An electron and a photon have the same d...

An electron and a photon have the same de Broglie wavelength. Which one of these has higher kinetic energy?

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To determine which has higher kinetic energy between an electron and a photon that have the same de Broglie wavelength, we can follow these steps: ### Step 1: Understand the de Broglie wavelength The de Broglie wavelength (\(\lambda\)) of a particle is given by the formula: \[ \lambda = \frac{h}{mv} \] where: ...

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If a proton and electron have the same de Broglie wavelength, then

kinetic energy of electron `lt` kinetic energy of proton

kinetic energy of electron = kinetic energy of proton

momentum of electron = momentum of proton

momentum of electron `lt` momentum of proton

An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

Zero

Infinity

Equal to the kinetic energy of the proton

Greater than the kinetic energy of the proton

An electron with speed v and a photon with speed c have the same de-Broglic wavelength. If the kinetic energy and momentum of electron is E_(e) and P_(e) and that of photon is E_(ph) and P_(ph) respectively, then correct statement is -

`(E_(e))/(E_(ph))=(2c)/(v)`

`(E_(e))/(E_(ph))=(v)/(2c)`

`(P_(e))/(P_(ph))=(2c)/(v)`

`(P_(e))/(P_(ph))=(v)/(2c)`