QBManager

Dual nature of radiation and Matter

142582 The minimum wavelength of $X$ -rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:

1

V^{2}

2

\sqrt{V}

3

\frac{1}{V}

4

\frac{1}{\sqrt{V}}

Explanation:

C Given: Electron (e) accelerated through potential difference $= V$
The maximum kinetic energy gained by electron $= eV$ $(e =$ charge of electron $)$
This energy is transferred to $x$ -ray during emission Energy of $x$ -ray $= h ν = \frac{h c}{λ}$
As K.E of $e^{-}$ is maximum, thus $λ$ is minimum.
$\Rightarrow$ So, energy of emitted $x$ -ray $=$ energy of $e^{-}$
$\frac{hc}{λ_{min}} = eV \Rightarrow λ_{min} = \frac{hc}{e} \times \frac{1}{V}$
$λ_{min} \propto \frac{1}{V}$
Hence, option (c) is correct.

NEET (UG) 07.05.2023

Dual nature of radiation and Matter

142584 The kinetic energy of an electron, $α$ -particle and a proton are given as $4 K, 2 K$ and $K$ respectively. The de-Broglie wavelength associated with electron $(λ_{e}) α$ -particle $(λ_{- α})$ and the proton $(λ_{p})$ are as follows :

1

λ_{u} = λ_{p} < λ_{e}

2

λ_{α} > λ_{p} > λ_{c}

3

λ_{a} < λ_{p} < λ_{e}

4

λ_{α} = λ_{p} > λ_{c}

Explanation:

C | | Electron | Alpha | Proton |
| :--- | :--- | :--- | :--- |
| Mass | $\frac{m}{1840}$ | $4 m$ | $m$ |
| Charge | $e$ | $2 e$ | $e$ |
| Kinetic \lt br> energy | $4 K$ | $2 K$ | $K$ |
| $λ = \frac{h}{\sqrt{2 mK}}$ | $\frac{h}{\sqrt{2 \cdot \frac{m}{1840} \cdot 4 K}}$ | $\frac{h}{\sqrt{2 \cdot 4 m \cdot 2 K}}$ | $\frac{h}{\sqrt{2 mK}}$ |

Shift-I

Dual nature of radiation and Matter

142585 The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume $m_{p} = m_{e} \times 1849$ )

1

1 : 43

2

1 : 30

3

1 : 62

4

2 : 43

Explanation:

A We know that,
De-Brogli wavelength,
$λ \propto \frac{1}{\sqrt{m}}$
For proton $λ_{p} = \frac{1}{\sqrt{m_{p}}}$
For electron $λ_{e} = \frac{1}{\sqrt{m_{e}}}$
Equation (i) dividing by equation (ii),
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{1849 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{1}{1849}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{43}$
$λ_{p} : λ_{e} = 1 : 43$

Shift-II

Dual nature of radiation and Matter

1 A - I, B - II, C -III, D - IV

2 A -I, B - III, C -IV, D - II

3 A - IV, B - I, C- II, D - III

4 A- IV, B - III, C - II, D - I

Shift-I

Dual nature of radiation and Matter

142587 Match List - I with List - II
| | List - I | | List - II |
| :--- | :--- | :--- | :--- |
| A. | Microwaves | I. | Physiotherapy |
| B. | UV rays | II. | Treatment of \lt br> cancer |
| C. | Infra-red light | III. | Lasik eye \lt br> surgery |
| D. | X-ray | IV. | Aircraft \lt br> navigation |
Choose the correct answer from the options given below:

1 A - IV, B - I, C - II, D - III

2 A - II, B - IV, C - III, D - I

3 A - IV, B - III, C - I, D - II

4 A - III, B - II, C - I, D - IV

Shift-II

Dual nature of radiation and Matter

142582 The minimum wavelength of $X$ -rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:

1

V^{2}

2

\sqrt{V}

3

\frac{1}{V}

4

\frac{1}{\sqrt{V}}

Explanation:

C Given: Electron (e) accelerated through potential difference $= V$
The maximum kinetic energy gained by electron $= eV$ $(e =$ charge of electron $)$
This energy is transferred to $x$ -ray during emission Energy of $x$ -ray $= h ν = \frac{h c}{λ}$
As K.E of $e^{-}$ is maximum, thus $λ$ is minimum.
$\Rightarrow$ So, energy of emitted $x$ -ray $=$ energy of $e^{-}$
$\frac{hc}{λ_{min}} = eV \Rightarrow λ_{min} = \frac{hc}{e} \times \frac{1}{V}$
$λ_{min} \propto \frac{1}{V}$
Hence, option (c) is correct.

