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Dual nature of radiation and Matter

142426 Graph shows the variation of de-Broglie wavelength $(λ)$ versus $\frac{1}{\sqrt{V}}$ , where ' $V$ ' is the accelerating potential for four particles carrying same charge but of masses $m_{1}, m_{2}, m_{3}$ , $m_{4}$ . Which particle has a smaller mass?

1

m_{1}

2

m_{3}

3

m_{4}

4

m_{2}

Explanation:

C From de-Broglie wavelength,
$λ = \frac{h}{p} And p = mv$
$∵ λ = \frac{h}{\sqrt{2 mqV}}$
$∴ λ \sqrt{V} = \frac{h}{\sqrt{2 mq}}$
$\frac{λ}{(\frac{1}{\sqrt{V}})} = \frac{h}{\sqrt{2 mq}}$
$∴$ Slope of the graph $= \frac{1}{\sqrt{2 mg}}$
The slope will be maximum for minimum mass.
$∴ m_{4}$ is minimum.

MHT-CET 2020

Dual nature of radiation and Matter

142428 According to de-Broglie hypothesis if an electron of mass ' $m$ ' is accelerated by potential difference ' $V$ ', then associated wavelength is ' $λ$ '. When a proton of mass ' $M$ ' is accelerated through potential difference of ' $9 V$ ', then the wavelength associated with it, is

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

\frac{λ}{3} \sqrt{\frac{M}{m}}

4

\frac{λ}{3} \sqrt{\frac{m}{M}}

Explanation:

D We know that,
Kinetic energy gained $(KE) = eV = \frac{p^{2}}{2 m}$
$p = \sqrt{2 meV}$
de-Broglie wavelength,
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 meV}}$
$λ^{'} = \frac{h}{\sqrt{2 meV}}$
$\frac{λ^{'}}{λ} = \sqrt{\frac{m}{M} \cdot \frac{V}{V^{'}}} = \sqrt{\frac{m}{M} \cdot \frac{1}{9}}$
$λ^{'} = \frac{1}{3} \sqrt{\frac{m}{M}} . λ$
$λ^{'} = \frac{λ}{3} \sqrt{\frac{m}{M}}$

MHT-CET 2020

Dual nature of radiation and Matter

142429 How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{- 10} m$ to $0.5 \times 10^{- 10} m$ ? $(E =$ energy of electron)

1

4 E

2

2 E

3

E

4

3 E

Explanation:

A We know that,
De-Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$\frac{1 \times 10^{- 10}}{0.5 \times 10^{- 10}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$2 = \sqrt{\frac{E_{2}}{E_{1}}}$
Squaring on both side
$(2)^{2} = {(\sqrt{\frac{E_{2}}{E_{1}}})}^{2}$
$4 = \frac{E_{2}}{E_{1}}$
$E_{2} = 4 E_{1}$

MHT-CET 2020

Dual nature of radiation and Matter

142430 According to de-Broglie hypothesis, the ratio of wavelength of an electron and that of photon having same energy ' $E$ ' is $(m =$ mass of electron, $c =$ velocity of light)

1

c \times \sqrt{\frac{2 m}{E}}

2

\frac{1}{c} \times \sqrt{\frac{2 m}{E}}

3

\frac{1}{c} \times \sqrt{\frac{E}{2 m}}

4

c \times \sqrt{\frac{E}{2 m}}

Explanation:

C We know that,
de-Broglie wavelength of electron,
$λ_{e} = \frac{h}{p_{e}}$
$p_{e} = \sqrt{2 mE}$
$λ_{e} = \frac{h}{\sqrt{2 mE}}$
Since, energy of photon E,
$E = \frac{hc}{λ_{p}}$
$∴ λ_{p} = \frac{hc}{E}$
Divide equation (i) by equation (ii), we get-
$\frac{λ_{e}}{λ_{p}} = \frac{h}{\sqrt{2 mE}} \cdot \frac{E}{hc}$
$\frac{λ_{e}}{λ_{p}} = \frac{1}{c} \cdot \sqrt{\frac{E}{2 m}}$

MHT-CET 2020

Dual nature of radiation and Matter

142426 Graph shows the variation of de-Broglie wavelength $(λ)$ versus $\frac{1}{\sqrt{V}}$ , where ' $V$ ' is the accelerating potential for four particles carrying same charge but of masses $m_{1}, m_{2}, m_{3}$ , $m_{4}$ . Which particle has a smaller mass?

1

m_{1}

2

m_{3}

3

m_{4}

4

m_{2}

Explanation:

C From de-Broglie wavelength,
$λ = \frac{h}{p} And p = mv$
$∵ λ = \frac{h}{\sqrt{2 mqV}}$
$∴ λ \sqrt{V} = \frac{h}{\sqrt{2 mq}}$
$\frac{λ}{(\frac{1}{\sqrt{V}})} = \frac{h}{\sqrt{2 mq}}$
$∴$ Slope of the graph $= \frac{1}{\sqrt{2 mg}}$
The slope will be maximum for minimum mass.
$∴ m_{4}$ is minimum.

MHT-CET 2020

Dual nature of radiation and Matter

142428 According to de-Broglie hypothesis if an electron of mass ' $m$ ' is accelerated by potential difference ' $V$ ', then associated wavelength is ' $λ$ '. When a proton of mass ' $M$ ' is accelerated through potential difference of ' $9 V$ ', then the wavelength associated with it, is

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

\frac{λ}{3} \sqrt{\frac{M}{m}}

4

\frac{λ}{3} \sqrt{\frac{m}{M}}

Explanation:

D We know that,
Kinetic energy gained $(KE) = eV = \frac{p^{2}}{2 m}$
$p = \sqrt{2 meV}$
de-Broglie wavelength,
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 meV}}$
$λ^{'} = \frac{h}{\sqrt{2 meV}}$
$\frac{λ^{'}}{λ} = \sqrt{\frac{m}{M} \cdot \frac{V}{V^{'}}} = \sqrt{\frac{m}{M} \cdot \frac{1}{9}}$
$λ^{'} = \frac{1}{3} \sqrt{\frac{m}{M}} . λ$
$λ^{'} = \frac{λ}{3} \sqrt{\frac{m}{M}}$

MHT-CET 2020

Dual nature of radiation and Matter

142429 How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{- 10} m$ to $0.5 \times 10^{- 10} m$ ? $(E =$ energy of electron)

1

4 E

2

2 E

3

E

4

3 E

Explanation:

A We know that,
De-Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$\frac{1 \times 10^{- 10}}{0.5 \times 10^{- 10}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$2 = \sqrt{\frac{E_{2}}{E_{1}}}$
Squaring on both side
$(2)^{2} = {(\sqrt{\frac{E_{2}}{E_{1}}})}^{2}$
$4 = \frac{E_{2}}{E_{1}}$
$E_{2} = 4 E_{1}$

MHT-CET 2020

Dual nature of radiation and Matter

142430 According to de-Broglie hypothesis, the ratio of wavelength of an electron and that of photon having same energy ' $E$ ' is $(m =$ mass of electron, $c =$ velocity of light)

1

c \times \sqrt{\frac{2 m}{E}}

2

\frac{1}{c} \times \sqrt{\frac{2 m}{E}}

3

\frac{1}{c} \times \sqrt{\frac{E}{2 m}}

4

c \times \sqrt{\frac{E}{2 m}}

Explanation:

C We know that,
de-Broglie wavelength of electron,
$λ_{e} = \frac{h}{p_{e}}$
$p_{e} = \sqrt{2 mE}$
$λ_{e} = \frac{h}{\sqrt{2 mE}}$
Since, energy of photon E,
$E = \frac{hc}{λ_{p}}$
$∴ λ_{p} = \frac{hc}{E}$
Divide equation (i) by equation (ii), we get-
$\frac{λ_{e}}{λ_{p}} = \frac{h}{\sqrt{2 mE}} \cdot \frac{E}{hc}$
$\frac{λ_{e}}{λ_{p}} = \frac{1}{c} \cdot \sqrt{\frac{E}{2 m}}$

MHT-CET 2020

Dual nature of radiation and Matter

142426 Graph shows the variation of de-Broglie wavelength $(λ)$ versus $\frac{1}{\sqrt{V}}$ , where ' $V$ ' is the accelerating potential for four particles carrying same charge but of masses $m_{1}, m_{2}, m_{3}$ , $m_{4}$ . Which particle has a smaller mass?

