\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9}} = 2.28 + K_{max}$
$K_{max} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 2.28$
$K_{max} = 2.48 eV - 2.28 eV$
$K_{max} = 0.2 eV$
According to the de-Broglie wavelength of electric-
$λ = \frac{h}{\sqrt{2 mK}}$
$λ = \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$λ = 2.8 \times 10^{- 9} m$
$λ \geq 2.8 \times 10^{- 9} m$

AIPMT - 2015

Dual nature of radiation and Matter

142327 The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency $v$ of the incident radiation is doubled. The work function of this metal is

1

\frac{h v}{4}

2

\frac{h v}{3}

3

\frac{h v}{2}

4

\frac{2 h v}{3}

Explanation:

D Given that,
$v_{initial} = 2 v$
${(v_{max})}_{initial} = 2 v_{max}$
Photoelectric equation-
${KE}_{max} = h v - ϕ_{0}$
$\frac{1}{2} {mv}_{max}^{2} = h v - ϕ_{0}$
$\frac{1}{2} m {(2 v_{max})}^{2} = 2 h v - ϕ_{0}$
$4 \times \frac{1}{2} {mv}_{max}^{2} = 2 h v - ϕ_{0}$
$4 (h v - ϕ_{0}) = 2 h v - ϕ_{0}$
$4 h v - 4 ϕ_{0} = 2 h v - ϕ_{0}$
$4 hv - 2 hv = 4 ϕ_{0} - ϕ_{0}$
$2 h v = 3 ϕ_{0}$
$ϕ_{0} = \frac{2}{3} hv$

AP EAMCET(Medical)-2007

Dual nature of radiation and Matter

142328 Light rays of wavelengths $6000 \AA$ and photon intensity $39.6 watt / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surfaced emit photoelectron, then the number of electrons emitted per second per unit area from the surface will be:
(Planck constant $= 6.64 \times 10^{- 34} J - s$ , velocity of light $= 3 \times 10^{- 8} m / s$ )

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{16}

Explanation:

C Given,
Wavelength of light $(λ) = 6000 \AA$
Intensity $= 39.6 W / m^{2}$
The number of photons incident per sec per unit area-
$\frac{n}{At} = \frac{I λ}{hc} = \frac{39.6 \times 6000 \times 10^{- 10}}{6.64 \times 10^{- 34} \times 3 \times 10^{8}}$
$⊔ 12 \times 10^{19}$
The number of electrons emitted $= 1 %$ of incident photon
$\frac{Ne}{At} = \frac{1}{100} \times (\frac{n}{At})$
$= \frac{1}{100} \times 12 \times 10^{19}$
$\frac{Ne}{At} = 12 \times 10^{17}$

AP EAMCET(Medical)-2004

Dual nature of radiation and Matter

142329 The work functions of the metals $A$ and $B$ are in the ratio $1 : 2$ . If the light of frequencies $f$ and $2 f$ are incident on metal surfaces of $A$ and $B$ respectively. The ratio of maximum kinetic energy of photo electrons emitted is [ $f$ is greater than threshold frequency of $A$ . $2 f$ is greater than threshold frequency of $B$ ]:

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

B Given, $\frac{ϕ_{A}}{ϕ_{B}} = \frac{1}{2} \Rightarrow ϕ_{B} = 2 ϕ_{A}$
We know that, $E_{max} = h v - ϕ$
${(E_{A})}_{max} = hf - (ϕ)_{A}$
${(E_{B})}_{max} = h \cdot 2 f - (ϕ)_{B}$
$= h \cdot 2 f - 2 ϕ_{A}$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{h \cdot f - ϕ_{A}}{h \cdot 2 f - 2 ϕ_{A}}$
$= \frac{1}{2} (\frac{hf - ϕ_{A}}{hf - ϕ_{A}})$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{1}{2}$
So, ${(E_{A})}_{max} : {(E_{B})}_{max} = 1 : 2$

