QBManager

Dual nature of radiation and Matter

142452 The energy of an electron having de-Broglic wavelength ' $λ$ ' is $(h =$ Plank's constant, $m =$ mass of electron)

1

\frac{h}{2 m λ}

2

\frac{h^{2}}{2 m λ^{2}}

3

\frac{h^{2}}{2 m^{2} λ^{2}}

4

\frac{h^{2}}{2 m^{2} λ}

Explanation:

B We know,
de-Broglie wavelength,
$λ = \frac{h}{p}$
$Or p = \frac{h}{λ}$
Kinetic energy $(E) = \frac{1}{2} {mv}^{2} = \frac{1}{2 m} (p^{2})$
$E = \frac{1}{2 m} \cdot \frac{h^{2}}{λ^{2}}$
$E = \frac{h^{2}}{2 m λ^{2}} (∵ p = \frac{h}{λ})$

MHT-CET 2018

Dual nature of radiation and Matter

142453 The masses of two particles having same kinetic energies are in the ratio 2:1. Then their de-Broglie wavelengths are in the ratio

1

2 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : \sqrt{2}

Explanation:

D Given that,
K.E. of $1^{st}$ object $=$ K.E. of $2^{nd}$ object
The ratio of their masses,
$\frac{m_{1}}{m_{2}} = \frac{2}{1}$
Form de-Broglie wavelength,
$λ = \frac{h}{p}$
Kinetic energy in terms of momentum,
$K.E = \frac{p^{2}}{2 m}$
$∴ K.E = \frac{h^{2}}{2 m λ^{2}}$
$λ^{2} = \frac{h^{2}}{2 m (K . E)}$
Since, $(K . E)_{1} = (K . E)_{2}$
$\frac{{(λ_{1})}^{2}}{(λ)^{2}} = \frac{(\frac{h^{2}}{2 m_{1} (K . E)_{1}})}{(\frac{h^{2}}{2 m_{2} (K \cdot E)_{1}})} = \frac{m_{2}}{m_{1}} = \frac{1}{2}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
$λ_{1} : λ_{2} = 1 : \sqrt{2}$

COMEDK 2018

Dual nature of radiation and Matter

142455 Two particles $A$ and $B$ of same mass have their total energies $E_{A}$ and $E_{B}$ in the ratio $E_{A} : E_{B}$ $= 1 : 2$ Their potential energies $U_{A}$ and $U_{B}$ are in the ratio $U_{A} : U_{B} = 1 : 2$ . If $λ_{A}$ and $λ_{B}$ are there de-Broglie wavelengths, then $λ_{A} : λ_{B}$ is

1

1 : 2

2

2 : 1

3

1 : \sqrt{2}

4

\sqrt{2} : 1

5

1 : 1

Explanation:

D Given that,
Energy ratio $= \frac{E_{A}}{E_{B}} = \frac{1}{2}$
Potential energy,
$\frac{U_{A}}{U_{B}} = \frac{1}{2}$
So, $E_{A} = x, E_{B} = 2 x$
$U_{A} = y, U_{B} = 2 y$
$∵ E_{A} = U_{A} + k_{A}$
$E_{B} = U_{B} + k_{B}$
Here $k_{A}$ and $k_{B}$ are kinetic energy of particles $A$ and $B$
So, $k_{A} = E_{A} - U_{A} = (x - y)$
$k_{B} = E_{B} - U_{B} = 2 (x - y)$ We know that,
de - Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mk}}$
$λ_{A} = \frac{h}{\sqrt{2 {mk}_{A}}} \Rightarrow λ_{B} = \frac{h}{\sqrt{2 {mk}_{B}}}$
$∴ \frac{λ_{A}}{λ_{B}} = \sqrt{\frac{k_{B}}{k_{A}}}$
From equation (i), (ii) and (iii)
$\frac{λ_{A}}{λ_{B}} = \sqrt{\frac{2 (x - y)}{(x - y)}} = \frac{\sqrt{2}}{1}$

