QBManager

Dual nature of radiation and Matter

142062 A light beam consists of two types of photons. In one type, each photon has energy $2 eV$ and in other type, each photon has energy $3 eV$ . The light beam is incident on a photoelectric material of work function $1 eV$ . The maximum kinetic energy of emitted photoelectron is

1

2 eV

2

3 eV

3

4 eV

4

1 eV

Explanation:

A Given that,
Work function of photoelectric material -
$ϕ_{o} = 1 eV$
According to Einstein photoelectric equation -
$(K . E)_{max} = E_{max} - ϕ_{o}$
$(K . E)_{max} = 3 - 1$
$(K.E)_{max} = 2 eV$

UPSEE - 2017

Dual nature of radiation and Matter

142063 When a light of wavelength $400 nm$ is incident on the cathode of a photocell, the stopping potential recorded is $6 V$ . If the wavelength of the incident light is increased to $600 nm$ . then the new stopping potential is

1

4.5 V

2

4 V

3

5.97 V

4

4.97 V

Explanation:

D Given that,
Primary wavelength of light $(λ_{1}) = 400 nm$
Secondary wavelength of light $(λ_{2}) = 600 nm$
$V_{1} = 6 V, V_{2} = ?$
From Einstein photoelectric equation,
$eV = \frac{hc}{λ} - ϕ_{0}$
${eV}_{1} = \frac{hc}{λ_{1}} - ϕ_{0}$
${eV}_{2} = \frac{hc}{λ_{2}} - ϕ_{0}$
From equation (i) and (ii), we get -
$e (V_{1} - V_{2}) = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$V_{1} - V_{2} = \frac{hc}{e} (\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}})$
$6 - V_{2} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{600 - 400}{600 \times 400}) \times 10^{9}$
$6 - V_{2} = 1.03$
$V_{2} = 4.97 V$

CG PET- 2017

Dual nature of radiation and Matter

142065 The photoelectric threshold of metal is $2000 \AA$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500 \AA$ is

1

2.0 eV

2

1.5 eV

3

15 eV

4

150 eV

Explanation:

A Given that,
$λ_{0} = 2000 \AA = 2000 \times 10^{- 10} m$
$λ = 1500 \AA = 1500 \times 10^{- 10} m$
Energy of the electron ejected from the surface
$E = \frac{hc}{e} (\frac{1}{λ} - \frac{1}{λ_{0}})$
$E = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{1500 \times 10^{- 10}} - \frac{1}{2000 \times 10^{- 10}}) \times \frac{1}{10^{- 10}}$
$E = 2 eV$

BITSAT-2010

Dual nature of radiation and Matter

142058 Assertion (A): Threshold frequency of a photo metal depends on the intensity of incident light. Reason (r): Greater is the intensity of incident light greater is the photoelectric current.

1 Both (A) and (R) are true and (R) is the correct explanation of (A)

2 Both (A) and (R( are true and (R) is not the correct explanation of (A)

3 (A) is not true, but (R) is not true

4 (A) is not true, but (R) is true

Shift-I

Dual nature of radiation and Matter

142062 A light beam consists of two types of photons. In one type, each photon has energy $2 eV$ and in other type, each photon has energy $3 eV$ . The light beam is incident on a photoelectric material of work function $1 eV$ . The maximum kinetic energy of emitted photoelectron is

1

2 eV

2

3 eV

3

4 eV

4

1 eV

Explanation:

A Given that,
Work function of photoelectric material -
$ϕ_{o} = 1 eV$
According to Einstein photoelectric equation -
$(K . E)_{max} = E_{max} - ϕ_{o}$
$(K . E)_{max} = 3 - 1$
$(K.E)_{max} = 2 eV$

UPSEE - 2017

Dual nature of radiation and Matter

142063 When a light of wavelength $400 nm$ is incident on the cathode of a photocell, the stopping potential recorded is $6 V$ . If the wavelength of the incident light is increased to $600 nm$ . then the new stopping potential is

1

4.5 V

2

4 V

3

5.97 V

4

4.97 V

Explanation:

D Given that,
Primary wavelength of light $(λ_{1}) = 400 nm$
Secondary wavelength of light $(λ_{2}) = 600 nm$
$V_{1} = 6 V, V_{2} = ?$
From Einstein photoelectric equation,
$eV = \frac{hc}{λ} - ϕ_{0}$
${eV}_{1} = \frac{hc}{λ_{1}} - ϕ_{0}$
${eV}_{2} = \frac{hc}{λ_{2}} - ϕ_{0}$
From equation (i) and (ii), we get -
$e (V_{1} - V_{2}) = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$V_{1} - V_{2} = \frac{hc}{e} (\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}})$
$6 - V_{2} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{600 - 400}{600 \times 400}) \times 10^{9}$
$6 - V_{2} = 1.03$
$V_{2} = 4.97 V$

CG PET- 2017

Dual nature of radiation and Matter

142065 The photoelectric threshold of metal is $2000 \AA$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500 \AA$ is

1

2.0 eV

2

1.5 eV

3

15 eV

4

150 eV

Explanation:

A Given that,
$λ_{0} = 2000 \AA = 2000 \times 10^{- 10} m$
$λ = 1500 \AA = 1500 \times 10^{- 10} m$
Energy of the electron ejected from the surface
$E = \frac{hc}{e} (\frac{1}{λ} - \frac{1}{λ_{0}})$
$E = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{1500 \times 10^{- 10}} - \frac{1}{2000 \times 10^{- 10}}) \times \frac{1}{10^{- 10}}$
$E = 2 eV$

BITSAT-2010

Dual nature of radiation and Matter

142058 Assertion (A): Threshold frequency of a photo metal depends on the intensity of incident light. Reason (r): Greater is the intensity of incident light greater is the photoelectric current.

1 Both (A) and (R) are true and (R) is the correct explanation of (A)

2 Both (A) and (R( are true and (R) is not the correct explanation of (A)

3 (A) is not true, but (R) is not true

4 (A) is not true, but (R) is true

Shift-I

Dual nature of radiation and Matter

142062 A light beam consists of two types of photons. In one type, each photon has energy $2 eV$ and in other type, each photon has energy $3 eV$ . The light beam is incident on a photoelectric material of work function $1 eV$ . The maximum kinetic energy of emitted photoelectron is

1

2 eV

2

3 eV

3

4 eV

4

1 eV

Explanation:

A Given that,
Work function of photoelectric material -
$ϕ_{o} = 1 eV$
According to Einstein photoelectric equation -
$(K . E)_{max} = E_{max} - ϕ_{o}$
$(K . E)_{max} = 3 - 1$
$(K.E)_{max} = 2 eV$

UPSEE - 2017

Dual nature of radiation and Matter

142063 When a light of wavelength $400 nm$ is incident on the cathode of a photocell, the stopping potential recorded is $6 V$ . If the wavelength of the incident light is increased to $600 nm$ . then the new stopping potential is

1

4.5 V

2

4 V

3

5.97 V

4

4.97 V

Explanation:

D Given that,
Primary wavelength of light $(λ_{1}) = 400 nm$
Secondary wavelength of light $(λ_{2}) = 600 nm$
$V_{1} = 6 V, V_{2} = ?$
From Einstein photoelectric equation,
$eV = \frac{hc}{λ} - ϕ_{0}$
${eV}_{1} = \frac{hc}{λ_{1}} - ϕ_{0}$
${eV}_{2} = \frac{hc}{λ_{2}} - ϕ_{0}$
From equation (i) and (ii), we get -
$e (V_{1} - V_{2}) = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$V_{1} - V_{2} = \frac{hc}{e} (\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}})$
$6 - V_{2} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{600 - 400}{600 \times 400}) \times 10^{9}$
$6 - V_{2} = 1.03$
$V_{2} = 4.97 V$

CG PET- 2017

Dual nature of radiation and Matter

142065 The photoelectric threshold of metal is $2000 \AA$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500 \AA$ is

1

2.0 eV

2

1.5 eV

3

15 eV

4

150 eV

Explanation:

A Given that,
$λ_{0} = 2000 \AA = 2000 \times 10^{- 10} m$
$λ = 1500 \AA = 1500 \times 10^{- 10} m$
Energy of the electron ejected from the surface
$E = \frac{hc}{e} (\frac{1}{λ} - \frac{1}{λ_{0}})$
$E = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{1500 \times 10^{- 10}} - \frac{1}{2000 \times 10^{- 10}}) \times \frac{1}{10^{- 10}}$
$E = 2 eV$

BITSAT-2010

Dual nature of radiation and Matter

142058 Assertion (A): Threshold frequency of a photo metal depends on the intensity of incident light. Reason (r): Greater is the intensity of incident light greater is the photoelectric current.

1 Both (A) and (R) are true and (R) is the correct explanation of (A)

2 Both (A) and (R( are true and (R) is not the correct explanation of (A)

3 (A) is not true, but (R) is not true

4 (A) is not true, but (R) is true

Shift-I

Dual nature of radiation and Matter

142062 A light beam consists of two types of photons. In one type, each photon has energy $2 eV$ and in other type, each photon has energy $3 eV$ . The light beam is incident on a photoelectric material of work function $1 eV$ . The maximum kinetic energy of emitted photoelectron is

1

2 eV

2

3 eV

3

4 eV

4

1 eV

Explanation:

A Given that,
Work function of photoelectric material -
$ϕ_{o} = 1 eV$
According to Einstein photoelectric equation -
$(K . E)_{max} = E_{max} - ϕ_{o}$
$(K . E)_{max} = 3 - 1$
$(K.E)_{max} = 2 eV$

UPSEE - 2017

Dual nature of radiation and Matter

142063 When a light of wavelength $400 nm$ is incident on the cathode of a photocell, the stopping potential recorded is $6 V$ . If the wavelength of the incident light is increased to $600 nm$ . then the new stopping potential is

1

4.5 V

2

4 V

3

5.97 V

4

4.97 V

Explanation:

D Given that,
Primary wavelength of light $(λ_{1}) = 400 nm$
Secondary wavelength of light $(λ_{2}) = 600 nm$
$V_{1} = 6 V, V_{2} = ?$
From Einstein photoelectric equation,
$eV = \frac{hc}{λ} - ϕ_{0}$
${eV}_{1} = \frac{hc}{λ_{1}} - ϕ_{0}$
${eV}_{2} = \frac{hc}{λ_{2}} - ϕ_{0}$
From equation (i) and (ii), we get -
$e (V_{1} - V_{2}) = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$V_{1} - V_{2} = \frac{hc}{e} (\frac{λ_{2} - λ_{1}}{λ_{1} λ_{2}})$
$6 - V_{2} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{600 - 400}{600 \times 400}) \times 10^{9}$
$6 - V_{2} = 1.03$
$V_{2} = 4.97 V$

CG PET- 2017

Dual nature of radiation and Matter

142065 The photoelectric threshold of metal is $2000 \AA$ . The energy of the electrons ejected from the surface by ultraviolet light of wavelength $1500 \AA$ is

1

2.0 eV

2

1.5 eV

3

15 eV

4

150 eV

Explanation:

A Given that,
$λ_{0} = 2000 \AA = 2000 \times 10^{- 10} m$
$λ = 1500 \AA = 1500 \times 10^{- 10} m$
Energy of the electron ejected from the surface
$E = \frac{hc}{e} (\frac{1}{λ} - \frac{1}{λ_{0}})$
$E = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{1500 \times 10^{- 10}} - \frac{1}{2000 \times 10^{- 10}}) \times \frac{1}{10^{- 10}}$
$E = 2 eV$

BITSAT-2010

Dual nature of radiation and Matter

142058 Assertion (A): Threshold frequency of a photo metal depends on the intensity of incident light. Reason (r): Greater is the intensity of incident light greater is the photoelectric current.

1 Both (A) and (R) are true and (R) is the correct explanation of (A)

2 Both (A) and (R( are true and (R) is not the correct explanation of (A)

3 (A) is not true, but (R) is not true

4 (A) is not true, but (R) is true

Shift-I

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