QBManager

Dual nature of radiation and Matter

142457 The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{, is}$

1

\frac{h}{\sqrt{mkT}}

2

\frac{h}{\sqrt{3 mkT}}

3

\frac{2 h}{\sqrt{3 mkT}}

4

\frac{2 h}{\sqrt{mkT}}

Explanation:

B Energy of neutron $=$ K.E. per molecule of water
$E_{n} = \frac{3}{2} kT$
de-Brogile wavelength
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 {mE}_{n}}}$
$= \frac{h}{\sqrt{2 m \times \frac{3}{2} kT}}$
$λ = \frac{h}{\sqrt{3 mkT}}$

NEET - 2017

Dual nature of radiation and Matter

142458 The de-Broglie wavelength of a tennis ball of mass $60 g$ moving with a velocity of $10 m / s$ is Given: Planck's constant $= 6.63 \times 10^{- 34} J s$

1

10^{- 16} m

2

10^{- 25} m

3

10^{- 31} m

4

10^{- 33} m

Explanation:

D Given that,
$m = 60 g = 60 \times 10^{- 3} kg$
$v = 10 m / s$
$h = 6.63 \times 10^{- 34} Js$
We know that,
de-Broglie wavelength-
$λ = \frac{h}{mv}$
$λ = \frac{6.6 \times 10^{- 34}}{60 \times 10^{- 3} \times 10}$
$λ = 1.1 \times 10^{- 33} m$
$λ ◻ 10^{- 33} m$

AMU-2017

Dual nature of radiation and Matter

142459 If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 %$ . Then the initial momentum of the electron is

1

\frac{20}{P}

2

20 P

3

\frac{P}{20}

4

30 P

Explanation:

B We know that,
de - Broglie wavelength
$λ = \frac{h}{p}$
$p = \frac{h}{λ}$
$| \frac{Δ p}{p} | = | \frac{Δ λ}{λ} |$
$\frac{p}{p_{initial}} = \frac{5}{100}$
$p_{initial} \times 5 = 100 P$
$p_{initial} = 20 P$

Shift-II

Dual nature of radiation and Matter

142460 The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is

1

2 : 3

2

2 \sqrt{2} : \sqrt{3}

3

\sqrt{3} : 2 \sqrt{2}

4

\sqrt{2} : \sqrt{3}

Explanation:

B Given that,
$T_{He} = 127 + 273 = 400 K$
$T_{H_{2}} = 27 + 273 = 300 K$
We know that,
$λ = \frac{h}{\sqrt{3 mKT}}$
$λ \propto \frac{1}{\sqrt{mT}}$
The ratio of de-Broglie wavelength of hydrogen and helium.
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{m_{He} \cdot T_{He}}{m_{H_{2}} \cdot T_{H_{2}}}}$
$= \sqrt{(\frac{4}{2}) \times \frac{400}{300}}$
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{8}{3}} = \frac{2 \sqrt{2}}{\sqrt{3}}$

Shift-II

Dual nature of radiation and Matter

142461 The wavelength $λ$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$ , speed of light $c$ and Planck constant.

1

\frac{λ mc}{h}

2

\frac{hmc}{λ}

3

\frac{2 hmc}{λ}

4

\frac{2 λ mc}{h}

Explanation:

D de - Broglie wavelength
$λ = \frac{h}{p}$
Momentum, $p = \frac{h}{λ}$
Kinetic energy of electron
${KE}_{e} = \frac{{mv}^{2}}{2} = \frac{p^{2}}{2 m}$
${KE}_{e} = \frac{{(\frac{h}{λ})}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Energy of photon $(E_{P}) = \frac{hc}{λ}$
Ratio of kinetic energy of electron $({KE}_{e})$ and photon $(E_{p})$
$\frac{{KE}_{e}}{E_{p}} = \frac{\frac{h^{2}}{2 m λ^{2}}}{\frac{hc}{λ}} = \frac{h}{2 m λ c}$
$\frac{E_{p}}{{KE}_{e}} = (\frac{2 λ mc}{h})$