Kerala CEE -2018

Dual nature of radiation and Matter

142456 If $λ_{1}$ and $λ_{2}$ denote the de-Broglie wavelengths of two particles with same masses but charges in the ratio of 1:2 after they are accelerated from rest through the same potential difference, then

1

λ_{1} = λ_{2}

2

λ_{1} < λ_{2}

3

λ_{1} > λ_{2}

4 None of these

Explanation:

C According to the de-Broglie wavelength-
$λ = \frac{h}{\sqrt{2 meV}}$
According to question,
$λ_{1} = \frac{h}{\sqrt{2 meV}}$
$λ_{2} = \frac{h}{\sqrt{2 {me}^{'} V}}$
From equation (i) and (ii),
$\frac{λ_{1}}{λ_{2}} = \frac{h}{\sqrt{2 meV}} \times \frac{\sqrt{2 {me}^{'} V}}{h}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{e^{'}}{e}} (∵ \frac{e}{e^{'}} = \frac{1}{2}) (given)$
$λ_{1} = \sqrt{2} λ_{2}$
Therefore, $λ_{1} > λ_{2}$

UPSEE - 2018

Dual nature of radiation and Matter

142452 The energy of an electron having de-Broglic wavelength ' $λ$ ' is $(h =$ Plank's constant, $m =$ mass of electron)

1

\frac{h}{2 m λ}

2

\frac{h^{2}}{2 m λ^{2}}

3

\frac{h^{2}}{2 m^{2} λ^{2}}

4

\frac{h^{2}}{2 m^{2} λ}

Explanation:

B We know,
de-Broglie wavelength,
$λ = \frac{h}{p}$
$Or p = \frac{h}{λ}$
Kinetic energy $(E) = \frac{1}{2} {mv}^{2} = \frac{1}{2 m} (p^{2})$
$E = \frac{1}{2 m} \cdot \frac{h^{2}}{λ^{2}}$
$E = \frac{h^{2}}{2 m λ^{2}} (∵ p = \frac{h}{λ})$

MHT-CET 2018

Dual nature of radiation and Matter

142453 The masses of two particles having same kinetic energies are in the ratio 2:1. Then their de-Broglie wavelengths are in the ratio

1

2 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : \sqrt{2}

Explanation:

D Given that,
K.E. of $1^{st}$ object $=$ K.E. of $2^{nd}$ object
The ratio of their masses,
$\frac{m_{1}}{m_{2}} = \frac{2}{1}$
Form de-Broglie wavelength,
$λ = \frac{h}{p}$
Kinetic energy in terms of momentum,
$K.E = \frac{p^{2}}{2 m}$
$∴ K.E = \frac{h^{2}}{2 m λ^{2}}$
$λ^{2} = \frac{h^{2}}{2 m (K . E)}$
Since, $(K . E)_{1} = (K . E)_{2}$
$\frac{{(λ_{1})}^{2}}{(λ)^{2}} = \frac{(\frac{h^{2}}{2 m_{1} (K . E)_{1}})}{(\frac{h^{2}}{2 m_{2} (K \cdot E)_{1}})} = \frac{m_{2}}{m_{1}} = \frac{1}{2}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
$λ_{1} : λ_{2} = 1 : \sqrt{2}$

COMEDK 2018

Dual nature of radiation and Matter

142455 Two particles $A$ and $B$ of same mass have their total energies $E_{A}$ and $E_{B}$ in the ratio $E_{A} : E_{B}$ $= 1 : 2$ Their potential energies $U_{A}$ and $U_{B}$ are in the ratio $U_{A} : U_{B} = 1 : 2$ . If $λ_{A}$ and $λ_{B}$ are there de-Broglie wavelengths, then $λ_{A} : λ_{B}$ is

1

1 : 2

2

2 : 1

3

1 : \sqrt{2}

4

\sqrt{2} : 1

5

1 : 1

Explanation:

