A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity \\(v\\) is given by\\(\lambda = \dfrac {h}{mv} \\) \\( (\because p = mv)\\)de-Broglie wavelength of a proton of mass \\(m_{1}\\) and kinetic energy \\(k\\) is given by\\(\lambda_{1} = \dfrac {h}{\sqrt {2m_{1}k}}\\) \\((\because p = \sqrt {2mk})\\)\\(\lambda_1= \dfrac {h}{\sqrt {2m_{1}qV}} .... (i)\\) \\( [\because k = qV]\\)For an alpha particle mass \\(m_{2}\\) carrying charge \\(q_{0}\\) is accelerated through potential \\(V\\), then\\(\lambda_{2} = \dfrac {h}{\sqrt {2m_{2}q_{0}V}}\\)\\(\because\\) For \\(\alpha - particle\ \ (^{4}_{2}He)\\) : \\( q_{0} = 2q\\) and \\(m_{2} = 4m_{1}\\)\\(\therefore \lambda_{2} = \dfrac {h}{\sqrt {2\times 4m_{1}\times 2q\times V}} .... (ii)\\)The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get\\(\dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {h}{\sqrt {2m_{1}qV}} \times \dfrac {\sqrt {2\times m_{1} \times 4\times 2qV}}{h} = \dfrac {4}{\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}}\\)We get \\(\dfrac{\lambda_1}{\lambda_2}= 2\sqrt {2}\\)

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

For electronDe - Broglie wavelength \\(\lambda_c = \dfrac{h}{p}\\)where p is momentum \\(p = mv\\)also by energy we have \\(E = \dfrac{1}{2} mv^2\\)\\(\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}\\)\\(\Rightarrow p = \sqrt{2 mE}\\)\\(\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}\\)For photon energy \\(\Rightarrow E = \dfrac{hc}{\lambda}\\)\\(\Rightarrow \lambda = \dfrac{h c}{E}\\)\\(\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}\\)\\(= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}\\)option (A) is correct.

An electromagnetic wave of wavelength \\(\lambda\\) is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength \\(\lambda'\\), then

Kinetic energy of emitted electron = Energy of incident photoni.e. \\(\dfrac{1}{2}mv^2 = h \nu\\)or \\(\dfrac{p^2}{2m} = \dfrac{hc}{\lambda}\\)\\(\left[ \because mv = p, \nu =\dfrac{c}{\lambda}\right]\\)or \\(p = \sqrt{\dfrac{2mhc}{\lambda}}\\)de-Broglie wavelength of emitted electrons \\(\lambda' = \dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2mhc}{\lambda}}}\\) or \\(\lambda' = \sqrt{\dfrac{h\lambda}{2mc}}\\) \\(\therefore \lambda = \dfrac{2mc}{h}\lambda'^2\\)

An electron is accelerated from rest through a potential difference of \\(V\\) volt. If the de Broglie wavelength of the electron is \\(1.227 \times 10^{-2} nm\\), the potential difference is :

The de-broglie wavelength of an electron is given as:\\(\lambda=\dfrac{1.227}{\sqrt{V}}\ nm\\)Substitute the wavelength in the above expression:\\(V=\left(\dfrac{1.227}{1.227\times10^{-2}}\right)^2\\)\\(V=10^4\ V\\)

An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has (i) the shortest and (ii) the largest, de Broglie wavelength?

De Broglie wavelength of particle of mass m in terms of kinetic energy \\((E_k)\\) is \\(\lambda=\dfrac{h}{\sqrt{2mE_k}} \propto \dfrac{1}{m}\\) for the same kinetic energy.(i) Out of given particles, the mass of alpha particle is maximum, so de Broglie wavelength associated with alpha particle is the shortest.(ii) As mass of electron is least, so electron has largest de-Broglie wavelength.

An \\(\alpha\\)- particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelength.

Given that;\\(V_{\alpha}=V_{Pr}\\)Since:- \\(\lambda=\dfrac{h}{p}\\)\\(\Rightarrow \lambda=\dfrac{h}{\sqrt{2mE}}\\)Where \\(E=kinetic\ Energy=eV\\) [V=acceleration potential]\\(\Rightarrow \lambda_{Pr}=\dfrac{h}{\sqrt{2m_p(2e)V_{Pr}}}\\)Whereas; \\(\lambda_{\alpha}=\dfrac{h}{\sqrt{m_{\alpha}(2e)V_{\alpha}}}\\)Where \\(m_{\alpha}=4m_{Pr}\\)and \\(V_{\alpha}=V_{Pr}=V\\)\\(\Rightarrow \lambda_{Pr}=\dfrac{h}{\sqrt{2m_p(e)V}}\\)...........(1)and \\(\lambda_{\alpha}=\dfrac{h}{\sqrt{2(4mp)(2e)V}}\\)...........(2)From (1) and (2)\\(\dfrac{\lambda_{Pr}}{\lambda_{\alpha}}=\dfrac{h}{\sqrt{2m_p(e)V}}\times \dfrac{\sqrt{2(4m_p)(2e)V}}{h}\\)\\(=\dfrac{\sqrt{2m_peV(8)}}{\sqrt{2m_peV}}\\)\\(=\sqrt{8}:1\\)