AP EAMCET(Medical)-2000

Dual nature of radiation and Matter

142331 Photons of energy $2.0 eV$ and wavelength $λ$ fall on a metal plate and release photoctrons with a maximum velocity v. By decreasing $λ$ by $25 %$ the maximum velocity of photoelectrons is doubled. The work function of the material of the metal plate in $eV$ is

1 2.22

2 1.985

3

2 / 35

4 1.80

Explanation:

D Given that,
Case-I
$\frac{hc}{λ} = 2 eV$
$\frac{hc}{λ} = ϕ + \frac{1}{2} ({mv}^{2})$
$\frac{hc}{λ} - ϕ = \frac{1}{2} {mv}^{2}$
Case-II
$\frac{hc}{(\frac{3 λ}{4})} = ϕ + \frac{1}{2} m (2 v)^{2}$
$\frac{hc}{(\frac{3 λ}{4})} - ϕ = \frac{1}{2} m (2 v)^{2}$
From equation (i) and (ii), we get-
$\frac{4 hc}{λ} (1 - \frac{1}{3}) = 3 ϕ$
$\frac{4 hc}{λ} (\frac{2}{3}) = 3 ϕ$
$\frac{4 \times 2 \times 2}{3} = 3 ϕ [\frac{hc}{λ} = 2 eV]$
$ϕ = \frac{16}{9}$

EAMCET-2000

Dual nature of radiation and Matter

142325 Light of wavelength $500 nm$ is incident on a metal with work function $2.28 eV$ . The deBroglie wavelength of the emitted electron is

1

< 2.8 \times 10^{- 16} m

2

< 2.8 \times 10^{- 9} m

3

\geq 2.8 \times 10^{- 9} m

4

\leq 2.8 \times 10^{- 12} m

Explanation:

C Given that,
$λ = 500 nm = 500 \times 10^{- 9} m$
$ϕ = 2.28 eV$
According to photoelectric equation-
$hv = ϕ_{o} + K_{max}$
$\frac{hc}{λ} = ϕ_{o} + K_{max}$
$\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9}} = 2.28 + K_{max}$
$K_{max} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 2.28$
$K_{max} = 2.48 eV - 2.28 eV$
$K_{max} = 0.2 eV$
According to the de-Broglie wavelength of electric-
$λ = \frac{h}{\sqrt{2 mK}}$
$λ = \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$λ = 2.8 \times 10^{- 9} m$
$λ \geq 2.8 \times 10^{- 9} m$

AIPMT - 2015

Dual nature of radiation and Matter

142327 The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency $v$ of the incident radiation is doubled. The work function of this metal is

1

\frac{h v}{4}

2

\frac{h v}{3}

3

\frac{h v}{2}

4

\frac{2 h v}{3}

Explanation:

D Given that,
$v_{initial} = 2 v$
${(v_{max})}_{initial} = 2 v_{max}$
Photoelectric equation-
${KE}_{max} = h v - ϕ_{0}$
$\frac{1}{2} {mv}_{max}^{2} = h v - ϕ_{0}$
$\frac{1}{2} m {(2 v_{max})}^{2} = 2 h v - ϕ_{0}$
$4 \times \frac{1}{2} {mv}_{max}^{2} = 2 h v - ϕ_{0}$
$4 (h v - ϕ_{0}) = 2 h v - ϕ_{0}$
$4 h v - 4 ϕ_{0} = 2 h v - ϕ_{0}$
$4 hv - 2 hv = 4 ϕ_{0} - ϕ_{0}$
$2 h v = 3 ϕ_{0}$
$ϕ_{0} = \frac{2}{3} hv$

AP EAMCET(Medical)-2007

Dual nature of radiation and Matter

142328 Light rays of wavelengths $6000 \AA$ and photon intensity $39.6 watt / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surfaced emit photoelectron, then the number of electrons emitted per second per unit area from the surface will be:
(Planck constant $= 6.64 \times 10^{- 34} J - s$ , velocity of light $= 3 \times 10^{- 8} m / s$ )