D Given that,
Energy ratio $= \frac{E_{A}}{E_{B}} = \frac{1}{2}$
Potential energy,
$\frac{U_{A}}{U_{B}} = \frac{1}{2}$
So, $E_{A} = x, E_{B} = 2 x$
$U_{A} = y, U_{B} = 2 y$
$∵ E_{A} = U_{A} + k_{A}$
$E_{B} = U_{B} + k_{B}$
Here $k_{A}$ and $k_{B}$ are kinetic energy of particles $A$ and $B$
So, $k_{A} = E_{A} - U_{A} = (x - y)$
$k_{B} = E_{B} - U_{B} = 2 (x - y)$ We know that,
de - Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mk}}$
$λ_{A} = \frac{h}{\sqrt{2 {mk}_{A}}} \Rightarrow λ_{B} = \frac{h}{\sqrt{2 {mk}_{B}}}$
$∴ \frac{λ_{A}}{λ_{B}} = \sqrt{\frac{k_{B}}{k_{A}}}$
From equation (i), (ii) and (iii)
$\frac{λ_{A}}{λ_{B}} = \sqrt{\frac{2 (x - y)}{(x - y)}} = \frac{\sqrt{2}}{1}$

Kerala CEE -2018

Dual nature of radiation and Matter

142456 If $λ_{1}$ and $λ_{2}$ denote the de-Broglie wavelengths of two particles with same masses but charges in the ratio of 1:2 after they are accelerated from rest through the same potential difference, then

1

λ_{1} = λ_{2}

2

λ_{1} < λ_{2}

3

λ_{1} > λ_{2}

4 None of these

Explanation:

C According to the de-Broglie wavelength-
$λ = \frac{h}{\sqrt{2 meV}}$
According to question,
$λ_{1} = \frac{h}{\sqrt{2 meV}}$
$λ_{2} = \frac{h}{\sqrt{2 {me}^{'} V}}$
From equation (i) and (ii),
$\frac{λ_{1}}{λ_{2}} = \frac{h}{\sqrt{2 meV}} \times \frac{\sqrt{2 {me}^{'} V}}{h}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{e^{'}}{e}} (∵ \frac{e}{e^{'}} = \frac{1}{2}) (given)$
$λ_{1} = \sqrt{2} λ_{2}$
Therefore, $λ_{1} > λ_{2}$

UPSEE - 2018

Dual nature of radiation and Matter

142452 The energy of an electron having de-Broglic wavelength ' $λ$ ' is $(h =$ Plank's constant, $m =$ mass of electron)

1

\frac{h}{2 m λ}

2

\frac{h^{2}}{2 m λ^{2}}

3

\frac{h^{2}}{2 m^{2} λ^{2}}

4

\frac{h^{2}}{2 m^{2} λ}

Explanation:

B We know,
de-Broglie wavelength,
$λ = \frac{h}{p}$
$Or p = \frac{h}{λ}$
Kinetic energy $(E) = \frac{1}{2} {mv}^{2} = \frac{1}{2 m} (p^{2})$
$E = \frac{1}{2 m} \cdot \frac{h^{2}}{λ^{2}}$
$E = \frac{h^{2}}{2 m λ^{2}} (∵ p = \frac{h}{λ})$

MHT-CET 2018

Dual nature of radiation and Matter

142453 The masses of two particles having same kinetic energies are in the ratio 2:1. Then their de-Broglie wavelengths are in the ratio

1

2 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : \sqrt{2}

Explanation:

D Given that,
K.E. of $1^{st}$ object $=$ K.E. of $2^{nd}$ object
The ratio of their masses,
$\frac{m_{1}}{m_{2}} = \frac{2}{1}$
Form de-Broglie wavelength,
$λ = \frac{h}{p}$
Kinetic energy in terms of momentum,
$K.E = \frac{p^{2}}{2 m}$
$∴ K.E = \frac{h^{2}}{2 m λ^{2}}$
$λ^{2} = \frac{h^{2}}{2 m (K . E)}$
Since, $(K . E)_{1} = (K . E)_{2}$
$\frac{{(λ_{1})}^{2}}{(λ)^{2}} = \frac{(\frac{h^{2}}{2 m_{1} (K . E)_{1}})}{(\frac{h^{2}}{2 m_{2} (K \cdot E)_{1}})} = \frac{m_{2}}{m_{1}} = \frac{1}{2}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
$λ_{1} : λ_{2} = 1 : \sqrt{2}$

COMEDK 2018

Dual nature of radiation and Matter

142455 Two particles $A$ and $B$ of same mass have their total energies $E_{A}$ and $E_{B}$ in the ratio $E_{A} : E_{B}$ $= 1 : 2$ Their potential energies $U_{A}$ and $U_{B}$ are in the ratio $U_{A} : U_{B} = 1 : 2$ . If $λ_{A}$ and $λ_{B}$ are there de-Broglie wavelengths, then $λ_{A} : λ_{B}$ is

1

1 : 2

2

2 : 1

3

1 : \sqrt{2}

4

\sqrt{2} : 1

5

1 : 1

Explanation:

D Given that,
Energy ratio $= \frac{E_{A}}{E_{B}} = \frac{1}{2}$
Potential energy,
$\frac{U_{A}}{U_{B}} = \frac{1}{2}$
So, $E_{A} = x, E_{B} = 2 x$
$U_{A} = y, U_{B} = 2 y$
$∵ E_{A} = U_{A} + k_{A}$
$E_{B} = U_{B} + k_{B}$
Here $k_{A}$ and $k_{B}$ are kinetic energy of particles $A$ and $B$
So, $k_{A} = E_{A} - U_{A} = (x - y)$
$k_{B} = E_{B} - U_{B} = 2 (x - y)$ We know that,
de - Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mk}}$
$λ_{A} = \frac{h}{\sqrt{2 {mk}_{A}}} \Rightarrow λ_{B} = \frac{h}{\sqrt{2 {mk}_{B}}}$
$∴ \frac{λ_{A}}{λ_{B}} = \sqrt{\frac{k_{B}}{k_{A}}}$
From equation (i), (ii) and (iii)
$\frac{λ_{A}}{λ_{B}} = \sqrt{\frac{2 (x - y)}{(x - y)}} = \frac{\sqrt{2}}{1}$

Kerala CEE -2018

Dual nature of radiation and Matter

142456 If $λ_{1}$ and $λ_{2}$ denote the de-Broglie wavelengths of two particles with same masses but charges in the ratio of 1:2 after they are accelerated from rest through the same potential difference, then

1

λ_{1} = λ_{2}

2

λ_{1} < λ_{2}

3

λ_{1} > λ_{2}

4 None of these

Explanation:

C According to the de-Broglie wavelength-
$λ = \frac{h}{\sqrt{2 meV}}$
According to question,
$λ_{1} = \frac{h}{\sqrt{2 meV}}$
$λ_{2} = \frac{h}{\sqrt{2 {me}^{'} V}}$
From equation (i) and (ii),
$\frac{λ_{1}}{λ_{2}} = \frac{h}{\sqrt{2 meV}} \times \frac{\sqrt{2 {me}^{'} V}}{h}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{e^{'}}{e}} (∵ \frac{e}{e^{'}} = \frac{1}{2}) (given)$
$λ_{1} = \sqrt{2} λ_{2}$
Therefore, $λ_{1} > λ_{2}$

UPSEE - 2018

Dual nature of radiation and Matter

142452 The energy of an electron having de-Broglic wavelength ' $λ$ ' is $(h =$ Plank's constant, $m =$ mass of electron)

1

\frac{h}{2 m λ}

2

\frac{h^{2}}{2 m λ^{2}}

3

\frac{h^{2}}{2 m^{2} λ^{2}}

4

\frac{h^{2}}{2 m^{2} λ}

Explanation:

B We know,
de-Broglie wavelength,
$λ = \frac{h}{p}$
$Or p = \frac{h}{λ}$
Kinetic energy $(E) = \frac{1}{2} {mv}^{2} = \frac{1}{2 m} (p^{2})$
$E = \frac{1}{2 m} \cdot \frac{h^{2}}{λ^{2}}$
$E = \frac{h^{2}}{2 m λ^{2}} (∵ p = \frac{h}{λ})$

MHT-CET 2018

Dual nature of radiation and Matter

142453 The masses of two particles having same kinetic energies are in the ratio 2:1. Then their de-Broglie wavelengths are in the ratio

1

2 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : \sqrt{2}

Explanation:

D Given that,
K.E. of $1^{st}$ object $=$ K.E. of $2^{nd}$ object
The ratio of their masses,
$\frac{m_{1}}{m_{2}} = \frac{2}{1}$
Form de-Broglie wavelength,
$λ = \frac{h}{p}$
Kinetic energy in terms of momentum,
$K.E = \frac{p^{2}}{2 m}$
$∴ K.E = \frac{h^{2}}{2 m λ^{2}}$
$λ^{2} = \frac{h^{2}}{2 m (K . E)}$
Since, $(K . E)_{1} = (K . E)_{2}$
$\frac{{(λ_{1})}^{2}}{(λ)^{2}} = \frac{(\frac{h^{2}}{2 m_{1} (K . E)_{1}})}{(\frac{h^{2}}{2 m_{2} (K \cdot E)_{1}})} = \frac{m_{2}}{m_{1}} = \frac{1}{2}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
$λ_{1} : λ_{2} = 1 : \sqrt{2}$

COMEDK 2018

Dual nature of radiation and Matter

142455 Two particles $A$ and $B$ of same mass have their total energies $E_{A}$ and $E_{B}$ in the ratio $E_{A} : E_{B}$ $= 1 : 2$ Their potential energies $U_{A}$ and $U_{B}$ are in the ratio $U_{A} : U_{B} = 1 : 2$ . If $λ_{A}$ and $λ_{B}$ are there de-Broglie wavelengths, then $λ_{A} : λ_{B}$ is

1

1 : 2

2

2 : 1

3

1 : \sqrt{2}

4

\sqrt{2} : 1

5

1 : 1

Explanation:

D Given that,
Energy ratio $= \frac{E_{A}}{E_{B}} = \frac{1}{2}$
Potential energy,
$\frac{U_{A}}{U_{B}} = \frac{1}{2}$
So, $E_{A} = x, E_{B} = 2 x$
$U_{A} = y, U_{B} = 2 y$
$∵ E_{A} = U_{A} + k_{A}$
$E_{B} = U_{B} + k_{B}$
Here $k_{A}$ and $k_{B}$ are kinetic energy of particles $A$ and $B$
So, $k_{A} = E_{A} - U_{A} = (x - y)$
$k_{B} = E_{B} - U_{B} = 2 (x - y)$ We know that,
de - Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mk}}$
$λ_{A} = \frac{h}{\sqrt{2 {mk}_{A}}} \Rightarrow λ_{B} = \frac{h}{\sqrt{2 {mk}_{B}}}$
$∴ \frac{λ_{A}}{λ_{B}} = \sqrt{\frac{k_{B}}{k_{A}}}$
From equation (i), (ii) and (iii)
$\frac{λ_{A}}{λ_{B}} = \sqrt{\frac{2 (x - y)}{(x - y)}} = \frac{\sqrt{2}}{1}$

Kerala CEE -2018

Dual nature of radiation and Matter

142456 If $λ_{1}$ and $λ_{2}$ denote the de-Broglie wavelengths of two particles with same masses but charges in the ratio of 1:2 after they are accelerated from rest through the same potential difference, then

1

λ_{1} = λ_{2}

2

λ_{1} < λ_{2}

3

λ_{1} > λ_{2}

4 None of these

Explanation:

C According to the de-Broglie wavelength-
$λ = \frac{h}{\sqrt{2 meV}}$
According to question,
$λ_{1} = \frac{h}{\sqrt{2 meV}}$
$λ_{2} = \frac{h}{\sqrt{2 {me}^{'} V}}$
From equation (i) and (ii),
$\frac{λ_{1}}{λ_{2}} = \frac{h}{\sqrt{2 meV}} \times \frac{\sqrt{2 {me}^{'} V}}{h}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{e^{'}}{e}} (∵ \frac{e}{e^{'}} = \frac{1}{2}) (given)$
$λ_{1} = \sqrt{2} λ_{2}$
Therefore, $λ_{1} > λ_{2}$

UPSEE - 2018

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