Jump to content

Recommended Posts

Posted

image.png.ee28c436246bdfbd0cb9d8cb5605fcb7.pngimage.png.85a80e300b8d2c9e0d4b76c49110ebb5.pngI need to calculate the wavelenght of a ball travelling that is 0.435 kg and travells at 36 m/s. To calculate this i used the equations of the first picture where i put the two first functions equal to each other as they are both equal to E. If i then substitute c in the second function i become h.v = m. wavelenght^2.v^2. And after some rearranging we become wavelenght = the sqaure root of (h/(v.m)). but if i looked further into my formulas there was this exact same formula but without the square root. Is there a reason that i cant equal the two first fuctions because i dont understand why it does not work.

Posted
19 minutes ago, bobot said:

image.png.ee28c436246bdfbd0cb9d8cb5605fcb7.pngimage.png.85a80e300b8d2c9e0d4b76c49110ebb5.pngI need to calculate the wavelenght of a ball travelling that is 0.435 kg and travells at 36 m/s. To calculate this i used the equations of the first picture where i put the two first functions equal to each other as they are both equal to E. If i then substitute c in the second function i become h.v = m. wavelenght^2.v^2. And after some rearranging we become wavelenght = the sqaure root of (h/(v.m)). but if i looked further into my formulas there was this exact same formula but without the square root. Is there a reason that i cant equal the two first fuctions because i dont understand why it does not work.

c is the speed of light specifically, and E=mcrefers to rest mass specifically, so it is not appropriate to your problem. All you need is the de Broglie relation of momentum to wavelength, which you show on the right of your picture.

(The full Einstein formula, bringing in momentum (p) for objects in motion relative to a frame of reference, is E= (mc2)2 + (pc)2. For a massless entity such as a photon this reduces to E=pc, which if you apply E=hv gives you de Broglie’s relation.)

Posted (edited)

Also, \( \nu \) in \( E = h\nu \) is the Greek letter nu, while  \( v \) in DeBroglie's relation \( \lambda=h/mv \) is the Latin letter v. Could that be related to your problem?

IOW, one is the frequency in Hertzs, while the other is the speed.

Edited by joigus
correction

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.