(C) For an electron with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.The de-Broglie wavelength

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

(C) For an electron with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.The de-Broglie wavelength of the electron is $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For a photon,the energy $E$ is related to its wavelength $\lambda_p$ by $E = \frac{hc}{\lambda_p}$,which implies $\lambda_p = \frac{hc}{E}$.The ratio of the wavelengths is $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c} \sqrt{\frac{E}{2m}}$.

Class 12 Physics Dual Nature of Radiation and matter Matter Waves and de Broglie Wavelength

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An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelength of the electron to the wavelength of the photon is ($c =$ velocity of light).

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A
$c \sqrt{\frac{E}{m}}$
B
$\frac{1}{c} \sqrt{\frac{2m}{E}}$
C
$\frac{1}{c} \sqrt{\frac{E}{2m}}$
D
$c \sqrt{\frac{m}{E}}$

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Dual Nature of Radiation and matter

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Matter Waves and de Broglie Wavelength

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An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation, determine the de-Broglie wavelength of the electron:

Medium

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The equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:

EasyJEE MAIN 2022

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An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ $(E_0 = \text{constant} > 0)$,with an initial velocity $\vec{V} = v_0\hat{i}$ $(v_0 = \text{constant} > 0)$. If $\lambda_0 = \frac{h}{mv_0}$,its de Broglie wavelength at time $t$ is . . . . . . .

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Neglecting the variation of mass with velocity,the wavelength associated with an electron having a kinetic energy $E$ is proportional to

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Assertion $(A):$ $A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of their de-Broglie wavelengths is unity.
Reason $(R):$ Here,we cannot apply the conservation of linear momentum.

MediumAIIMS 2015

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An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelength of the electron to the wavelength of the photon is ($c =$ velocity of light).

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An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation, determine the de-Broglie wavelength of the electron:

The equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ $(E_0 = \text{constant} > 0)$,with an initial velocity $\vec{V} = v_0\hat{i}$ $(v_0 = \text{constant} > 0)$. If $\lambda_0 = \frac{h}{mv_0}$,its de Broglie wavelength at time $t$ is . . . . . . .

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Assertion $(A):$ $A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of their de-Broglie wavelengths is unity.
Reason $(R):$ Here,we cannot apply the conservation of linear momentum.