QBManager

Dual nature of radiation and Matter

142332 The photoelectric work function of a metal surface is $2 eV$ . When light of frequency $1.5 \times$ $10^{15} Hz$ is incident on it, the maximum kinetic energy of the photo electrons. approximately is

1

8 eV

2

6 eV

3

2 eV

4

4 eV

Explanation:

D Given,
$ϕ = 2 eV$
$v = 1.5 \times 10^{15} Hz$
${KE}_{max} = ?$
We know that,
${KE}_{max} = h v - ϕ$
$(KE)_{max} = \frac{6.6 \times 10^{- 34} \times 1.5 \times 10^{15}}{1.6 \times 10^{- 19}} eV - 2 eV$
$(KE)_{max} = (6.19 - 2) eV = 4.19 eV$
$(KE)_{max} \approx 4 eV$

EAMCET-1998

Dual nature of radiation and Matter

142333 When a metal surface is illuminated by a light of wavelength $400 nm$ and $250 nm$ . The maximum velocities of the photo electrons ejected are $v$ and $2 v$ respectively. The work function of the metal is $(h =$ planck's constant, $c =$ velocity of light in air)

1

2 hc \times 10^{6} J

2

1.5 hc \times 10^{6} J

3 he

\times 10^{6} J

4

0.5 hc \times 10^{6} J

Explanation:

A Given,
$λ_{1} = 400 nm = 400 \times 10^{- 9} m$
$λ_{2} = 250 nm = 250 \times 10^{- 9} m$
${(v_{1})}_{max} = v$
${(v_{2})}_{max} = 2 v$
According to Einstein s photoelectric equation-
$E = ϕ + \frac{1}{2} {mv}_{max}^{2}$
$\frac{1}{2} {mv}_{max}^{2} = \frac{hc}{λ} - ϕ$
Case-I
$\frac{1}{2} {mv}^{2} = \frac{hc}{λ_{1}} - ϕ$
Case-II
$\frac{1}{2} m (2 v)^{2} = \frac{hc}{λ_{2}} - ϕ$
From equation (i) and (ii), we get-
$ϕ = \frac{hc}{3} [\frac{4}{λ_{1}} - \frac{1}{λ_{2}}]$
$ϕ = \frac{hc}{3} [\frac{4}{400 \times 10^{- 9}} - \frac{1}{250 \times 10^{- 9}}]$
$ϕ = 2 hc \times 10^{6} J$

EAMCET-2000

Dual nature of radiation and Matter

142334 The work function of the nickel is $5 eV$ . When a light of wavelength $2000 \AA$ falls on it, it emits photoelectrons in the circuit. Then, the potential difference necessary to stop the fastest electrons emitted is (Given, $h = 6.67 \times 10^{- 34} J$ -s)

1

1.0 V

2

1.75 V

3

1.25 V

4

0.75 V

Explanation:

C Given that,
Work function $(ϕ) = 5 eV$
Wavelength of light $(λ) = 2000 \AA$
Planck's constant $(h) = 6.6 \times 10^{- 34} J - s$
$K_{max} = qv = {eV}_{0}$
$K_{max} = E - ϕ = \frac{hc}{λ} - 5 eV$
${eV}_{0} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 5 eV$
${eV}_{0} = 6.18 eV - 5 eV$
${eV}_{0} = 6.18 - 5 eV$
${eV}_{0} = 1.18 eV$
$V_{0} \approx 1.25 V$

EAMCET-2007

Dual nature of radiation and Matter

142335 If the work function of a potential is $6.875 eV$ , its threshold wavelength will be (Take $= c =$ $3 \times 10^{8} m / s$ )

1

3600 \AA

2

2400 \AA

3

1800 \AA

4

1200 \AA

Explanation:

C Given that,
Work function $(ϕ) = 6.875 eV = 6.875 \times 1.6 \times 10^{- 19} V$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that, $ϕ = \frac{hc}{λ}$
$λ = \frac{hc}{ϕ}$
Where, $h =$ plank constant $= 6.67 \times 10^{- 34}$
So, $λ = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{6.875 \times 1.6 \times 10^{- 19}} = 1819 \times 10^{- 10}$
$λ \approx 1800 \AA$

