The de Broglie wavelength

plstevens

The mass of an electron is 9.11*10^-31 kg. If the de Broglie wavelength for an electron in an hydrogen atom is 3.31*10^-10 m, how fast is the electron moving relative to the speed of light? The speed of light is 3.00*10^8 m/s.

here's what I did: i solved for velocity=6.626*10^-34J/(9.11*10^-31kg)(3.31*10^-10)
v=2.1974*10^-74
and i tried to gain the percent by dividing the speed of light by velocity.

where did i go wrong?

cepheid

Your calculation is wrong. Don't just blindly do calculations. Think...does the number your calculator has spewed out actually make any sense? When it's something ridiculous like 10^-74 m/s, the answer is emphatically NO. Kind of slow for a particle, don't you think?

I get v = (0.00732)c

Marco_84

how did u calculate.... watch the exponents first......

the magnitude is 10^7 m/s...

[tex]\frac{6}{9\times3}\times\frac{10^{-34}}{10^{-31}\times10^{-10}}\approx\frac{2}{9}10^7 m/s[/tex]
this suggest us that it is better to treat the electron relativistically if we want to penetrate deep in its properties...
regards
marco

Marco_84

thanx Dirac :)

plstevens

so hows do i get the percentage here's what i'm doing: 3.00*10^8 m/s /100 = 0.00732/x. x=2.4*10^8, but i know this isn't right so, what shall i do?

cepheid

I'm not sure what percentage you are talking about, since it's not mentioned in the original post.

For the velocity of the particle, I get:

[tex] v = 2.197 \times 10^6 \ \ \ \frac{\textrm{m}}{\textrm{s}} [/tex]

The question asks how fast the particle is moving relative to the speed of light. Well, their ratio is

[tex] \frac{v}{c} = \frac{2.197 \times 10^6 \ \ \ \textrm{m/s}}{3.00 \times 10^8 \ \ \ \textrm{m/s}} = 0.00732 [/tex]

So, expressed in units of the speed of light, the velocity is

[tex] v = 0.00732c [/tex]

The particle is moving at 0.00732 times the speed of light. Obviously, as a percentage, that's 0.732%. So I guess if you wanted to, you could say that the particle is moving at 0.732% of the speed of light. It's a completely equivalent statement though. It doesn't add any extra meaning.