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{16}

Explanation:

C Given,
Wavelength of light $(λ) = 6000 \AA$
Intensity $= 39.6 W / m^{2}$
The number of photons incident per sec per unit area-
$\frac{n}{At} = \frac{I λ}{hc} = \frac{39.6 \times 6000 \times 10^{- 10}}{6.64 \times 10^{- 34} \times 3 \times 10^{8}}$
$⊔ 12 \times 10^{19}$
The number of electrons emitted $= 1 %$ of incident photon
$\frac{Ne}{At} = \frac{1}{100} \times (\frac{n}{At})$
$= \frac{1}{100} \times 12 \times 10^{19}$
$\frac{Ne}{At} = 12 \times 10^{17}$

AP EAMCET(Medical)-2004

Dual nature of radiation and Matter

142329 The work functions of the metals $A$ and $B$ are in the ratio $1 : 2$ . If the light of frequencies $f$ and $2 f$ are incident on metal surfaces of $A$ and $B$ respectively. The ratio of maximum kinetic energy of photo electrons emitted is [ $f$ is greater than threshold frequency of $A$ . $2 f$ is greater than threshold frequency of $B$ ]:

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

B Given, $\frac{ϕ_{A}}{ϕ_{B}} = \frac{1}{2} \Rightarrow ϕ_{B} = 2 ϕ_{A}$
We know that, $E_{max} = h v - ϕ$
${(E_{A})}_{max} = hf - (ϕ)_{A}$
${(E_{B})}_{max} = h \cdot 2 f - (ϕ)_{B}$
$= h \cdot 2 f - 2 ϕ_{A}$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{h \cdot f - ϕ_{A}}{h \cdot 2 f - 2 ϕ_{A}}$
$= \frac{1}{2} (\frac{hf - ϕ_{A}}{hf - ϕ_{A}})$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{1}{2}$
So, ${(E_{A})}_{max} : {(E_{B})}_{max} = 1 : 2$

AP EAMCET(Medical)-2000

Dual nature of radiation and Matter

142331 Photons of energy $2.0 eV$ and wavelength $λ$ fall on a metal plate and release photoctrons with a maximum velocity v. By decreasing $λ$ by $25 %$ the maximum velocity of photoelectrons is doubled. The work function of the material of the metal plate in $eV$ is

1 2.22

2 1.985

3

2 / 35

4 1.80

Explanation:

D Given that,
Case-I
$\frac{hc}{λ} = 2 eV$
$\frac{hc}{λ} = ϕ + \frac{1}{2} ({mv}^{2})$
$\frac{hc}{λ} - ϕ = \frac{1}{2} {mv}^{2}$
Case-II
$\frac{hc}{(\frac{3 λ}{4})} = ϕ + \frac{1}{2} m (2 v)^{2}$
$\frac{hc}{(\frac{3 λ}{4})} - ϕ = \frac{1}{2} m (2 v)^{2}$
From equation (i) and (ii), we get-
$\frac{4 hc}{λ} (1 - \frac{1}{3}) = 3 ϕ$
$\frac{4 hc}{λ} (\frac{2}{3}) = 3 ϕ$
$\frac{4 \times 2 \times 2}{3} = 3 ϕ [\frac{hc}{λ} = 2 eV]$
$ϕ = \frac{16}{9}$

EAMCET-2000

Dual nature of radiation and Matter

142325 Light of wavelength $500 nm$ is incident on a metal with work function $2.28 eV$ . The deBroglie wavelength of the emitted electron is

1

< 2.8 \times 10^{- 16} m

2

< 2.8 \times 10^{- 9} m

3

\geq 2.8 \times 10^{- 9} m

4

\leq 2.8 \times 10^{- 12} m

Explanation:

C Given that,
$λ = 500 nm = 500 \times 10^{- 9} m$
$ϕ = 2.28 eV$
According to photoelectric equation-
$hv = ϕ_{o} + K_{max}$
$\frac{hc}{λ} = ϕ_{o} + K_{max}$
$\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9}} = 2.28 + K_{max}$
$K_{max} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 2.28$
$K_{max} = 2.48 eV - 2.28 eV$
$K_{max} = 0.2 eV$
According to the de-Broglie wavelength of electric-
$λ = \frac{h}{\sqrt{2 mK}}$
$λ = \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$λ = 2.8 \times 10^{- 9} m$
$λ \geq 2.8 \times 10^{- 9} m$

AIPMT - 2015

Dual nature of radiation and Matter

142327 The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency $v$ of the incident radiation is doubled. The work function of this metal is

1

\frac{h v}{4}

2

\frac{h v}{3}

3

\frac{h v}{2}

4

\frac{2 h v}{3}

Explanation:

D Given that,
$v_{initial} = 2 v$
${(v_{max})}_{initial} = 2 v_{max}$
Photoelectric equation-
${KE}_{max} = h v - ϕ_{0}$
$\frac{1}{2} {mv}_{max}^{2} = h v - ϕ_{0}$
$\frac{1}{2} m {(2 v_{max})}^{2} = 2 h v - ϕ_{0}$
$4 \times \frac{1}{2} {mv}_{max}^{2} = 2 h v - ϕ_{0}$
$4 (h v - ϕ_{0}) = 2 h v - ϕ_{0}$
$4 h v - 4 ϕ_{0} = 2 h v - ϕ_{0}$
$4 hv - 2 hv = 4 ϕ_{0} - ϕ_{0}$
$2 h v = 3 ϕ_{0}$
$ϕ_{0} = \frac{2}{3} hv$

AP EAMCET(Medical)-2007

Dual nature of radiation and Matter

142328 Light rays of wavelengths $6000 \AA$ and photon intensity $39.6 watt / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surfaced emit photoelectron, then the number of electrons emitted per second per unit area from the surface will be:
(Planck constant $= 6.64 \times 10^{- 34} J - s$ , velocity of light $= 3 \times 10^{- 8} m / s$ )

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{16}

Explanation:

C Given,
Wavelength of light $(λ) = 6000 \AA$
Intensity $= 39.6 W / m^{2}$
The number of photons incident per sec per unit area-
$\frac{n}{At} = \frac{I λ}{hc} = \frac{39.6 \times 6000 \times 10^{- 10}}{6.64 \times 10^{- 34} \times 3 \times 10^{8}}$
$⊔ 12 \times 10^{19}$
The number of electrons emitted $= 1 %$ of incident photon
$\frac{Ne}{At} = \frac{1}{100} \times (\frac{n}{At})$
$= \frac{1}{100} \times 12 \times 10^{19}$
$\frac{Ne}{At} = 12 \times 10^{17}$

AP EAMCET(Medical)-2004

Dual nature of radiation and Matter

142329 The work functions of the metals $A$ and $B$ are in the ratio $1 : 2$ . If the light of frequencies $f$ and $2 f$ are incident on metal surfaces of $A$ and $B$ respectively. The ratio of maximum kinetic energy of photo electrons emitted is [ $f$ is greater than threshold frequency of $A$ . $2 f$ is greater than threshold frequency of $B$ ]:

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

B Given, $\frac{ϕ_{A}}{ϕ_{B}} = \frac{1}{2} \Rightarrow ϕ_{B} = 2 ϕ_{A}$
We know that, $E_{max} = h v - ϕ$
${(E_{A})}_{max} = hf - (ϕ)_{A}$
${(E_{B})}_{max} = h \cdot 2 f - (ϕ)_{B}$
$= h \cdot 2 f - 2 ϕ_{A}$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{h \cdot f - ϕ_{A}}{h \cdot 2 f - 2 ϕ_{A}}$
$= \frac{1}{2} (\frac{hf - ϕ_{A}}{hf - ϕ_{A}})$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{1}{2}$
So, ${(E_{A})}_{max} : {(E_{B})}_{max} = 1 : 2$

AP EAMCET(Medical)-2000

Dual nature of radiation and Matter

142331 Photons of energy $2.0 eV$ and wavelength $λ$ fall on a metal plate and release photoctrons with a maximum velocity v. By decreasing $λ$ by $25 %$ the maximum velocity of photoelectrons is doubled. The work function of the material of the metal plate in $eV$ is

1 2.22

2 1.985

3

2 / 35

4 1.80

Explanation:

D Given that,
Case-I
$\frac{hc}{λ} = 2 eV$
$\frac{hc}{λ} = ϕ + \frac{1}{2} ({mv}^{2})$
$\frac{hc}{λ} - ϕ = \frac{1}{2} {mv}^{2}$
Case-II
$\frac{hc}{(\frac{3 λ}{4})} = ϕ + \frac{1}{2} m (2 v)^{2}$
$\frac{hc}{(\frac{3 λ}{4})} - ϕ = \frac{1}{2} m (2 v)^{2}$
From equation (i) and (ii), we get-
$\frac{4 hc}{λ} (1 - \frac{1}{3}) = 3 ϕ$
$\frac{4 hc}{λ} (\frac{2}{3}) = 3 ϕ$
$\frac{4 \times 2 \times 2}{3} = 3 ϕ [\frac{hc}{λ} = 2 eV]$
$ϕ = \frac{16}{9}$

EAMCET-2000

Dual nature of radiation and Matter

142325 Light of wavelength $500 nm$ is incident on a metal with work function $2.28 eV$ . The deBroglie wavelength of the emitted electron is

1

< 2.8 \times 10^{- 16} m

2

< 2.8 \times 10^{- 9} m

3

\geq 2.8 \times 10^{- 9} m

4

\leq 2.8 \times 10^{- 12} m

Explanation:

C Given that,
$λ = 500 nm = 500 \times 10^{- 9} m$
$ϕ = 2.28 eV$
According to photoelectric equation-
$hv = ϕ_{o} + K_{max}$
$\frac{hc}{λ} = ϕ_{o} + K_{max}$
$\frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9}} = 2.28 + K_{max}$
$K_{max} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{500 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 2.28$
$K_{max} = 2.48 eV - 2.28 eV$
$K_{max} = 0.2 eV$
According to the de-Broglie wavelength of electric-
$λ = \frac{h}{\sqrt{2 mK}}$
$λ = \frac{6.62 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 0.2 \times 1.6 \times 10^{- 19}}}$
$λ = 2.8 \times 10^{- 9} m$
$λ \geq 2.8 \times 10^{- 9} m$

AIPMT - 2015

Dual nature of radiation and Matter

142327 The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency $v$ of the incident radiation is doubled. The work function of this metal is

1

\frac{h v}{4}

2

\frac{h v}{3}

3

\frac{h v}{2}

4

\frac{2 h v}{3}

Explanation:

D Given that,
$v_{initial} = 2 v$
${(v_{max})}_{initial} = 2 v_{max}$
Photoelectric equation-
${KE}_{max} = h v - ϕ_{0}$
$\frac{1}{2} {mv}_{max}^{2} = h v - ϕ_{0}$
$\frac{1}{2} m {(2 v_{max})}^{2} = 2 h v - ϕ_{0}$
$4 \times \frac{1}{2} {mv}_{max}^{2} = 2 h v - ϕ_{0}$
$4 (h v - ϕ_{0}) = 2 h v - ϕ_{0}$
$4 h v - 4 ϕ_{0} = 2 h v - ϕ_{0}$
$4 hv - 2 hv = 4 ϕ_{0} - ϕ_{0}$
$2 h v = 3 ϕ_{0}$
$ϕ_{0} = \frac{2}{3} hv$