JIPMER-2017

Dual nature of radiation and Matter

142457 The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{, is}$

1

\frac{h}{\sqrt{mkT}}

2

\frac{h}{\sqrt{3 mkT}}

3

\frac{2 h}{\sqrt{3 mkT}}

4

\frac{2 h}{\sqrt{mkT}}

Explanation:

B Energy of neutron $=$ K.E. per molecule of water
$E_{n} = \frac{3}{2} kT$
de-Brogile wavelength
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 {mE}_{n}}}$
$= \frac{h}{\sqrt{2 m \times \frac{3}{2} kT}}$
$λ = \frac{h}{\sqrt{3 mkT}}$

NEET - 2017

Dual nature of radiation and Matter

142458 The de-Broglie wavelength of a tennis ball of mass $60 g$ moving with a velocity of $10 m / s$ is Given: Planck's constant $= 6.63 \times 10^{- 34} J s$

1

10^{- 16} m

2

10^{- 25} m

3

10^{- 31} m

4

10^{- 33} m

Explanation:

D Given that,
$m = 60 g = 60 \times 10^{- 3} kg$
$v = 10 m / s$
$h = 6.63 \times 10^{- 34} Js$
We know that,
de-Broglie wavelength-
$λ = \frac{h}{mv}$
$λ = \frac{6.6 \times 10^{- 34}}{60 \times 10^{- 3} \times 10}$
$λ = 1.1 \times 10^{- 33} m$
$λ ◻ 10^{- 33} m$

AMU-2017

Dual nature of radiation and Matter

142459 If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 %$ . Then the initial momentum of the electron is

1

\frac{20}{P}

2

20 P

3

\frac{P}{20}

4

30 P

Explanation:

B We know that,
de - Broglie wavelength
$λ = \frac{h}{p}$
$p = \frac{h}{λ}$
$| \frac{Δ p}{p} | = | \frac{Δ λ}{λ} |$
$\frac{p}{p_{initial}} = \frac{5}{100}$
$p_{initial} \times 5 = 100 P$
$p_{initial} = 20 P$

Shift-II

Dual nature of radiation and Matter

142460 The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is

1

2 : 3

2

2 \sqrt{2} : \sqrt{3}

3

\sqrt{3} : 2 \sqrt{2}

4

\sqrt{2} : \sqrt{3}

Explanation:

B Given that,
$T_{He} = 127 + 273 = 400 K$
$T_{H_{2}} = 27 + 273 = 300 K$
We know that,
$λ = \frac{h}{\sqrt{3 mKT}}$
$λ \propto \frac{1}{\sqrt{mT}}$
The ratio of de-Broglie wavelength of hydrogen and helium.
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{m_{He} \cdot T_{He}}{m_{H_{2}} \cdot T_{H_{2}}}}$
$= \sqrt{(\frac{4}{2}) \times \frac{400}{300}}$
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{8}{3}} = \frac{2 \sqrt{2}}{\sqrt{3}}$

Shift-II

Dual nature of radiation and Matter

142461 The wavelength $λ$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$ , speed of light $c$ and Planck constant.

1

\frac{λ mc}{h}

2

\frac{hmc}{λ}

3

\frac{2 hmc}{λ}

4

\frac{2 λ mc}{h}

Explanation:

D de - Broglie wavelength
$λ = \frac{h}{p}$
Momentum, $p = \frac{h}{λ}$
Kinetic energy of electron
${KE}_{e} = \frac{{mv}^{2}}{2} = \frac{p^{2}}{2 m}$
${KE}_{e} = \frac{{(\frac{h}{λ})}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Energy of photon $(E_{P}) = \frac{hc}{λ}$
Ratio of kinetic energy of electron $({KE}_{e})$ and photon $(E_{p})$
$\frac{{KE}_{e}}{E_{p}} = \frac{\frac{h^{2}}{2 m λ^{2}}}{\frac{hc}{λ}} = \frac{h}{2 m λ c}$
$\frac{E_{p}}{{KE}_{e}} = (\frac{2 λ mc}{h})$

JIPMER-2017

Dual nature of radiation and Matter

142457 The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{, is}$

1

\frac{h}{\sqrt{mkT}}

2

\frac{h}{\sqrt{3 mkT}}

3

\frac{2 h}{\sqrt{3 mkT}}

4

\frac{2 h}{\sqrt{mkT}}

Explanation:

B Energy of neutron $=$ K.E. per molecule of water
$E_{n} = \frac{3}{2} kT$
de-Brogile wavelength
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 {mE}_{n}}}$
$= \frac{h}{\sqrt{2 m \times \frac{3}{2} kT}}$
$λ = \frac{h}{\sqrt{3 mkT}}$

NEET - 2017

Dual nature of radiation and Matter

142458 The de-Broglie wavelength of a tennis ball of mass $60 g$ moving with a velocity of $10 m / s$ is Given: Planck's constant $= 6.63 \times 10^{- 34} J s$

1

10^{- 16} m

2

10^{- 25} m

3

10^{- 31} m

4

10^{- 33} m

Explanation:

D Given that,
$m = 60 g = 60 \times 10^{- 3} kg$
$v = 10 m / s$
$h = 6.63 \times 10^{- 34} Js$
We know that,
de-Broglie wavelength-
$λ = \frac{h}{mv}$
$λ = \frac{6.6 \times 10^{- 34}}{60 \times 10^{- 3} \times 10}$
$λ = 1.1 \times 10^{- 33} m$
$λ ◻ 10^{- 33} m$

AMU-2017

Dual nature of radiation and Matter

142459 If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 %$ . Then the initial momentum of the electron is

1

\frac{20}{P}

2

20 P

3

\frac{P}{20}

4

30 P

Explanation:

B We know that,
de - Broglie wavelength
$λ = \frac{h}{p}$
$p = \frac{h}{λ}$
$| \frac{Δ p}{p} | = | \frac{Δ λ}{λ} |$
$\frac{p}{p_{initial}} = \frac{5}{100}$
$p_{initial} \times 5 = 100 P$
$p_{initial} = 20 P$

Shift-II

Dual nature of radiation and Matter

142460 The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is

1

2 : 3

2

2 \sqrt{2} : \sqrt{3}

3

\sqrt{3} : 2 \sqrt{2}

4

\sqrt{2} : \sqrt{3}

Explanation:

B Given that,
$T_{He} = 127 + 273 = 400 K$
$T_{H_{2}} = 27 + 273 = 300 K$
We know that,
$λ = \frac{h}{\sqrt{3 mKT}}$
$λ \propto \frac{1}{\sqrt{mT}}$
The ratio of de-Broglie wavelength of hydrogen and helium.
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{m_{He} \cdot T_{He}}{m_{H_{2}} \cdot T_{H_{2}}}}$
$= \sqrt{(\frac{4}{2}) \times \frac{400}{300}}$
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{8}{3}} = \frac{2 \sqrt{2}}{\sqrt{3}}$

Shift-II

Dual nature of radiation and Matter

142461 The wavelength $λ$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$ , speed of light $c$ and Planck constant.

1

\frac{λ mc}{h}

2

\frac{hmc}{λ}

3

\frac{2 hmc}{λ}

4

\frac{2 λ mc}{h}

Explanation:

D de - Broglie wavelength
$λ = \frac{h}{p}$
Momentum, $p = \frac{h}{λ}$
Kinetic energy of electron
${KE}_{e} = \frac{{mv}^{2}}{2} = \frac{p^{2}}{2 m}$
${KE}_{e} = \frac{{(\frac{h}{λ})}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Energy of photon $(E_{P}) = \frac{hc}{λ}$
Ratio of kinetic energy of electron $({KE}_{e})$ and photon $(E_{p})$
$\frac{{KE}_{e}}{E_{p}} = \frac{\frac{h^{2}}{2 m λ^{2}}}{\frac{hc}{λ}} = \frac{h}{2 m λ c}$
$\frac{E_{p}}{{KE}_{e}} = (\frac{2 λ mc}{h})$

JIPMER-2017

Dual nature of radiation and Matter

142457 The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{, is}$

1

\frac{h}{\sqrt{mkT}}

2

\frac{h}{\sqrt{3 mkT}}

3

\frac{2 h}{\sqrt{3 mkT}}

4

\frac{2 h}{\sqrt{mkT}}

Explanation:

B Energy of neutron $=$ K.E. per molecule of water
$E_{n} = \frac{3}{2} kT$
de-Brogile wavelength
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 {mE}_{n}}}$
$= \frac{h}{\sqrt{2 m \times \frac{3}{2} kT}}$
$λ = \frac{h}{\sqrt{3 mkT}}$

NEET - 2017

Dual nature of radiation and Matter

142458 The de-Broglie wavelength of a tennis ball of mass $60 g$ moving with a velocity of $10 m / s$ is Given: Planck's constant $= 6.63 \times 10^{- 34} J s$

1

10^{- 16} m

2

10^{- 25} m

3

10^{- 31} m

4

10^{- 33} m

Explanation:

D Given that,
$m = 60 g = 60 \times 10^{- 3} kg$
$v = 10 m / s$
$h = 6.63 \times 10^{- 34} Js$
We know that,
de-Broglie wavelength-
$λ = \frac{h}{mv}$
$λ = \frac{6.6 \times 10^{- 34}}{60 \times 10^{- 3} \times 10}$
$λ = 1.1 \times 10^{- 33} m$
$λ ◻ 10^{- 33} m$

AMU-2017

Dual nature of radiation and Matter

142459 If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 %$ . Then the initial momentum of the electron is

1

\frac{20}{P}

2

20 P

3

\frac{P}{20}

4

30 P

Explanation:

B We know that,
de - Broglie wavelength
$λ = \frac{h}{p}$
$p = \frac{h}{λ}$
$| \frac{Δ p}{p} | = | \frac{Δ λ}{λ} |$
$\frac{p}{p_{initial}} = \frac{5}{100}$
$p_{initial} \times 5 = 100 P$
$p_{initial} = 20 P$

Shift-II

Dual nature of radiation and Matter

142460 The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is

1

2 : 3

2

2 \sqrt{2} : \sqrt{3}

3

\sqrt{3} : 2 \sqrt{2}

4

\sqrt{2} : \sqrt{3}

Explanation:

B Given that,
$T_{He} = 127 + 273 = 400 K$
$T_{H_{2}} = 27 + 273 = 300 K$
We know that,
$λ = \frac{h}{\sqrt{3 mKT}}$
$λ \propto \frac{1}{\sqrt{mT}}$
The ratio of de-Broglie wavelength of hydrogen and helium.
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{m_{He} \cdot T_{He}}{m_{H_{2}} \cdot T_{H_{2}}}}$
$= \sqrt{(\frac{4}{2}) \times \frac{400}{300}}$
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{8}{3}} = \frac{2 \sqrt{2}}{\sqrt{3}}$

Shift-II

Dual nature of radiation and Matter

142461 The wavelength $λ$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$ , speed of light $c$ and Planck constant.

1

\frac{λ mc}{h}

2

\frac{hmc}{λ}

3

\frac{2 hmc}{λ}

4

\frac{2 λ mc}{h}

Explanation:

D de - Broglie wavelength
$λ = \frac{h}{p}$
Momentum, $p = \frac{h}{λ}$
Kinetic energy of electron
${KE}_{e} = \frac{{mv}^{2}}{2} = \frac{p^{2}}{2 m}$
${KE}_{e} = \frac{{(\frac{h}{λ})}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Energy of photon $(E_{P}) = \frac{hc}{λ}$
Ratio of kinetic energy of electron $({KE}_{e})$ and photon $(E_{p})$
$\frac{{KE}_{e}}{E_{p}} = \frac{\frac{h^{2}}{2 m λ^{2}}}{\frac{hc}{λ}} = \frac{h}{2 m λ c}$
$\frac{E_{p}}{{KE}_{e}} = (\frac{2 λ mc}{h})$

JIPMER-2017

Dual nature of radiation and Matter

142457 The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{, is}$

1

\frac{h}{\sqrt{mkT}}

2

\frac{h}{\sqrt{3 mkT}}

3

\frac{2 h}{\sqrt{3 mkT}}

4

\frac{2 h}{\sqrt{mkT}}

Explanation:

B Energy of neutron $=$ K.E. per molecule of water
$E_{n} = \frac{3}{2} kT$
de-Brogile wavelength
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 {mE}_{n}}}$
$= \frac{h}{\sqrt{2 m \times \frac{3}{2} kT}}$
$λ = \frac{h}{\sqrt{3 mkT}}$

NEET - 2017

Dual nature of radiation and Matter

142458 The de-Broglie wavelength of a tennis ball of mass $60 g$ moving with a velocity of $10 m / s$ is Given: Planck's constant $= 6.63 \times 10^{- 34} J s$

1

10^{- 16} m

2

10^{- 25} m

3

10^{- 31} m

4

10^{- 33} m

Explanation:

D Given that,
$m = 60 g = 60 \times 10^{- 3} kg$
$v = 10 m / s$
$h = 6.63 \times 10^{- 34} Js$
We know that,
de-Broglie wavelength-
$λ = \frac{h}{mv}$
$λ = \frac{6.6 \times 10^{- 34}}{60 \times 10^{- 3} \times 10}$
$λ = 1.1 \times 10^{- 33} m$
$λ ◻ 10^{- 33} m$

AMU-2017

Dual nature of radiation and Matter

142459 If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 %$ . Then the initial momentum of the electron is

1

\frac{20}{P}

2

20 P

3

\frac{P}{20}

4

30 P

Explanation:

B We know that,
de - Broglie wavelength
$λ = \frac{h}{p}$
$p = \frac{h}{λ}$
$| \frac{Δ p}{p} | = | \frac{Δ λ}{λ} |$
$\frac{p}{p_{initial}} = \frac{5}{100}$
$p_{initial} \times 5 = 100 P$
$p_{initial} = 20 P$

Shift-II

Dual nature of radiation and Matter

142460 The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is

1

2 : 3

2

2 \sqrt{2} : \sqrt{3}

3

\sqrt{3} : 2 \sqrt{2}

4

\sqrt{2} : \sqrt{3}

Explanation:

B Given that,
$T_{He} = 127 + 273 = 400 K$
$T_{H_{2}} = 27 + 273 = 300 K$
We know that,
$λ = \frac{h}{\sqrt{3 mKT}}$
$λ \propto \frac{1}{\sqrt{mT}}$
The ratio of de-Broglie wavelength of hydrogen and helium.
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{m_{He} \cdot T_{He}}{m_{H_{2}} \cdot T_{H_{2}}}}$
$= \sqrt{(\frac{4}{2}) \times \frac{400}{300}}$
$\frac{λ_{H_{2}}}{λ_{He}} = \sqrt{\frac{8}{3}} = \frac{2 \sqrt{2}}{\sqrt{3}}$

Shift-II

Dual nature of radiation and Matter

142461 The wavelength $λ$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$ , speed of light $c$ and Planck constant.

1

\frac{λ mc}{h}

2

\frac{hmc}{λ}

3

\frac{2 hmc}{λ}

4

\frac{2 λ mc}{h}

Explanation:

D de - Broglie wavelength
$λ = \frac{h}{p}$
Momentum, $p = \frac{h}{λ}$
Kinetic energy of electron
${KE}_{e} = \frac{{mv}^{2}}{2} = \frac{p^{2}}{2 m}$
${KE}_{e} = \frac{{(\frac{h}{λ})}^{2}}{2 m} = \frac{h^{2}}{2 m λ^{2}}$
Energy of photon $(E_{P}) = \frac{hc}{λ}$
Ratio of kinetic energy of electron $({KE}_{e})$ and photon $(E_{p})$
$\frac{{KE}_{e}}{E_{p}} = \frac{\frac{h^{2}}{2 m λ^{2}}}{\frac{hc}{λ}} = \frac{h}{2 m λ c}$
$\frac{E_{p}}{{KE}_{e}} = (\frac{2 λ mc}{h})$

JIPMER-2017