# The Brogile wavelength of an Alpha particle

• mrausum
So the correct equation to use is lamda = h/mvIn summary, to calculate the de Broglie wavelength of an alpha particle with an energy of 5.78 MeV, the equation lamda = h/mv should be used, where h is Planck's constant, m is the mass of the alpha particle, and v is its velocity. This wavelength can then be compared to the known nuclear diameter of approximately 2x10^-14 m.

## Homework Statement

An alpha particle of energy 5.78 MeV is emitted from a particular nucleus.
Calculate the de Broglie wavelength of the alpha particle. How does it compare
with the nuclear diameter, which is known to be approximately 2x10−14 m.

E=h/p and E=hf

## The Attempt at a Solution

I was thinking of using E=hc/lamda but this would be wrong because alpha particles don't travel at c.

The energy of 5.78MeV is entirely kinetic.

so how do i do the question?

mrausum said:
so how do i do the question?

Use Ek=1/2mv2

well it hasn't given the mass. But are you saying work out v from E=1/2mv^2, then f from E/h. Then wavelength from lamda=v/f?

mrausum said:
well it hasn't given the mass. But are you saying work out v from E=1/2mv^2, then f from E/h. Then wavelength from lamda=v/f?

find v, then use

$$\lambda = \frac{h}{mv}$$

I am sure you can look up the mass of an alpha particle. I am not sure how else you can solve this problem without my suggestion though.

how do you get lambda = h/mv
i know E = hc/lambda = h/mv
so, if i combine both equations, shouldn't i suppose to get lambda = cmv ?
did i miss something or what.

HjGanap said:
how do you get lambda = h/mv
i know E = hc/lambda = h/mv
so, if i combine both equations, shouldn't i suppose to get lambda = cmv ?
did i miss something or what.

p = h/lamda
lamda = h/p=h/mv

You can't use E = hc/lamda because alpha particles don't travel at the speed of light.