de Broglie wavelength \lambda = \dfrac{h}{p} where p = mv \implies \lambda = \dfrac{h}{mv} Thus implies \lambda \propto \dfrac{1}{m} (\because v and h are constant ) Hence Reason statement is correct.Mass of microscopic and sub-microscopic particle is very small. Thus according to de broglie equation, these particles tend to behave like a wave with certain finite wavelength which suggests the wave-particle dual nature of such particles like electrons.Thus de Broglie equation has significance for microscopic and sub -microscopic particles. Thus both Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

De Broglie wavelength \lambda associated with neutrons is related with absolute temperature T as:

De broglie wavelength \lambda=\dfrac{h}{mv} where v is the velocity of the particle.Let E be the kinetic energy of the neutrons.Thus E=\dfrac{1}{2}mv^2 \implies v=\sqrt{\dfrac{2E}{m}} Therefore \lambda=\dfrac{h}{\sqrt{2mE}} Since energy of neutrons E\propto T \lambda\propto \dfrac{1}{\sqrt{T}}

As eV_{s} = hv - W V_{s} = \left (\dfrac {h}{e}\right ) v - \left (\dfrac {W}{e}\right ) For V_{s} versus v graph, slope = \dfrac {h}{e} = constant .

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Dual Nature Of Radiation And Matter

Q: If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is:

Let the initial de Broglie wavelength be \lambda_0=\dfrac{h}{mv_0} The kinetic energy is increased 16 times.Thus KE'=16KE_0 \implies \dfrac{1}{2}mv^2=16\times \dfrac{1}{2}mv_0^2 \implies v=4v_0 Thus the new de Broglie wavelength= \lambda=\dfrac{h}{mv}=\dfrac{h}{m(4v_0)}=\dfrac{\lambda_0}{4} Thus the percentage change in wavelength: =\dfrac{\lambda_0-\lambda}{\lambda_0}\times 100\\=75 %

Q: If velocity of a particle is three times of that of electron and ratio of de Broglie wavelength of particle to that of electron is 1.814 \times 10^{-4} . The particle will be :

de broglie wavelength, \lambda = \dfrac {h}{p} \Rightarrow \lambda = \dfrac {h}{mv} \Rightarrow \dfrac {\lambda_{p}}{\lambda_{e}} = \dfrac {m_{e}v_{e}}{m_{p}v_{p}} \Rightarrow m_{p} = \dfrac {m_{e}v_{e}}{v_{p}}\times \dfrac {\lambda_{e}}{\lambda_{p}} Here, m_{e} = 9.1\times 10^{-31} kg, v_{p} = 3v_{e} and \dfrac {\lambda_{p}}{\lambda_{e}} = 1.814 \times 10^{-4} \Rightarrow m_{p} = \dfrac {9.1\times 10^{-31}}{1.814\times 10^{-4} \times 3} = 1.672\times 10^{-27} kg Thus, the particle is neutron.

Physics
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JEE Mains JEE Advanced NEET AIIMS BITSAT Easy Qs Med Qs Hard Qs

Assertion

The de Broglie equation has significance for any microscopic or sub-microscopic particles.

Reason

The de Broglie wavelength is inversely proportional to the mass of the object if velocity is constant.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

Easy

AIIMS

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If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is:

25%

75%

60%

50%

Medium

AIIMS

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De Broglie wavelength $λ$ associated with neutrons is related with absolute temperature T as:

λ \propto T

λ \propto \frac{1}{T}

λ \propto \frac{1}{T}

λ \propto T^{2}

Easy

AIIMS

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If we assume kinetic energy of a proton is equal to energy of the photon, the ratio of de Broglie wave length of proton to photon is proportional to :

E

E^{- 1 / 2}

E^{1 / 2}

E^{3 / 2}

Medium

AIIMS

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Assertion

As work function of a material increase by some mechanism, it requires greater energy to excite the electrons from its surface.

Reason

A plot of stopping potential $(V_{s})$ versus frequency $(v)$ for different metals, has greater slope for metals with greater work functions.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

Medium

AIIMS

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If velocity of a particle is three times of that of electron and ratio of de Broglie wavelength of particle to that of electron is $1.814 \times 1 0^{- 4}$ . The particle will be :

Neutron

Deutron

Alpha

Tritium

Medium

AIIMS

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Dual Nature Of Radiation And Matter

Physics 93 Views

The de Broglie equation has significance for any microscopic or sub-microscopic particles.

The de Broglie wavelength is inversely proportional to the mass of the object if velocity is constant.

If the kinetic energy of a particle is increased by 16 times, the percentage change in the de Broglie wavelength of the particle is:

De Broglie wavelength λ associated with neutrons is related with absolute temperature T as:

If we assume kinetic energy of a proton is equal to energy of the photon, the ratio of de Broglie wave length of proton to photon is proportional to :

As work function of a material increase by some mechanism, it requires greater energy to excite the electrons from its surface.

A plot of stopping potential (Vs​) versus frequency (v) for different metals, has greater slope for metals with greater work functions.

If velocity of a particle is three times of that of electron and ratio of de Broglie wavelength of particle to that of electron is 1.814×10−4. The particle will be :

Physics
93 Views

De Broglie wavelength $λ$ associated with neutrons is related with absolute temperature T as:

A plot of stopping potential $(V_{s})$ versus frequency $(v)$ for different metals, has greater slope for metals with greater work functions.

If velocity of a particle is three times of that of electron and ratio of de Broglie wavelength of particle to that of electron is $1.814 \times 1 0^{- 4}$ . The particle will be :