Question

An α particle is accelerated by 25V. Calculate its debroglie wavelength

Answer 1

First of all, the kinetic energy of the alpha particle would be 50 eV, right? The mass energy of an alpha particle is about 4 GeV, which is vastly greater than its kinetic energy, which means that this problem can be treated nonrelativistically.

\[ \lambda = \frac{h}{p} \]
so we just need to find p. But nonrelativistically,
\[ p = \sqrt{2\cdot m_\alpha \cdot T} \] so
\[ \lambda = \frac{h}{\sqrt{2\cdot m_\alpha \cdot T} } \]

So now all you have to do is plug in the right numbers. Be careful of your units, though, it's very important to be consistent.

Answer 2

To be consistent I suppose I should do it relativistically and then show it's basically the same.... Relativistically speaking,
\[ E^2 = p^2c^2 + m^2 c^4 \]
But
\[E = \gamma mc^2 \]
so
\[p^2 c^2 = (\gamma^2-1)m^2c^4\]
or
\[p^2 = (\gamma^2-1)m^2c^2 \]

\[ \gamma^2 -1 = \frac{1}{1-v^2/c^2} - 1 = \frac{v^2/c^2}{1 - v^2/c^2}\]
so
\[ p^2 = \frac{mv^2}{1-v^2/c^2} \cdot m = 2\cdot m \cdot T\cdot \gamma^2\]
yielding
\[ p = \gamma \sqrt{2mT}\]
which is what we had before except with an extra factor of gamma. However,
\[ \gamma mc^2 = mc^2 + T\]
so
\[ \gamma = 1 + \frac{T}{mc^2} = 1 + \frac{50 eV}{4\cdot 10^9 eV} \approx1+10^{-8} = 1.00000001\]

Answer 3

okay i guess i have had enough of physics and chem today
what is your profession jemurray?

Answer 4

I am a physics grad student.

Answer 5

you mean engineer?

Answer 6

...no...

Answer 7

physicist?'

Answer 8

Right. Grad student = graduate student. I'm a Ph.D. candidate.

Answer 9

hehe im in ninth . i live in south asia:india

Answer 10

Ah, I see... I assume then that the second part of my post was not necessary, but it's good to know I suppose.