Advertisements
Advertisements
प्रश्न
The ratio between the de Broglie wavelength associated with proton accelerated through a potential of 512 V and that of alpha particle accelerated through a potential of X volts is found to be one. Find the value of X.
Advertisements
उत्तर
de-Broglie wavelength of accelerated charge particle
λ = `"h"/sqrt(2"mqV")`
`λ ∝ "h"/sqrt("mqV")`
Ratio of wavelength of proton and alpha particle
`λ_"p"/λ_α = sqrt(("m"_α"q"_α"V"_α)/("m"_"p""q"_"p""V"_"p")) = sqrt(("m"_α/"m"_"p") ("q"_α/"q"_"p") ("V"_α/"V"_"p"))`
Here, `"m"_α/"m"_"p"` = 4; `"q"_α/"q"_"p"` = 2; `"V"_α/"V"_"p" = "X"/512`; `λ_"p"/λ_α` = 1
1 = `sqrt(4 xx 2 xx ("X"/512))`
= `sqrt("X"/64)`
= `"X"/64`
X = 64 V
APPEARS IN
संबंधित प्रश्न
Write the expression for the de Broglie wavelength associated with a charged particle of charge q and mass m, when it is accelerated through a potential V.
Why we do not see the wave properties of a baseball?
A proton and an electron have the same kinetic energy. Which one has a greater de Broglie wavelength? Justify.
An electron and an alpha particle have the same kinetic energy. How are the de Broglie wavelengths associated with them related?
What is Bremsstrahlung?
Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Derive an expression for de Broglie wavelength of electrons.
Briefly explain the principle and working of electron microscope.
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
An electron is accelerated through a potential difference of 81 V. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this wavelength correspond?