1

m_{1}

2

m_{3}

3

m_{4}

4

m_{2}

Explanation:

C From de-Broglie wavelength,
$λ = \frac{h}{p} And p = mv$
$∵ λ = \frac{h}{\sqrt{2 mqV}}$
$∴ λ \sqrt{V} = \frac{h}{\sqrt{2 mq}}$
$\frac{λ}{(\frac{1}{\sqrt{V}})} = \frac{h}{\sqrt{2 mq}}$
$∴$ Slope of the graph $= \frac{1}{\sqrt{2 mg}}$
The slope will be maximum for minimum mass.
$∴ m_{4}$ is minimum.

MHT-CET 2020

Dual nature of radiation and Matter

142428 According to de-Broglie hypothesis if an electron of mass ' $m$ ' is accelerated by potential difference ' $V$ ', then associated wavelength is ' $λ$ '. When a proton of mass ' $M$ ' is accelerated through potential difference of ' $9 V$ ', then the wavelength associated with it, is

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

\frac{λ}{3} \sqrt{\frac{M}{m}}

4

\frac{λ}{3} \sqrt{\frac{m}{M}}

Explanation:

D We know that,
Kinetic energy gained $(KE) = eV = \frac{p^{2}}{2 m}$
$p = \sqrt{2 meV}$
de-Broglie wavelength,
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 meV}}$
$λ^{'} = \frac{h}{\sqrt{2 meV}}$
$\frac{λ^{'}}{λ} = \sqrt{\frac{m}{M} \cdot \frac{V}{V^{'}}} = \sqrt{\frac{m}{M} \cdot \frac{1}{9}}$
$λ^{'} = \frac{1}{3} \sqrt{\frac{m}{M}} . λ$
$λ^{'} = \frac{λ}{3} \sqrt{\frac{m}{M}}$

MHT-CET 2020

Dual nature of radiation and Matter

142429 How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{- 10} m$ to $0.5 \times 10^{- 10} m$ ? $(E =$ energy of electron)

1

4 E

2

2 E

3

E

4

3 E

Explanation:

A We know that,
De-Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$\frac{1 \times 10^{- 10}}{0.5 \times 10^{- 10}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$2 = \sqrt{\frac{E_{2}}{E_{1}}}$
Squaring on both side
$(2)^{2} = {(\sqrt{\frac{E_{2}}{E_{1}}})}^{2}$
$4 = \frac{E_{2}}{E_{1}}$
$E_{2} = 4 E_{1}$

MHT-CET 2020

Dual nature of radiation and Matter

142430 According to de-Broglie hypothesis, the ratio of wavelength of an electron and that of photon having same energy ' $E$ ' is $(m =$ mass of electron, $c =$ velocity of light)

1

c \times \sqrt{\frac{2 m}{E}}

2

\frac{1}{c} \times \sqrt{\frac{2 m}{E}}

3

\frac{1}{c} \times \sqrt{\frac{E}{2 m}}

4

c \times \sqrt{\frac{E}{2 m}}

Explanation:

C We know that,
de-Broglie wavelength of electron,
$λ_{e} = \frac{h}{p_{e}}$
$p_{e} = \sqrt{2 mE}$
$λ_{e} = \frac{h}{\sqrt{2 mE}}$
Since, energy of photon E,
$E = \frac{hc}{λ_{p}}$
$∴ λ_{p} = \frac{hc}{E}$
Divide equation (i) by equation (ii), we get-
$\frac{λ_{e}}{λ_{p}} = \frac{h}{\sqrt{2 mE}} \cdot \frac{E}{hc}$
$\frac{λ_{e}}{λ_{p}} = \frac{1}{c} \cdot \sqrt{\frac{E}{2 m}}$

MHT-CET 2020

Dual nature of radiation and Matter

142426 Graph shows the variation of de-Broglie wavelength $(λ)$ versus $\frac{1}{\sqrt{V}}$ , where ' $V$ ' is the accelerating potential for four particles carrying same charge but of masses $m_{1}, m_{2}, m_{3}$ , $m_{4}$ . Which particle has a smaller mass?