NEET (UG) 07.05.2023

Dual nature of radiation and Matter

142584 The kinetic energy of an electron, $α$ -particle and a proton are given as $4 K, 2 K$ and $K$ respectively. The de-Broglie wavelength associated with electron $(λ_{e}) α$ -particle $(λ_{- α})$ and the proton $(λ_{p})$ are as follows :

1

λ_{u} = λ_{p} < λ_{e}

2

λ_{α} > λ_{p} > λ_{c}

3

λ_{a} < λ_{p} < λ_{e}

4

λ_{α} = λ_{p} > λ_{c}

Explanation:

C | | Electron | Alpha | Proton |
| :--- | :--- | :--- | :--- |
| Mass | $\frac{m}{1840}$ | $4 m$ | $m$ |
| Charge | $e$ | $2 e$ | $e$ |
| Kinetic \lt br> energy | $4 K$ | $2 K$ | $K$ |
| $λ = \frac{h}{\sqrt{2 mK}}$ | $\frac{h}{\sqrt{2 \cdot \frac{m}{1840} \cdot 4 K}}$ | $\frac{h}{\sqrt{2 \cdot 4 m \cdot 2 K}}$ | $\frac{h}{\sqrt{2 mK}}$ |

Shift-I

Dual nature of radiation and Matter

142585 The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume $m_{p} = m_{e} \times 1849$ )

1

1 : 43

2

1 : 30

3

1 : 62

4

2 : 43

Explanation:

A We know that,
De-Brogli wavelength,
$λ \propto \frac{1}{\sqrt{m}}$
For proton $λ_{p} = \frac{1}{\sqrt{m_{p}}}$
For electron $λ_{e} = \frac{1}{\sqrt{m_{e}}}$
Equation (i) dividing by equation (ii),
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{1849 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{1}{1849}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{43}$
$λ_{p} : λ_{e} = 1 : 43$

Shift-II

Dual nature of radiation and Matter

1 A - I, B - II, C -III, D - IV

2 A -I, B - III, C -IV, D - II

3 A - IV, B - I, C- II, D - III

4 A- IV, B - III, C - II, D - I

Shift-I

Dual nature of radiation and Matter

142587 Match List - I with List - II
| | List - I | | List - II |
| :--- | :--- | :--- | :--- |
| A. | Microwaves | I. | Physiotherapy |
| B. | UV rays | II. | Treatment of \lt br> cancer |
| C. | Infra-red light | III. | Lasik eye \lt br> surgery |
| D. | X-ray | IV. | Aircraft \lt br> navigation |
Choose the correct answer from the options given below:

1 A - IV, B - I, C - II, D - III

2 A - II, B - IV, C - III, D - I

3 A - IV, B - III, C - I, D - II

4 A - III, B - II, C - I, D - IV

Shift-II

Dual nature of radiation and Matter

142582 The minimum wavelength of $X$ -rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:

1

V^{2}

2

\sqrt{V}

3

\frac{1}{V}

4

\frac{1}{\sqrt{V}}

Explanation:

C Given: Electron (e) accelerated through potential difference $= V$
The maximum kinetic energy gained by electron $= eV$ $(e =$ charge of electron $)$
This energy is transferred to $x$ -ray during emission Energy of $x$ -ray $= h ν = \frac{h c}{λ}$
As K.E of $e^{-}$ is maximum, thus $λ$ is minimum.
$\Rightarrow$ So, energy of emitted $x$ -ray $=$ energy of $e^{-}$
$\frac{hc}{λ_{min}} = eV \Rightarrow λ_{min} = \frac{hc}{e} \times \frac{1}{V}$
$λ_{min} \propto \frac{1}{V}$
Hence, option (c) is correct.

NEET (UG) 07.05.2023

Dual nature of radiation and Matter

142584 The kinetic energy of an electron, $α$ -particle and a proton are given as $4 K, 2 K$ and $K$ respectively. The de-Broglie wavelength associated with electron $(λ_{e}) α$ -particle $(λ_{- α})$ and the proton $(λ_{p})$ are as follows :

1

λ_{u} = λ_{p} < λ_{e}

2

λ_{α} > λ_{p} > λ_{c}

3

λ_{a} < λ_{p} < λ_{e}

4

λ_{α} = λ_{p} > λ_{c}

Explanation:

C | | Electron | Alpha | Proton |
| :--- | :--- | :--- | :--- |
| Mass | $\frac{m}{1840}$ | $4 m$ | $m$ |
| Charge | $e$ | $2 e$ | $e$ |
| Kinetic \lt br> energy | $4 K$ | $2 K$ | $K$ |
| $λ = \frac{h}{\sqrt{2 mK}}$ | $\frac{h}{\sqrt{2 \cdot \frac{m}{1840} \cdot 4 K}}$ | $\frac{h}{\sqrt{2 \cdot 4 m \cdot 2 K}}$ | $\frac{h}{\sqrt{2 mK}}$ |

Shift-I

Dual nature of radiation and Matter

142585 The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume $m_{p} = m_{e} \times 1849$ )

1

1 : 43

2

1 : 30

3

1 : 62

4

2 : 43

Explanation:

A We know that,
De-Brogli wavelength,
$λ \propto \frac{1}{\sqrt{m}}$
For proton $λ_{p} = \frac{1}{\sqrt{m_{p}}}$
For electron $λ_{e} = \frac{1}{\sqrt{m_{e}}}$
Equation (i) dividing by equation (ii),
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{1849 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{1}{1849}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{43}$
$λ_{p} : λ_{e} = 1 : 43$

Shift-II

Dual nature of radiation and Matter

1 A - I, B - II, C -III, D - IV

2 A -I, B - III, C -IV, D - II

3 A - IV, B - I, C- II, D - III

4 A- IV, B - III, C - II, D - I

Shift-I

Dual nature of radiation and Matter

142587 Match List - I with List - II
| | List - I | | List - II |
| :--- | :--- | :--- | :--- |
| A. | Microwaves | I. | Physiotherapy |
| B. | UV rays | II. | Treatment of \lt br> cancer |
| C. | Infra-red light | III. | Lasik eye \lt br> surgery |
| D. | X-ray | IV. | Aircraft \lt br> navigation |
Choose the correct answer from the options given below:

1 A - IV, B - I, C - II, D - III

2 A - II, B - IV, C - III, D - I

3 A - IV, B - III, C - I, D - II

4 A - III, B - II, C - I, D - IV

Shift-II

Dual nature of radiation and Matter

142582 The minimum wavelength of $X$ -rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:

1

V^{2}

2

\sqrt{V}

3

\frac{1}{V}

4

\frac{1}{\sqrt{V}}

Explanation:

C Given: Electron (e) accelerated through potential difference $= V$
The maximum kinetic energy gained by electron $= eV$ $(e =$ charge of electron $)$
This energy is transferred to $x$ -ray during emission Energy of $x$ -ray $= h ν = \frac{h c}{λ}$
As K.E of $e^{-}$ is maximum, thus $λ$ is minimum.
$\Rightarrow$ So, energy of emitted $x$ -ray $=$ energy of $e^{-}$
$\frac{hc}{λ_{min}} = eV \Rightarrow λ_{min} = \frac{hc}{e} \times \frac{1}{V}$
$λ_{min} \propto \frac{1}{V}$
Hence, option (c) is correct.

NEET (UG) 07.05.2023

Dual nature of radiation and Matter

142584 The kinetic energy of an electron, $α$ -particle and a proton are given as $4 K, 2 K$ and $K$ respectively. The de-Broglie wavelength associated with electron $(λ_{e}) α$ -particle $(λ_{- α})$ and the proton $(λ_{p})$ are as follows :

1

λ_{u} = λ_{p} < λ_{e}

2

λ_{α} > λ_{p} > λ_{c}

3

λ_{a} < λ_{p} < λ_{e}

4

λ_{α} = λ_{p} > λ_{c}

Explanation:

C | | Electron | Alpha | Proton |
| :--- | :--- | :--- | :--- |
| Mass | $\frac{m}{1840}$ | $4 m$ | $m$ |
| Charge | $e$ | $2 e$ | $e$ |
| Kinetic \lt br> energy | $4 K$ | $2 K$ | $K$ |
| $λ = \frac{h}{\sqrt{2 mK}}$ | $\frac{h}{\sqrt{2 \cdot \frac{m}{1840} \cdot 4 K}}$ | $\frac{h}{\sqrt{2 \cdot 4 m \cdot 2 K}}$ | $\frac{h}{\sqrt{2 mK}}$ |

Shift-I

Dual nature of radiation and Matter

142585 The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume $m_{p} = m_{e} \times 1849$ )

1

1 : 43

2

1 : 30

3

1 : 62

4

2 : 43

Explanation:

A We know that,
De-Brogli wavelength,
$λ \propto \frac{1}{\sqrt{m}}$
For proton $λ_{p} = \frac{1}{\sqrt{m_{p}}}$
For electron $λ_{e} = \frac{1}{\sqrt{m_{e}}}$
Equation (i) dividing by equation (ii),
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{1849 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{1}{1849}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{43}$
$λ_{p} : λ_{e} = 1 : 43$

Shift-II

Dual nature of radiation and Matter

1 A - I, B - II, C -III, D - IV

2 A -I, B - III, C -IV, D - II

3 A - IV, B - I, C- II, D - III

4 A- IV, B - III, C - II, D - I

Shift-I

Dual nature of radiation and Matter

142587 Match List - I with List - II
| | List - I | | List - II |
| :--- | :--- | :--- | :--- |
| A. | Microwaves | I. | Physiotherapy |
| B. | UV rays | II. | Treatment of \lt br> cancer |
| C. | Infra-red light | III. | Lasik eye \lt br> surgery |
| D. | X-ray | IV. | Aircraft \lt br> navigation |
Choose the correct answer from the options given below:

1 A - IV, B - I, C - II, D - III

2 A - II, B - IV, C - III, D - I

3 A - IV, B - III, C - I, D - II

4 A - III, B - II, C - I, D - IV

Shift-II

Dual nature of radiation and Matter

142582 The minimum wavelength of $X$ -rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:

1

V^{2}

2

\sqrt{V}

3

\frac{1}{V}

4

\frac{1}{\sqrt{V}}

Explanation:

C Given: Electron (e) accelerated through potential difference $= V$
The maximum kinetic energy gained by electron $= eV$ $(e =$ charge of electron $)$
This energy is transferred to $x$ -ray during emission Energy of $x$ -ray $= h ν = \frac{h c}{λ}$
As K.E of $e^{-}$ is maximum, thus $λ$ is minimum.
$\Rightarrow$ So, energy of emitted $x$ -ray $=$ energy of $e^{-}$
$\frac{hc}{λ_{min}} = eV \Rightarrow λ_{min} = \frac{hc}{e} \times \frac{1}{V}$
$λ_{min} \propto \frac{1}{V}$
Hence, option (c) is correct.

NEET (UG) 07.05.2023

Dual nature of radiation and Matter

142584 The kinetic energy of an electron, $α$ -particle and a proton are given as $4 K, 2 K$ and $K$ respectively. The de-Broglie wavelength associated with electron $(λ_{e}) α$ -particle $(λ_{- α})$ and the proton $(λ_{p})$ are as follows :

1

λ_{u} = λ_{p} < λ_{e}

2

λ_{α} > λ_{p} > λ_{c}

3

λ_{a} < λ_{p} < λ_{e}

4

λ_{α} = λ_{p} > λ_{c}

Explanation:

C | | Electron | Alpha | Proton |
| :--- | :--- | :--- | :--- |
| Mass | $\frac{m}{1840}$ | $4 m$ | $m$ |
| Charge | $e$ | $2 e$ | $e$ |
| Kinetic \lt br> energy | $4 K$ | $2 K$ | $K$ |
| $λ = \frac{h}{\sqrt{2 mK}}$ | $\frac{h}{\sqrt{2 \cdot \frac{m}{1840} \cdot 4 K}}$ | $\frac{h}{\sqrt{2 \cdot 4 m \cdot 2 K}}$ | $\frac{h}{\sqrt{2 mK}}$ |

Shift-I

Dual nature of radiation and Matter

142585 The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume $m_{p} = m_{e} \times 1849$ )

1

1 : 43

2

1 : 30

3

1 : 62

4

2 : 43

Explanation:

A We know that,
De-Brogli wavelength,
$λ \propto \frac{1}{\sqrt{m}}$
For proton $λ_{p} = \frac{1}{\sqrt{m_{p}}}$
For electron $λ_{e} = \frac{1}{\sqrt{m_{e}}}$
Equation (i) dividing by equation (ii),
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{1849 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{1}{1849}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{43}$
$λ_{p} : λ_{e} = 1 : 43$

Shift-II

Dual nature of radiation and Matter

1 A - I, B - II, C -III, D - IV

2 A -I, B - III, C -IV, D - II

3 A - IV, B - I, C- II, D - III

4 A- IV, B - III, C - II, D - I

Shift-I

Dual nature of radiation and Matter

142587 Match List - I with List - II
| | List - I | | List - II |
| :--- | :--- | :--- | :--- |
| A. | Microwaves | I. | Physiotherapy |
| B. | UV rays | II. | Treatment of \lt br> cancer |
| C. | Infra-red light | III. | Lasik eye \lt br> surgery |
| D. | X-ray | IV. | Aircraft \lt br> navigation |
Choose the correct answer from the options given below:

1 A - IV, B - I, C - II, D - III

2 A - II, B - IV, C - III, D - I

3 A - IV, B - III, C - I, D - II

4 A - III, B - II, C - I, D - IV

Shift-II