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Dual Nature of Radiation and Matter
de Broglie Wavelength
14 questions from 13 papers, each with its answer.
- questions
- 14
- papers
- 13
- years asked, 2017–2026
- 4 of 10
- of Dual Nature of Radiation and Matter
- 17.07%
Appears in
Concepts tested
- de Broglie wavelength 2
- Electron and photon of equal energy
- Electron diffraction
- Equal de Broglie wavelengths
- Fractional change in wavelength
- Graph between 1/lambda^2 and kinetic energy
- Momentum from circular motion in a magnetic field
- p proportional to 1/lambda
- Photon vs Particle de Broglie Wavelength
- Wavelength and kinetic energy
- Wavelength inversely proportional to mass
- Wavelength of accelerated electron
- λ vs p graph
All Questions
2025 1 question
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A photon and an electron (mass $m$) have the same energy $E$. The ratio $(\lambda_{\text{photon}} / \lambda_{\text{electron}})$ of their de Broglie wavelengths is ($c$ is the speed of light):
Photon wavelength is $\lambda_p = \frac{hc}{E}$ and electron wavelength is $\lambda_e = \frac{h}{\sqrt{2mE}}$. Their ratio is $\frac{\lambda_p}{\lambda_e} = c\sqrt{\frac{2m}{E}}$.
For a photon: $E = \frac{h c}{\lambda_p} \implies \lambda_p = \frac{h c}{E}$. For a non-relativistic electron of energy $E$: $p = \sqrt{2mE} \implies \lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$. Taking the ratio: $\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{h c / E}{h / \sqrt{2mE}} = c \frac{\sqrt{2mE}}{E} = c \sqrt{\frac{2m}{E}}$.
2024 1 question
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The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle):de Broglie wavelength $\lambda = \frac{h}{\sqrt{2mE}}$ $\Rightarrow \frac{1}{\lambda^2} = \frac{2mE}{h^2}$, i.e. $\frac{1}{\lambda^2} \propto E$ So the graph of $\frac{1}{\lambda^2}$ vs E is a straight line passing through the origin.
2024 Re-NEET 1 question
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An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_e$ and $\lambda_\alpha$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:De-Broglie wavelength $\lambda=\dfrac{h}{\sqrt{2mqV}}$, so $\lambda\propto\dfrac{1}{\sqrt{mq}}$. Since $m_\alpha\gg m_e$, $\lambda_e>\lambda_\alpha$.
2022 1 question
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The graph which shows the variation of the de Broglie wavelength ($\lambda$) of a particle and its associated momentum ($p$) is :de Broglie wavelength $\lambda = \frac{h}{p}$, so $\lambda \propto \frac{1}{p}$ The graph of $\lambda$ vs p is a rectangular hyperbola.
2019 Odisha 1 question
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A proton and an $\alpha$-particle are accelerated from rest to the same energy. The de Broglie wavelengths $\lambda_p$ and $\lambda_\alpha$ are in the ratio,$\lambda=\frac{h}{\sqrt{2mK}}$; with equal K, $\frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha}{m_p}}=\sqrt{\frac{4m}{m}}=2:1$.
2015 AIPMT-I 1 question
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Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?$\lambda = \frac{h}{p} \Rightarrow p \propto \frac{1}{\lambda}$
Momentum is inversely proportional to wavelength, so the p–$\lambda$ graph is a rectangular hyperbola: momentum decreases as wavelength increases.
2014 AIPMT 1 question
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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:$\lambda = \frac{h}{\sqrt{2mK}}$
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{16} = 4 \Rightarrow \lambda_2 = \frac{\lambda_1}{4}$
% change $= \frac{1 - 4}{4}\times100 = -75\%$, i.e. a decrease of 75%.
2013 NEET 2 questions
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The wavelength $\lambda_e$ of an electron and $\lambda_P$ of a photon of same energy E are related by:Photon: $\lambda_P = \frac{h}{p} = \frac{hc}{E}$
Electron: $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \Rightarrow E = \frac{h^2}{2m\lambda_e^2}$
So $\lambda_P \propto \lambda_e^2$ -
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?As the speed of the electrons increases, their de Broglie wavelength decreases.
The angular width of the central maximum ($\propto \lambda$) therefore decreases.
2012 AIPMT-MAINS 1 question
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If the momentum of an electron is changed by P, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of electron will be$\lambda = \frac{h}{p}$, so for small changes $\frac{\Delta p}{p} = \frac{\Delta\lambda}{\lambda}$ (in magnitude)
Here $\Delta p = P$ and $\frac{\Delta\lambda}{\lambda} = \frac{0.5}{100}$
$\frac{P}{p} = \frac{0.5}{100} \Rightarrow p = 200P$
2012 AIPMT-PRE 1 question
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An $\alpha$-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/$m^2$. The de Broglie wavelength associated with the particle will beRadius of the path: $r = \frac{p}{qB} = \frac{p}{2eB} \Rightarrow p = 2eBr$
$p = 2\times1.6\times10^{-19}\times0.25\times0.83\times10^{-2}$ kg m/s
$\lambda = \frac{h}{p} = \frac{6.6\times10^{-34}}{0.83\times10^{-2}\times2\times1.6\times10^{-19}\times0.25}$
$\lambda = 9.94\times10^{-13}$ m ≈ 0.01 Å
2011 AIPMT-PRE 1 question
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Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons wouldFor an electron accelerated through a potential difference V, kinetic energy = eV and momentum $p = \sqrt{2meV}$.
de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$, so $\lambda \propto \frac{1}{\sqrt{V}}$
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{25}{100}} = \frac{1}{2}$
$\lambda_2 = \frac{\lambda_1}{2}$
So the wavelength decreases by 2 times.
Note: the source prints '100 km'; it is 100 kV.
2008 AIPMT 1 question
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A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of $3 \times 10^{6}$ m s$^{-1}$. The velocity of the particle is :
(mass of electron $= 9.1 \times 10^{-31}$ kg)de Broglie wavelength associated with the electron moving with velocity $v_e$: $\lambda_e = \dfrac{h}{m_e v_e}$
$\lambda_e = \dfrac{h}{9.1 \times 10^{-31} \times 3 \times 10^{6}}$
Wavelength of the particle of mass 1 mg $= 1 \times 10^{-6}$ kg moving with velocity $v$: $\lambda_p = \dfrac{h}{1 \times 10^{-6} \times v}$
As given, $\lambda_e = \lambda_p$
$\Rightarrow \dfrac{h}{9.1 \times 10^{-31} \times 3 \times 10^{6}} = \dfrac{h}{10^{-6} \times v}$
$\Rightarrow 10^{-6} \times v = 9.1 \times 10^{-31} \times 3 \times 10^{6}$
$\Rightarrow v = \dfrac{27.3 \times 10^{-25}}{10^{-6}}$ m/s
$v = 2.73 \times 10^{-18}$ m/s $\approx 2.7 \times 10^{-18}$ m s$^{-1}$
2002 AIPMT 1 question
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If particles are moving with same velocity, then De-Broglie wavelength is maximum for : -de Broglie wavelength: $\lambda = \dfrac{h}{mv}$
For the same velocity, $\lambda \propto \dfrac{1}{m}$; the lightest particle has the longest wavelength.
Masses: $\alpha$-particle $\approx$ 4 u, proton $\approx$ 1 u, neutron $\approx$ 1 u, $\beta$-particle (electron) $\approx \dfrac{1}{1837}$ u.
The $\beta$-particle (electron) has the smallest mass.
So the de Broglie wavelength is maximum for the $\beta$-particle.