1

m_{1}

2

m_{3}

3

m_{4}

4

m_{2}

Explanation:

C From de-Broglie wavelength,
$λ = \frac{h}{p} And p = mv$
$∵ λ = \frac{h}{\sqrt{2 mqV}}$
$∴ λ \sqrt{V} = \frac{h}{\sqrt{2 mq}}$
$\frac{λ}{(\frac{1}{\sqrt{V}})} = \frac{h}{\sqrt{2 mq}}$
$∴$ Slope of the graph $= \frac{1}{\sqrt{2 mg}}$
The slope will be maximum for minimum mass.
$∴ m_{4}$ is minimum.

MHT-CET 2020

Dual nature of radiation and Matter

142428 According to de-Broglie hypothesis if an electron of mass ' $m$ ' is accelerated by potential difference ' $V$ ', then associated wavelength is ' $λ$ '. When a proton of mass ' $M$ ' is accelerated through potential difference of ' $9 V$ ', then the wavelength associated with it, is

1

λ \sqrt{\frac{M}{m}}

2

λ \sqrt{\frac{m}{M}}

3

\frac{λ}{3} \sqrt{\frac{M}{m}}

4

\frac{λ}{3} \sqrt{\frac{m}{M}}

Explanation:

D We know that,
Kinetic energy gained $(KE) = eV = \frac{p^{2}}{2 m}$
$p = \sqrt{2 meV}$
de-Broglie wavelength,
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 meV}}$
$λ^{'} = \frac{h}{\sqrt{2 meV}}$
$\frac{λ^{'}}{λ} = \sqrt{\frac{m}{M} \cdot \frac{V}{V^{'}}} = \sqrt{\frac{m}{M} \cdot \frac{1}{9}}$
$λ^{'} = \frac{1}{3} \sqrt{\frac{m}{M}} . λ$
$λ^{'} = \frac{λ}{3} \sqrt{\frac{m}{M}}$

MHT-CET 2020

Dual nature of radiation and Matter

142429 How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{- 10} m$ to $0.5 \times 10^{- 10} m$ ? $(E =$ energy of electron)

1

4 E

2

2 E

3

E

4

3 E

Explanation:

A We know that,
De-Broglie wavelength,
$λ = \frac{h}{\sqrt{2 mE}}$
$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$\frac{1 \times 10^{- 10}}{0.5 \times 10^{- 10}} = \sqrt{\frac{E_{2}}{E_{1}}}$
$2 = \sqrt{\frac{E_{2}}{E_{1}}}$
Squaring on both side
$(2)^{2} = {(\sqrt{\frac{E_{2}}{E_{1}}})}^{2}$
$4 = \frac{E_{2}}{E_{1}}$
$E_{2} = 4 E_{1}$

MHT-CET 2020

Dual nature of radiation and Matter

142430 According to de-Broglie hypothesis, the ratio of wavelength of an electron and that of photon having same energy ' $E$ ' is $(m =$ mass of electron, $c =$ velocity of light)

1

c \times \sqrt{\frac{2 m}{E}}

2

\frac{1}{c} \times \sqrt{\frac{2 m}{E}}

3

\frac{1}{c} \times \sqrt{\frac{E}{2 m}}

4

c \times \sqrt{\frac{E}{2 m}}

Explanation:

C We know that,
de-Broglie wavelength of electron,
$λ_{e} = \frac{h}{p_{e}}$
$p_{e} = \sqrt{2 mE}$
$λ_{e} = \frac{h}{\sqrt{2 mE}}$
Since, energy of photon E,
$E = \frac{hc}{λ_{p}}$
$∴ λ_{p} = \frac{hc}{E}$
Divide equation (i) by equation (ii), we get-
$\frac{λ_{e}}{λ_{p}} = \frac{h}{\sqrt{2 mE}} \cdot \frac{E}{hc}$
$\frac{λ_{e}}{λ_{p}} = \frac{1}{c} \cdot \sqrt{\frac{E}{2 m}}$

MHT-CET 2020