Light of wavelength \\(500\ nm\\) is incident on a metal with a work function \\(2.28\ eV\\). The de Borglie wavelength of the emitted electron is:

From photoelectric effect, \\(h\nu=W_0+eV\\)or \\(\dfrac{hc}{\lambda}=W_0+eV\\)or \\(\displaystyle \frac{(6.6\times 10^{-34})\times (3\times 10^8)}{500\times 10^{-9}}=2.28\times 1.6\times 10^{-19}+eV\\)or \\(eV=0.31 \times 10^{-19}\\) or \\(V=0.2\\)Here V should be \\(V\leq 0.2\\)The de Broglie wavelength of electron , \\(\lambda=\dfrac{12.27\times 10^{-10}}{V}\\)so, \\(\lambda\geq 2.8\times 10^{-9} m\\)

Derive an expression for de Broglie wavelength of matter waves.

de Broglie equated the energy equations of Planck (wave) and Einstein (particle).\\(E=hv\\) (planck's relation)\\(E=m{ c }^{ 2 }\\) (Einstein's mass - energy relation)\\(hv=m{ c }^{ 2 }\\)\\(\dfrac { hc }{ \lambda } =m{ c }^{ 2 }\\) (or)\\(\lambda =\dfrac { h }{ mc }\\)For a particle moving with a velocity \\(V\\),\\(\lambda =\dfrac { h }{ mV } =\dfrac { h }{ p }\\)where \\(p=mV\\) is the momentum of the particle.

What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 Volt?

Given : \\(V =100\\) voltsde-Broglie wavelength of electron \\(\lambda = \dfrac{h}{\sqrt{2meV}}\\)\\(\therefore\\) \\(\lambda = \dfrac{6.6\times 10^{-34}}{\sqrt{2(9.1\times 10^{-31})(1.6\times 10^{-19})(100)}}\\)Or \\(\lambda = \dfrac{6.6\times 10^{-34}}{5.4\times 10^{-10}}\\) m\\(\implies \\) \\(\lambda = 1.227\times 10^{-10} m = 1.227\\) \\(\mathring{A}\\)

Wave Nature of Matter | Formulas, Definition, Examples

definition

De-broglie wavelength

De Broglie's wavelength is the wavelength associated with a massive particle,hypothesized by De Broglie that explains Bohr's quantised orbits by bringing in the wave-particle duality. It is written as

λ = \frac{h}{m v}

(de broglie wavelength)

definition

Heisenberg uncertainty principle

The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision.

Δ x Δ p > \frac{ℏ}{2}

Δ E Δ t > \frac{ℏ}{2}

formula

Uncertainty in position or momentum

Δ x Δ p > \frac{ℏ}{2}

where

Δ x

is the uncertainty in position,

Δ p

is the uncertainty in momentum and

ℏ

is the reduced Planck's constant

example

Mass of a particle using de-Broglie wavelength

The de-Broglie wavelength associated with a particle moving with a velocity of

10^{2}

m/s is

6.6 \times 10^{- 24} m

.Then the mass of the particle is

λ = \frac{h}{m v}

\Rightarrow m = \frac{h}{υ λ} = \frac{6 . 6 \times 1 0 ^{- 34}}{6 . 6 \times 1 0 ^{- 24} \times 1 0 ^{2}} = 10^{- 12} k g

Wave Nature of Matter

definition

De-broglie wavelength

definition

Heisenberg uncertainty principle

formula

Uncertainty in position or momentum

example

Mass of a particle using de-Broglie wavelength

REVISE WITH CONCEPTS

Wave Nature of Matter - Problem L1

Quick Summary With Stories

De Broglie Wavelength

Heisenberg Uncertainty Principle

A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

An electromagnetic wave of wavelength $λ$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength $λ^{'}$ , then

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 1 0^{- 2} n m$ , the potential difference is :

An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has (i) the shortest and (ii) the largest, de Broglie wavelength?

An $α$ - particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelength.

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is:

Derive an expression for de Broglie wavelength of matter waves.

What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 Volt?

definition

De-broglie wavelength

definition

Heisenberg uncertainty principle

formula

Uncertainty in position or momentum

example

Mass of a particle using de-Broglie wavelength

REVISE WITH CONCEPTS

Wave Nature of Matter - Problem L1

Quick Summary With Stories

De Broglie Wavelength

Heisenberg Uncertainty Principle

A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength λ′, then

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10−2nm, the potential difference is :

An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has (i) the shortest and (ii) the largest, de Broglie wavelength?

An α- particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelength.

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

Light of wavelength 500 nm is incident on a metal with a work function 2.28 eV. The de Borglie wavelength of the emitted electron is:

Derive an expression for de Broglie wavelength of matter waves.

What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 Volt?

An electromagnetic wave of wavelength $λ$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength $λ^{'}$ , then

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 1 0^{- 2} n m$ , the potential difference is :

An $α$ - particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelength.

Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is: