QBManager

Dual nature of radiation and Matter

142088 The threshold frequency for certain metal is $3.3$ $\times 10^{14} Hz$ . If light of frequency $8.2 \times 10^{14} Hz$ is incident on the metal, the cut-off voltage of the photoelectric current will be -

1

4.9 V

2

3.0 V

3

2.0 V

4

1.0 V

Explanation:

C Given that,
$v_{o} = 3.3 \times 10^{14} Hz$
$v = 8.2 \times 10^{14} Hz$
We know that,
${eV}_{o} = h (v - v_{o})$
$V_{o} = \frac{h (v - v_{o})}{e}$
$V_{o} = \frac{6.62 \times 10^{- 34} (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{- 19}}$
$V_{o} = 2.029$
$V_{o} \approx 2.0 V$

BCECE-2006

Dual nature of radiation and Matter

142089 Light of wavelength $λ$ , strikes a photoelectric surface and electrons are ejected with an energy $E$ . If $E$ is to be increased to exactly twice its original value, the wavelength changes to $λ^{'}$ , where-

1

λ^{'}

is less than

\frac{λ}{2}

2

λ^{'}

is greater than

\frac{λ}{2}

3

λ^{'}

is greater than

\frac{λ}{2}

but less than

λ

4

λ^{'}

is exactly equal to

\frac{λ}{2}

Explanation:

C According to Einstein photoelectric equation
$\frac{hc}{λ} = E + ϕ_{o}$
$\frac{hc}{λ^{'}} = 2 E + ϕ_{o}$
From equation (i) and (ii), we get-
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$\frac{λ^{'}}{λ} < 1$
$\frac{λ^{'}}{λ} < λ$
Also
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$= \frac{1}{2} [\frac{E + ϕ_{o}}{E + ϕ_{o} / 2}]$
$\frac{λ^{'}}{λ} > \frac{1}{2}$
So, from (iii) and (iv)
$λ > λ^{'} > \frac{λ}{2}$

BCECE-2010

Dual nature of radiation and Matter

142092 The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$ . The corresponding change in stopping potential is

1

0.03 V

2

0.3 V

3

3 V

4

3.3 V

Explanation:

B Given that,
$λ_{1} = 2000 \AA = 2 \times 10^{- 7} m$
$λ_{2} = 2100 \AA = 2.1 \times 10^{- 7} m$
We know that,
$V_{1} - V_{2} = \frac{hc}{e} (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{2} - \frac{1}{2.1}) \times 10^{7}$
$= 0.295 = 0.3 V$

Manipal UGET-2016

Dual nature of radiation and Matter

142093 Light of wavelength ' $λ$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will

1 increase

2 decrease

3 be zero

4 become exactly half

Explanation:

A Incident photon (E) $= h v$
$E = \frac{hc}{λ}$
$E \propto \frac{1}{λ}$
If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect,
$\frac{hc}{λ} = {KE}_{max} + ϕ_{o}$
$\frac{hc}{λ} = {eV}_{o} + ϕ_{o}$
${eV}_{o} = hc [\frac{1}{λ} - \frac{1}{λ_{o}}]$
$V_{o} \propto (\frac{1}{λ} - \frac{1}{λ_{o}})$
Hence, if wavelength decreases, stopping potential increases.

MHT-CET 2016

Dual nature of radiation and Matter

142088 The threshold frequency for certain metal is $3.3$ $\times 10^{14} Hz$ . If light of frequency $8.2 \times 10^{14} Hz$ is incident on the metal, the cut-off voltage of the photoelectric current will be -

1

4.9 V

2

3.0 V

3

2.0 V

4

1.0 V

Explanation:

C Given that,
$v_{o} = 3.3 \times 10^{14} Hz$
$v = 8.2 \times 10^{14} Hz$
We know that,
${eV}_{o} = h (v - v_{o})$
$V_{o} = \frac{h (v - v_{o})}{e}$
$V_{o} = \frac{6.62 \times 10^{- 34} (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{- 19}}$
$V_{o} = 2.029$
$V_{o} \approx 2.0 V$

BCECE-2006

Dual nature of radiation and Matter

142089 Light of wavelength $λ$ , strikes a photoelectric surface and electrons are ejected with an energy $E$ . If $E$ is to be increased to exactly twice its original value, the wavelength changes to $λ^{'}$ , where-