Manipal UGET-2013

Dual nature of radiation and Matter

142332 The photoelectric work function of a metal surface is $2 eV$ . When light of frequency $1.5 \times$ $10^{15} Hz$ is incident on it, the maximum kinetic energy of the photo electrons. approximately is

1

8 eV

2

6 eV

3

2 eV

4

4 eV

Explanation:

D Given,
$ϕ = 2 eV$
$v = 1.5 \times 10^{15} Hz$
${KE}_{max} = ?$
We know that,
${KE}_{max} = h v - ϕ$
$(KE)_{max} = \frac{6.6 \times 10^{- 34} \times 1.5 \times 10^{15}}{1.6 \times 10^{- 19}} eV - 2 eV$
$(KE)_{max} = (6.19 - 2) eV = 4.19 eV$
$(KE)_{max} \approx 4 eV$

EAMCET-1998

Dual nature of radiation and Matter

142333 When a metal surface is illuminated by a light of wavelength $400 nm$ and $250 nm$ . The maximum velocities of the photo electrons ejected are $v$ and $2 v$ respectively. The work function of the metal is $(h =$ planck's constant, $c =$ velocity of light in air)

1

2 hc \times 10^{6} J

2

1.5 hc \times 10^{6} J

3 he

\times 10^{6} J

4

0.5 hc \times 10^{6} J

Explanation:

A Given,
$λ_{1} = 400 nm = 400 \times 10^{- 9} m$
$λ_{2} = 250 nm = 250 \times 10^{- 9} m$
${(v_{1})}_{max} = v$
${(v_{2})}_{max} = 2 v$
According to Einstein s photoelectric equation-
$E = ϕ + \frac{1}{2} {mv}_{max}^{2}$
$\frac{1}{2} {mv}_{max}^{2} = \frac{hc}{λ} - ϕ$
Case-I
$\frac{1}{2} {mv}^{2} = \frac{hc}{λ_{1}} - ϕ$
Case-II
$\frac{1}{2} m (2 v)^{2} = \frac{hc}{λ_{2}} - ϕ$
From equation (i) and (ii), we get-
$ϕ = \frac{hc}{3} [\frac{4}{λ_{1}} - \frac{1}{λ_{2}}]$
$ϕ = \frac{hc}{3} [\frac{4}{400 \times 10^{- 9}} - \frac{1}{250 \times 10^{- 9}}]$
$ϕ = 2 hc \times 10^{6} J$

EAMCET-2000

Dual nature of radiation and Matter

142334 The work function of the nickel is $5 eV$ . When a light of wavelength $2000 \AA$ falls on it, it emits photoelectrons in the circuit. Then, the potential difference necessary to stop the fastest electrons emitted is (Given, $h = 6.67 \times 10^{- 34} J$ -s)

1

1.0 V

2

1.75 V

3

1.25 V

4

0.75 V

Explanation:

C Given that,
Work function $(ϕ) = 5 eV$
Wavelength of light $(λ) = 2000 \AA$
Planck's constant $(h) = 6.6 \times 10^{- 34} J - s$
$K_{max} = qv = {eV}_{0}$
$K_{max} = E - ϕ = \frac{hc}{λ} - 5 eV$
${eV}_{0} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 5 eV$
${eV}_{0} = 6.18 eV - 5 eV$
${eV}_{0} = 6.18 - 5 eV$
${eV}_{0} = 1.18 eV$
$V_{0} \approx 1.25 V$

EAMCET-2007

Dual nature of radiation and Matter

142335 If the work function of a potential is $6.875 eV$ , its threshold wavelength will be (Take $= c =$ $3 \times 10^{8} m / s$ )

1

3600 \AA

2

2400 \AA

3

1800 \AA

4

1200 \AA

Explanation:

C Given that,
Work function $(ϕ) = 6.875 eV = 6.875 \times 1.6 \times 10^{- 19} V$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that, $ϕ = \frac{hc}{λ}$
$λ = \frac{hc}{ϕ}$
Where, $h =$ plank constant $= 6.67 \times 10^{- 34}$
So, $λ = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{6.875 \times 1.6 \times 10^{- 19}} = 1819 \times 10^{- 10}$
$λ \approx 1800 \AA$

Manipal UGET-2013

Dual nature of radiation and Matter

142332 The photoelectric work function of a metal surface is $2 eV$ . When light of frequency $1.5 \times$ $10^{15} Hz$ is incident on it, the maximum kinetic energy of the photo electrons. approximately is

1

8 eV

2

6 eV

3

2 eV

4

4 eV

Explanation:

D Given,
$ϕ = 2 eV$
$v = 1.5 \times 10^{15} Hz$
${KE}_{max} = ?$
We know that,
${KE}_{max} = h v - ϕ$
$(KE)_{max} = \frac{6.6 \times 10^{- 34} \times 1.5 \times 10^{15}}{1.6 \times 10^{- 19}} eV - 2 eV$
$(KE)_{max} = (6.19 - 2) eV = 4.19 eV$
$(KE)_{max} \approx 4 eV$

EAMCET-1998

Dual nature of radiation and Matter

142333 When a metal surface is illuminated by a light of wavelength $400 nm$ and $250 nm$ . The maximum velocities of the photo electrons ejected are $v$ and $2 v$ respectively. The work function of the metal is $(h =$ planck's constant, $c =$ velocity of light in air)

1

2 hc \times 10^{6} J

2

1.5 hc \times 10^{6} J

3 he

\times 10^{6} J

4

0.5 hc \times 10^{6} J

Explanation:

A Given,
$λ_{1} = 400 nm = 400 \times 10^{- 9} m$
$λ_{2} = 250 nm = 250 \times 10^{- 9} m$
${(v_{1})}_{max} = v$
${(v_{2})}_{max} = 2 v$
According to Einstein s photoelectric equation-
$E = ϕ + \frac{1}{2} {mv}_{max}^{2}$
$\frac{1}{2} {mv}_{max}^{2} = \frac{hc}{λ} - ϕ$
Case-I
$\frac{1}{2} {mv}^{2} = \frac{hc}{λ_{1}} - ϕ$
Case-II
$\frac{1}{2} m (2 v)^{2} = \frac{hc}{λ_{2}} - ϕ$
From equation (i) and (ii), we get-
$ϕ = \frac{hc}{3} [\frac{4}{λ_{1}} - \frac{1}{λ_{2}}]$
$ϕ = \frac{hc}{3} [\frac{4}{400 \times 10^{- 9}} - \frac{1}{250 \times 10^{- 9}}]$
$ϕ = 2 hc \times 10^{6} J$

EAMCET-2000

Dual nature of radiation and Matter

142334 The work function of the nickel is $5 eV$ . When a light of wavelength $2000 \AA$ falls on it, it emits photoelectrons in the circuit. Then, the potential difference necessary to stop the fastest electrons emitted is (Given, $h = 6.67 \times 10^{- 34} J$ -s)

1

1.0 V

2

1.75 V

3

1.25 V

4

0.75 V

Explanation:

C Given that,
Work function $(ϕ) = 5 eV$
Wavelength of light $(λ) = 2000 \AA$
Planck's constant $(h) = 6.6 \times 10^{- 34} J - s$
$K_{max} = qv = {eV}_{0}$
$K_{max} = E - ϕ = \frac{hc}{λ} - 5 eV$
${eV}_{0} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 5 eV$
${eV}_{0} = 6.18 eV - 5 eV$
${eV}_{0} = 6.18 - 5 eV$
${eV}_{0} = 1.18 eV$
$V_{0} \approx 1.25 V$

EAMCET-2007

Dual nature of radiation and Matter

142335 If the work function of a potential is $6.875 eV$ , its threshold wavelength will be (Take $= c =$ $3 \times 10^{8} m / s$ )

1

3600 \AA

2

2400 \AA

3

1800 \AA

4

1200 \AA

Explanation:

C Given that,
Work function $(ϕ) = 6.875 eV = 6.875 \times 1.6 \times 10^{- 19} V$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that, $ϕ = \frac{hc}{λ}$
$λ = \frac{hc}{ϕ}$
Where, $h =$ plank constant $= 6.67 \times 10^{- 34}$
So, $λ = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{6.875 \times 1.6 \times 10^{- 19}} = 1819 \times 10^{- 10}$
$λ \approx 1800 \AA$

Manipal UGET-2013

Dual nature of radiation and Matter

142332 The photoelectric work function of a metal surface is $2 eV$ . When light of frequency $1.5 \times$ $10^{15} Hz$ is incident on it, the maximum kinetic energy of the photo electrons. approximately is

1

8 eV

2

6 eV

3

2 eV

4

4 eV

Explanation:

D Given,
$ϕ = 2 eV$
$v = 1.5 \times 10^{15} Hz$
${KE}_{max} = ?$
We know that,
${KE}_{max} = h v - ϕ$
$(KE)_{max} = \frac{6.6 \times 10^{- 34} \times 1.5 \times 10^{15}}{1.6 \times 10^{- 19}} eV - 2 eV$
$(KE)_{max} = (6.19 - 2) eV = 4.19 eV$
$(KE)_{max} \approx 4 eV$

EAMCET-1998

Dual nature of radiation and Matter

142333 When a metal surface is illuminated by a light of wavelength $400 nm$ and $250 nm$ . The maximum velocities of the photo electrons ejected are $v$ and $2 v$ respectively. The work function of the metal is $(h =$ planck's constant, $c =$ velocity of light in air)

1

2 hc \times 10^{6} J

2

1.5 hc \times 10^{6} J

3 he

\times 10^{6} J

4

0.5 hc \times 10^{6} J

Explanation:

A Given,
$λ_{1} = 400 nm = 400 \times 10^{- 9} m$
$λ_{2} = 250 nm = 250 \times 10^{- 9} m$
${(v_{1})}_{max} = v$
${(v_{2})}_{max} = 2 v$
According to Einstein s photoelectric equation-
$E = ϕ + \frac{1}{2} {mv}_{max}^{2}$
$\frac{1}{2} {mv}_{max}^{2} = \frac{hc}{λ} - ϕ$
Case-I
$\frac{1}{2} {mv}^{2} = \frac{hc}{λ_{1}} - ϕ$
Case-II
$\frac{1}{2} m (2 v)^{2} = \frac{hc}{λ_{2}} - ϕ$
From equation (i) and (ii), we get-
$ϕ = \frac{hc}{3} [\frac{4}{λ_{1}} - \frac{1}{λ_{2}}]$
$ϕ = \frac{hc}{3} [\frac{4}{400 \times 10^{- 9}} - \frac{1}{250 \times 10^{- 9}}]$
$ϕ = 2 hc \times 10^{6} J$

EAMCET-2000

Dual nature of radiation and Matter

142334 The work function of the nickel is $5 eV$ . When a light of wavelength $2000 \AA$ falls on it, it emits photoelectrons in the circuit. Then, the potential difference necessary to stop the fastest electrons emitted is (Given, $h = 6.67 \times 10^{- 34} J$ -s)

1

1.0 V

2

1.75 V

3

1.25 V

4

0.75 V

Explanation:

C Given that,
Work function $(ϕ) = 5 eV$
Wavelength of light $(λ) = 2000 \AA$
Planck's constant $(h) = 6.6 \times 10^{- 34} J - s$
$K_{max} = qv = {eV}_{0}$
$K_{max} = E - ϕ = \frac{hc}{λ} - 5 eV$
${eV}_{0} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 5 eV$
${eV}_{0} = 6.18 eV - 5 eV$
${eV}_{0} = 6.18 - 5 eV$
${eV}_{0} = 1.18 eV$
$V_{0} \approx 1.25 V$

EAMCET-2007

Dual nature of radiation and Matter

142335 If the work function of a potential is $6.875 eV$ , its threshold wavelength will be (Take $= c =$ $3 \times 10^{8} m / s$ )

1

3600 \AA

2

2400 \AA

3

1800 \AA

4

1200 \AA

Explanation:

C Given that,
Work function $(ϕ) = 6.875 eV = 6.875 \times 1.6 \times 10^{- 19} V$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that, $ϕ = \frac{hc}{λ}$
$λ = \frac{hc}{ϕ}$
Where, $h =$ plank constant $= 6.67 \times 10^{- 34}$
So, $λ = \frac{6.67 \times 10^{- 34} \times 3 \times 10^{8}}{6.875 \times 1.6 \times 10^{- 19}} = 1819 \times 10^{- 10}$
$λ \approx 1800 \AA$

Manipal UGET-2013

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