A proton is accelerated through 225 v. its de broglie wavelength is
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The given deBroglie equation is
λ=h/pλ=h/p
But we know that from Kinetic energy equation, in Classical mechanics that p=2mK−−−−√p=2mK where mm implies mass of object, KK implies kinetic energy of the object.
For charged particles, that Kinetic energy is related to the voltage through which the proton is accelerated, so K=eVK=eV Overall substitution leads to
λ=h2meV√λ=h2meV
We know that h=6.625∗10−34h=6.625∗10−34 2=22=2 m=1.6∗10−27m=1.6∗10−27 e=1.6∗10−19e=1.6∗10−19 and as given V=225V=225.
Substitution and further simplification gives us λ=λ=1.95190935∗10−12m1.95190935∗10−12m, or 1.95pm1.95pm
That's about the wavelength of X-rays.
λ=h/pλ=h/p
But we know that from Kinetic energy equation, in Classical mechanics that p=2mK−−−−√p=2mK where mm implies mass of object, KK implies kinetic energy of the object.
For charged particles, that Kinetic energy is related to the voltage through which the proton is accelerated, so K=eVK=eV Overall substitution leads to
λ=h2meV√λ=h2meV
We know that h=6.625∗10−34h=6.625∗10−34 2=22=2 m=1.6∗10−27m=1.6∗10−27 e=1.6∗10−19e=1.6∗10−19 and as given V=225V=225.
Substitution and further simplification gives us λ=λ=1.95190935∗10−12m1.95190935∗10−12m, or 1.95pm1.95pm
That's about the wavelength of X-rays.
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