1

λ^{'}

is less than

\frac{λ}{2}

2

λ^{'}

is greater than

\frac{λ}{2}

3

λ^{'}

is greater than

\frac{λ}{2}

but less than

λ

4

λ^{'}

is exactly equal to

\frac{λ}{2}

Explanation:

C According to Einstein photoelectric equation
$\frac{hc}{λ} = E + ϕ_{o}$
$\frac{hc}{λ^{'}} = 2 E + ϕ_{o}$
From equation (i) and (ii), we get-
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$\frac{λ^{'}}{λ} < 1$
$\frac{λ^{'}}{λ} < λ$
Also
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$= \frac{1}{2} [\frac{E + ϕ_{o}}{E + ϕ_{o} / 2}]$
$\frac{λ^{'}}{λ} > \frac{1}{2}$
So, from (iii) and (iv)
$λ > λ^{'} > \frac{λ}{2}$

BCECE-2010

Dual nature of radiation and Matter

142092 The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$ . The corresponding change in stopping potential is

1

0.03 V

2

0.3 V

3

3 V

4

3.3 V

Explanation:

B Given that,
$λ_{1} = 2000 \AA = 2 \times 10^{- 7} m$
$λ_{2} = 2100 \AA = 2.1 \times 10^{- 7} m$
We know that,
$V_{1} - V_{2} = \frac{hc}{e} (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{2} - \frac{1}{2.1}) \times 10^{7}$
$= 0.295 = 0.3 V$

Manipal UGET-2016

Dual nature of radiation and Matter

142093 Light of wavelength ' $λ$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will

1 increase

2 decrease

3 be zero

4 become exactly half

Explanation:

A Incident photon (E) $= h v$
$E = \frac{hc}{λ}$
$E \propto \frac{1}{λ}$
If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect,
$\frac{hc}{λ} = {KE}_{max} + ϕ_{o}$
$\frac{hc}{λ} = {eV}_{o} + ϕ_{o}$
${eV}_{o} = hc [\frac{1}{λ} - \frac{1}{λ_{o}}]$
$V_{o} \propto (\frac{1}{λ} - \frac{1}{λ_{o}})$
Hence, if wavelength decreases, stopping potential increases.

MHT-CET 2016

Dual nature of radiation and Matter

142088 The threshold frequency for certain metal is $3.3$ $\times 10^{14} Hz$ . If light of frequency $8.2 \times 10^{14} Hz$ is incident on the metal, the cut-off voltage of the photoelectric current will be -

1

4.9 V

2

3.0 V

3

2.0 V

4

1.0 V

Explanation:

C Given that,
$v_{o} = 3.3 \times 10^{14} Hz$
$v = 8.2 \times 10^{14} Hz$
We know that,
${eV}_{o} = h (v - v_{o})$
$V_{o} = \frac{h (v - v_{o})}{e}$
$V_{o} = \frac{6.62 \times 10^{- 34} (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{- 19}}$
$V_{o} = 2.029$
$V_{o} \approx 2.0 V$

BCECE-2006

Dual nature of radiation and Matter

142089 Light of wavelength $λ$ , strikes a photoelectric surface and electrons are ejected with an energy $E$ . If $E$ is to be increased to exactly twice its original value, the wavelength changes to $λ^{'}$ , where-

1

λ^{'}

is less than

\frac{λ}{2}

2

λ^{'}

is greater than

\frac{λ}{2}

3

λ^{'}

is greater than

\frac{λ}{2}

but less than

λ

4

λ^{'}

is exactly equal to

\frac{λ}{2}

Explanation:

C According to Einstein photoelectric equation
$\frac{hc}{λ} = E + ϕ_{o}$
$\frac{hc}{λ^{'}} = 2 E + ϕ_{o}$
From equation (i) and (ii), we get-
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$\frac{λ^{'}}{λ} < 1$
$\frac{λ^{'}}{λ} < λ$
Also
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$= \frac{1}{2} [\frac{E + ϕ_{o}}{E + ϕ_{o} / 2}]$
$\frac{λ^{'}}{λ} > \frac{1}{2}$
So, from (iii) and (iv)
$λ > λ^{'} > \frac{λ}{2}$

BCECE-2010

Dual nature of radiation and Matter

142092 The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$ . The corresponding change in stopping potential is

1

0.03 V

2

0.3 V

3

3 V

4

3.3 V

Explanation:

B Given that,
$λ_{1} = 2000 \AA = 2 \times 10^{- 7} m$
$λ_{2} = 2100 \AA = 2.1 \times 10^{- 7} m$
We know that,
$V_{1} - V_{2} = \frac{hc}{e} (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{2} - \frac{1}{2.1}) \times 10^{7}$
$= 0.295 = 0.3 V$

Manipal UGET-2016

Dual nature of radiation and Matter

142093 Light of wavelength ' $λ$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will

1 increase

2 decrease

3 be zero

4 become exactly half

Explanation:

A Incident photon (E) $= h v$
$E = \frac{hc}{λ}$
$E \propto \frac{1}{λ}$
If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect,
$\frac{hc}{λ} = {KE}_{max} + ϕ_{o}$
$\frac{hc}{λ} = {eV}_{o} + ϕ_{o}$
${eV}_{o} = hc [\frac{1}{λ} - \frac{1}{λ_{o}}]$
$V_{o} \propto (\frac{1}{λ} - \frac{1}{λ_{o}})$
Hence, if wavelength decreases, stopping potential increases.

MHT-CET 2016

Dual nature of radiation and Matter

142088 The threshold frequency for certain metal is $3.3$ $\times 10^{14} Hz$ . If light of frequency $8.2 \times 10^{14} Hz$ is incident on the metal, the cut-off voltage of the photoelectric current will be -

1

4.9 V

2

3.0 V

3

2.0 V

4

1.0 V

Explanation:

C Given that,
$v_{o} = 3.3 \times 10^{14} Hz$
$v = 8.2 \times 10^{14} Hz$
We know that,
${eV}_{o} = h (v - v_{o})$
$V_{o} = \frac{h (v - v_{o})}{e}$
$V_{o} = \frac{6.62 \times 10^{- 34} (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{- 19}}$
$V_{o} = 2.029$
$V_{o} \approx 2.0 V$

BCECE-2006

Dual nature of radiation and Matter

142089 Light of wavelength $λ$ , strikes a photoelectric surface and electrons are ejected with an energy $E$ . If $E$ is to be increased to exactly twice its original value, the wavelength changes to $λ^{'}$ , where-

1

λ^{'}

is less than

\frac{λ}{2}

2

λ^{'}

is greater than

\frac{λ}{2}

3

λ^{'}

is greater than

\frac{λ}{2}

but less than

λ

4

λ^{'}

is exactly equal to

\frac{λ}{2}

Explanation:

C According to Einstein photoelectric equation
$\frac{hc}{λ} = E + ϕ_{o}$
$\frac{hc}{λ^{'}} = 2 E + ϕ_{o}$
From equation (i) and (ii), we get-
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$\frac{λ^{'}}{λ} < 1$
$\frac{λ^{'}}{λ} < λ$
Also
$\frac{λ^{'}}{λ} = (\frac{E + ϕ_{o}}{2 E + ϕ_{o}})$
$= \frac{1}{2} [\frac{E + ϕ_{o}}{E + ϕ_{o} / 2}]$
$\frac{λ^{'}}{λ} > \frac{1}{2}$
So, from (iii) and (iv)
$λ > λ^{'} > \frac{λ}{2}$

BCECE-2010

Dual nature of radiation and Matter

142092 The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$ . The corresponding change in stopping potential is

1

0.03 V

2

0.3 V

3

3 V

4

3.3 V

Explanation:

B Given that,
$λ_{1} = 2000 \AA = 2 \times 10^{- 7} m$
$λ_{2} = 2100 \AA = 2.1 \times 10^{- 7} m$
We know that,
$V_{1} - V_{2} = \frac{hc}{e} (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$
$= \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} (\frac{1}{2} - \frac{1}{2.1}) \times 10^{7}$
$= 0.295 = 0.3 V$

Manipal UGET-2016

Dual nature of radiation and Matter

142093 Light of wavelength ' $λ$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will

1 increase

2 decrease

3 be zero

4 become exactly half

Explanation:

A Incident photon (E) $= h v$
$E = \frac{hc}{λ}$
$E \propto \frac{1}{λ}$
If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect,
$\frac{hc}{λ} = {KE}_{max} + ϕ_{o}$
$\frac{hc}{λ} = {eV}_{o} + ϕ_{o}$
${eV}_{o} = hc [\frac{1}{λ} - \frac{1}{λ_{o}}]$
$V_{o} \propto (\frac{1}{λ} - \frac{1}{λ_{o}})$
Hence, if wavelength decreases, stopping potential increases.

MHT-CET 2016