QBManager

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357587 Photon and electron are given same energy $(10^{- 20} J)$ . Wavelength associated with photon and electron are $λ_{p}$ and $λ_{e}$ , the correct statement will be

1

λ_{p} > λ_{e}

2

λ_{p} < λ_{e}

3

λ_{p} = λ_{e}

4

\frac{λ_{e}}{λ_{p}} = c

Explanation:

We have, $λ_{p} = h c / E; E =$ energy given to photon
$λ_{e} = \frac{h}{\sqrt{2 M_{e} E}}$
$E =$ energy given to electron
$\Rightarrow \frac{λ_{p}}{λ_{c}} = c \sqrt{\frac{2 M_{e}}{E}} > 1 \Rightarrow λ_{p} > λ_{e}$

AIIMS - 2012

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357588 Monochromatic light of wavelength $667 n m$ is produced by a helium neon laser. The power emitted is $9 m W$ . The number of photons arriving per second on the average at a target irradiated by this beam is

1

9 \times 10^{17}

2

9 \times 10^{15}

3

3 \times 10^{19}

4

3 \times 10^{16}

Explanation:

Here, $λ = 667 \times 10^{- 9} m, P = 9 \times 10^{- 3} W$
$\begin{aligned} Power = \frac{energy (E)}{time (t)} \\ = \frac{N h c}{λ} \end{aligned}$
where $N$ - total number of photons emitted per second
So, $N = \frac{P \times λ}{h c}$
$= \frac{9 \times 10^{- 3} \times 667 \times 10^{- 9}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 3 \times 10^{16} / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357589 A photocell is illuminated by a small bright source placed 1 $m$ away. When the same source is placed $\frac{1}{2} m$ away, the number of electrons emitted would

1 increase by a factor of 2

2 increase by a factor of 4

3 increase by a factor of 8

4 increase by a factor of 16

Explanation:

Number of electrons emitted is inversely proportional to the square of distance, $i . e .,$ $n \propto \frac{1}{r^{2}}$ .
$∴ n_{1} \propto \frac{1}{{(1)}^{2}} (1)$
Also, $n_{2} \propto \frac{1}{{(\frac{1}{2})}^{2}} (2)$
Dividing eqs. (1) and (2), we get
$\frac{n_{1}}{n_{2}} = \frac{1}{4} \Rightarrow n_{2} = 4 n_{1}$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357590 If the mass of neutron $= 1.7 \times 10^{- 27} k g$ . Then the de - broglie wavelength of neutron of energy $3 e V$ is:

1

1.6 \times 10^{- 10} m

2

1.6 \times 10^{- 11} m

3

1.4 \times 10^{- 10} m

4

1.4 \times 10^{- 11} m

Explanation:

For a neutron, $m = 1.7 \times 10^{- 17} k g$
$\begin{aligned} λ = \frac{h}{\sqrt{2 m E}} \\ λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.7 \times 10^{- 29} \times 3 \times 1.6 \times 10^{- 19}}} \end{aligned}$
$= \frac{6.6 \times 10^{- 34}}{10^{- 23} \sqrt{2 \times 1.7 \times 4.8}}$
$= \frac{6.6 \times 10^{11}}{\sqrt{3.4 \times 4.8}} = 1.633 \times 10^{11} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357591 Assertion :
Mass of moving photon varies inversely as the wavelength.
Reason :
Energy of the particle with mass is
$[$ Mass $\times (Speed of light)^{2}]$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Photon has zero rest-mass. It has mass when moving with $c$ as given by de Broglie formula $λ = \frac{h}{m c}$ . Then $E = m c^{2}$ is a separate fact. So correct option is (2).

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357587 Photon and electron are given same energy $(10^{- 20} J)$ . Wavelength associated with photon and electron are $λ_{p}$ and $λ_{e}$ , the correct statement will be

1

λ_{p} > λ_{e}

2

λ_{p} < λ_{e}

3

λ_{p} = λ_{e}

4

\frac{λ_{e}}{λ_{p}} = c

Explanation:

We have, $λ_{p} = h c / E; E =$ energy given to photon
$λ_{e} = \frac{h}{\sqrt{2 M_{e} E}}$
$E =$ energy given to electron
$\Rightarrow \frac{λ_{p}}{λ_{c}} = c \sqrt{\frac{2 M_{e}}{E}} > 1 \Rightarrow λ_{p} > λ_{e}$

AIIMS - 2012

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357588 Monochromatic light of wavelength $667 n m$ is produced by a helium neon laser. The power emitted is $9 m W$ . The number of photons arriving per second on the average at a target irradiated by this beam is

1

9 \times 10^{17}

2

9 \times 10^{15}

3

3 \times 10^{19}

4

3 \times 10^{16}

Explanation:

Here, $λ = 667 \times 10^{- 9} m, P = 9 \times 10^{- 3} W$
$\begin{aligned} Power = \frac{energy (E)}{time (t)} \\ = \frac{N h c}{λ} \end{aligned}$
where $N$ - total number of photons emitted per second
So, $N = \frac{P \times λ}{h c}$
$= \frac{9 \times 10^{- 3} \times 667 \times 10^{- 9}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 3 \times 10^{16} / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357589 A photocell is illuminated by a small bright source placed 1 $m$ away. When the same source is placed $\frac{1}{2} m$ away, the number of electrons emitted would

1 increase by a factor of 2

2 increase by a factor of 4

3 increase by a factor of 8

4 increase by a factor of 16

Explanation:

Number of electrons emitted is inversely proportional to the square of distance, $i . e .,$ $n \propto \frac{1}{r^{2}}$ .
$∴ n_{1} \propto \frac{1}{{(1)}^{2}} (1)$
Also, $n_{2} \propto \frac{1}{{(\frac{1}{2})}^{2}} (2)$
Dividing eqs. (1) and (2), we get
$\frac{n_{1}}{n_{2}} = \frac{1}{4} \Rightarrow n_{2} = 4 n_{1}$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357590 If the mass of neutron $= 1.7 \times 10^{- 27} k g$ . Then the de - broglie wavelength of neutron of energy $3 e V$ is:

1

1.6 \times 10^{- 10} m

2

1.6 \times 10^{- 11} m

3

1.4 \times 10^{- 10} m

4

1.4 \times 10^{- 11} m

Explanation:

For a neutron, $m = 1.7 \times 10^{- 17} k g$
$\begin{aligned} λ = \frac{h}{\sqrt{2 m E}} \\ λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.7 \times 10^{- 29} \times 3 \times 1.6 \times 10^{- 19}}} \end{aligned}$
$= \frac{6.6 \times 10^{- 34}}{10^{- 23} \sqrt{2 \times 1.7 \times 4.8}}$
$= \frac{6.6 \times 10^{11}}{\sqrt{3.4 \times 4.8}} = 1.633 \times 10^{11} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357591 Assertion :
Mass of moving photon varies inversely as the wavelength.
Reason :
Energy of the particle with mass is
$[$ Mass $\times (Speed of light)^{2}]$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Photon has zero rest-mass. It has mass when moving with $c$ as given by de Broglie formula $λ = \frac{h}{m c}$ . Then $E = m c^{2}$ is a separate fact. So correct option is (2).

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357587 Photon and electron are given same energy $(10^{- 20} J)$ . Wavelength associated with photon and electron are $λ_{p}$ and $λ_{e}$ , the correct statement will be

1

λ_{p} > λ_{e}

2

λ_{p} < λ_{e}

3

λ_{p} = λ_{e}

4

\frac{λ_{e}}{λ_{p}} = c

Explanation:

We have, $λ_{p} = h c / E; E =$ energy given to photon
$λ_{e} = \frac{h}{\sqrt{2 M_{e} E}}$
$E =$ energy given to electron
$\Rightarrow \frac{λ_{p}}{λ_{c}} = c \sqrt{\frac{2 M_{e}}{E}} > 1 \Rightarrow λ_{p} > λ_{e}$

AIIMS - 2012

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357588 Monochromatic light of wavelength $667 n m$ is produced by a helium neon laser. The power emitted is $9 m W$ . The number of photons arriving per second on the average at a target irradiated by this beam is

1

9 \times 10^{17}

2

9 \times 10^{15}

3

3 \times 10^{19}

4

3 \times 10^{16}

Explanation:

Here, $λ = 667 \times 10^{- 9} m, P = 9 \times 10^{- 3} W$
$\begin{aligned} Power = \frac{energy (E)}{time (t)} \\ = \frac{N h c}{λ} \end{aligned}$
where $N$ - total number of photons emitted per second
So, $N = \frac{P \times λ}{h c}$
$= \frac{9 \times 10^{- 3} \times 667 \times 10^{- 9}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 3 \times 10^{16} / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357589 A photocell is illuminated by a small bright source placed 1 $m$ away. When the same source is placed $\frac{1}{2} m$ away, the number of electrons emitted would

1 increase by a factor of 2

2 increase by a factor of 4

3 increase by a factor of 8

4 increase by a factor of 16

Explanation:

Number of electrons emitted is inversely proportional to the square of distance, $i . e .,$ $n \propto \frac{1}{r^{2}}$ .
$∴ n_{1} \propto \frac{1}{{(1)}^{2}} (1)$
Also, $n_{2} \propto \frac{1}{{(\frac{1}{2})}^{2}} (2)$
Dividing eqs. (1) and (2), we get
$\frac{n_{1}}{n_{2}} = \frac{1}{4} \Rightarrow n_{2} = 4 n_{1}$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357590 If the mass of neutron $= 1.7 \times 10^{- 27} k g$ . Then the de - broglie wavelength of neutron of energy $3 e V$ is:

1

1.6 \times 10^{- 10} m

2

1.6 \times 10^{- 11} m

3

1.4 \times 10^{- 10} m

4

1.4 \times 10^{- 11} m

Explanation:

For a neutron, $m = 1.7 \times 10^{- 17} k g$
$\begin{aligned} λ = \frac{h}{\sqrt{2 m E}} \\ λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.7 \times 10^{- 29} \times 3 \times 1.6 \times 10^{- 19}}} \end{aligned}$
$= \frac{6.6 \times 10^{- 34}}{10^{- 23} \sqrt{2 \times 1.7 \times 4.8}}$
$= \frac{6.6 \times 10^{11}}{\sqrt{3.4 \times 4.8}} = 1.633 \times 10^{11} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357591 Assertion :
Mass of moving photon varies inversely as the wavelength.
Reason :
Energy of the particle with mass is
$[$ Mass $\times (Speed of light)^{2}]$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Photon has zero rest-mass. It has mass when moving with $c$ as given by de Broglie formula $λ = \frac{h}{m c}$ . Then $E = m c^{2}$ is a separate fact. So correct option is (2).

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357587 Photon and electron are given same energy $(10^{- 20} J)$ . Wavelength associated with photon and electron are $λ_{p}$ and $λ_{e}$ , the correct statement will be

1

λ_{p} > λ_{e}

2

λ_{p} < λ_{e}

3

λ_{p} = λ_{e}

4

\frac{λ_{e}}{λ_{p}} = c

Explanation:

We have, $λ_{p} = h c / E; E =$ energy given to photon
$λ_{e} = \frac{h}{\sqrt{2 M_{e} E}}$
$E =$ energy given to electron
$\Rightarrow \frac{λ_{p}}{λ_{c}} = c \sqrt{\frac{2 M_{e}}{E}} > 1 \Rightarrow λ_{p} > λ_{e}$

AIIMS - 2012

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357588 Monochromatic light of wavelength $667 n m$ is produced by a helium neon laser. The power emitted is $9 m W$ . The number of photons arriving per second on the average at a target irradiated by this beam is

1

9 \times 10^{17}

2

9 \times 10^{15}

3

3 \times 10^{19}

4

3 \times 10^{16}

Explanation:

Here, $λ = 667 \times 10^{- 9} m, P = 9 \times 10^{- 3} W$
$\begin{aligned} Power = \frac{energy (E)}{time (t)} \\ = \frac{N h c}{λ} \end{aligned}$
where $N$ - total number of photons emitted per second
So, $N = \frac{P \times λ}{h c}$
$= \frac{9 \times 10^{- 3} \times 667 \times 10^{- 9}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 3 \times 10^{16} / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357589 A photocell is illuminated by a small bright source placed 1 $m$ away. When the same source is placed $\frac{1}{2} m$ away, the number of electrons emitted would

1 increase by a factor of 2

2 increase by a factor of 4

3 increase by a factor of 8

4 increase by a factor of 16

Explanation:

Number of electrons emitted is inversely proportional to the square of distance, $i . e .,$ $n \propto \frac{1}{r^{2}}$ .
$∴ n_{1} \propto \frac{1}{{(1)}^{2}} (1)$
Also, $n_{2} \propto \frac{1}{{(\frac{1}{2})}^{2}} (2)$
Dividing eqs. (1) and (2), we get
$\frac{n_{1}}{n_{2}} = \frac{1}{4} \Rightarrow n_{2} = 4 n_{1}$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357590 If the mass of neutron $= 1.7 \times 10^{- 27} k g$ . Then the de - broglie wavelength of neutron of energy $3 e V$ is:

1

1.6 \times 10^{- 10} m

2

1.6 \times 10^{- 11} m

3

1.4 \times 10^{- 10} m

4

1.4 \times 10^{- 11} m

Explanation:

For a neutron, $m = 1.7 \times 10^{- 17} k g$
$\begin{aligned} λ = \frac{h}{\sqrt{2 m E}} \\ λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.7 \times 10^{- 29} \times 3 \times 1.6 \times 10^{- 19}}} \end{aligned}$
$= \frac{6.6 \times 10^{- 34}}{10^{- 23} \sqrt{2 \times 1.7 \times 4.8}}$
$= \frac{6.6 \times 10^{11}}{\sqrt{3.4 \times 4.8}} = 1.633 \times 10^{11} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357591 Assertion :
Mass of moving photon varies inversely as the wavelength.
Reason :
Energy of the particle with mass is
$[$ Mass $\times (Speed of light)^{2}]$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Photon has zero rest-mass. It has mass when moving with $c$ as given by de Broglie formula $λ = \frac{h}{m c}$ . Then $E = m c^{2}$ is a separate fact. So correct option is (2).

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357587 Photon and electron are given same energy $(10^{- 20} J)$ . Wavelength associated with photon and electron are $λ_{p}$ and $λ_{e}$ , the correct statement will be

1

λ_{p} > λ_{e}

2

λ_{p} < λ_{e}

3

λ_{p} = λ_{e}

4

\frac{λ_{e}}{λ_{p}} = c

Explanation:

We have, $λ_{p} = h c / E; E =$ energy given to photon
$λ_{e} = \frac{h}{\sqrt{2 M_{e} E}}$
$E =$ energy given to electron
$\Rightarrow \frac{λ_{p}}{λ_{c}} = c \sqrt{\frac{2 M_{e}}{E}} > 1 \Rightarrow λ_{p} > λ_{e}$

AIIMS - 2012

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357588 Monochromatic light of wavelength $667 n m$ is produced by a helium neon laser. The power emitted is $9 m W$ . The number of photons arriving per second on the average at a target irradiated by this beam is

1

9 \times 10^{17}

2

9 \times 10^{15}

3

3 \times 10^{19}

4

3 \times 10^{16}

Explanation:

Here, $λ = 667 \times 10^{- 9} m, P = 9 \times 10^{- 3} W$
$\begin{aligned} Power = \frac{energy (E)}{time (t)} \\ = \frac{N h c}{λ} \end{aligned}$
where $N$ - total number of photons emitted per second
So, $N = \frac{P \times λ}{h c}$
$= \frac{9 \times 10^{- 3} \times 667 \times 10^{- 9}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 3 \times 10^{16} / s$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357589 A photocell is illuminated by a small bright source placed 1 $m$ away. When the same source is placed $\frac{1}{2} m$ away, the number of electrons emitted would

1 increase by a factor of 2

2 increase by a factor of 4

3 increase by a factor of 8

4 increase by a factor of 16

Explanation:

Number of electrons emitted is inversely proportional to the square of distance, $i . e .,$ $n \propto \frac{1}{r^{2}}$ .
$∴ n_{1} \propto \frac{1}{{(1)}^{2}} (1)$
Also, $n_{2} \propto \frac{1}{{(\frac{1}{2})}^{2}} (2)$
Dividing eqs. (1) and (2), we get
$\frac{n_{1}}{n_{2}} = \frac{1}{4} \Rightarrow n_{2} = 4 n_{1}$ .
So correct option is (2)

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357590 If the mass of neutron $= 1.7 \times 10^{- 27} k g$ . Then the de - broglie wavelength of neutron of energy $3 e V$ is:

1

1.6 \times 10^{- 10} m

2

1.6 \times 10^{- 11} m

3

1.4 \times 10^{- 10} m

4

1.4 \times 10^{- 11} m

Explanation:

For a neutron, $m = 1.7 \times 10^{- 17} k g$
$\begin{aligned} λ = \frac{h}{\sqrt{2 m E}} \\ λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.7 \times 10^{- 29} \times 3 \times 1.6 \times 10^{- 19}}} \end{aligned}$
$= \frac{6.6 \times 10^{- 34}}{10^{- 23} \sqrt{2 \times 1.7 \times 4.8}}$
$= \frac{6.6 \times 10^{11}}{\sqrt{3.4 \times 4.8}} = 1.633 \times 10^{11} m$

PHXII11:DUAL NATURE OF RADIATION AND MATTER

357591 Assertion :
Mass of moving photon varies inversely as the wavelength.
Reason :
Energy of the particle with mass is
$[$ Mass $\times (Speed of light)^{2}]$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Photon has zero rest-mass. It has mass when moving with $c$ as given by de Broglie formula $λ = \frac{h}{m c}$ . Then $E = m c^{2}$ is a separate fact. So correct option is (2).