AP EAMCET(Medical)-2007

Dual nature of radiation and Matter

142328 Light rays of wavelengths $6000 \AA$ and photon intensity $39.6 watt / m^{2}$ is incident on a metal surface. If only one percent of photons incident on the surfaced emit photoelectron, then the number of electrons emitted per second per unit area from the surface will be:
(Planck constant $= 6.64 \times 10^{- 34} J - s$ , velocity of light $= 3 \times 10^{- 8} m / s$ )

1

12 \times 10^{18}

2

10 \times 10^{18}

3

12 \times 10^{17}

4

12 \times 10^{16}

Explanation:

C Given,
Wavelength of light $(λ) = 6000 \AA$
Intensity $= 39.6 W / m^{2}$
The number of photons incident per sec per unit area-
$\frac{n}{At} = \frac{I λ}{hc} = \frac{39.6 \times 6000 \times 10^{- 10}}{6.64 \times 10^{- 34} \times 3 \times 10^{8}}$
$⊔ 12 \times 10^{19}$
The number of electrons emitted $= 1 %$ of incident photon
$\frac{Ne}{At} = \frac{1}{100} \times (\frac{n}{At})$
$= \frac{1}{100} \times 12 \times 10^{19}$
$\frac{Ne}{At} = 12 \times 10^{17}$

AP EAMCET(Medical)-2004

Dual nature of radiation and Matter

142329 The work functions of the metals $A$ and $B$ are in the ratio $1 : 2$ . If the light of frequencies $f$ and $2 f$ are incident on metal surfaces of $A$ and $B$ respectively. The ratio of maximum kinetic energy of photo electrons emitted is [ $f$ is greater than threshold frequency of $A$ . $2 f$ is greater than threshold frequency of $B$ ]:

1

1 : 1

2

1 : 2

3

1 : 3

4

1 : 4

Explanation:

B Given, $\frac{ϕ_{A}}{ϕ_{B}} = \frac{1}{2} \Rightarrow ϕ_{B} = 2 ϕ_{A}$
We know that, $E_{max} = h v - ϕ$
${(E_{A})}_{max} = hf - (ϕ)_{A}$
${(E_{B})}_{max} = h \cdot 2 f - (ϕ)_{B}$
$= h \cdot 2 f - 2 ϕ_{A}$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{h \cdot f - ϕ_{A}}{h \cdot 2 f - 2 ϕ_{A}}$
$= \frac{1}{2} (\frac{hf - ϕ_{A}}{hf - ϕ_{A}})$
$\frac{{(E_{A})}_{max}}{{(E_{B})}_{max}} = \frac{1}{2}$
So, ${(E_{A})}_{max} : {(E_{B})}_{max} = 1 : 2$

AP EAMCET(Medical)-2000

Dual nature of radiation and Matter

142331 Photons of energy $2.0 eV$ and wavelength $λ$ fall on a metal plate and release photoctrons with a maximum velocity v. By decreasing $λ$ by $25 %$ the maximum velocity of photoelectrons is doubled. The work function of the material of the metal plate in $eV$ is

1 2.22

2 1.985

3

2 / 35

4 1.80

Explanation:

D Given that,
Case-I
$\frac{hc}{λ} = 2 eV$
$\frac{hc}{λ} = ϕ + \frac{1}{2} ({mv}^{2})$
$\frac{hc}{λ} - ϕ = \frac{1}{2} {mv}^{2}$
Case-II
$\frac{hc}{(\frac{3 λ}{4})} = ϕ + \frac{1}{2} m (2 v)^{2}$
$\frac{hc}{(\frac{3 λ}{4})} - ϕ = \frac{1}{2} m (2 v)^{2}$
From equation (i) and (ii), we get-
$\frac{4 hc}{λ} (1 - \frac{1}{3}) = 3 ϕ$
$\frac{4 hc}{λ} (\frac{2}{3}) = 3 ϕ$
$\frac{4 \times 2 \times 2}{3} = 3 ϕ [\frac{hc}{λ} = 2 eV]$
$ϕ = \frac{16}{9}$

EAMCET-2000

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: