Physics graphs

Watch this thread
papie
Badges: 16
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 9 months ago
#1
will post question below
0
reply
papie
Badges: 16
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report Thread starter 9 months ago
#2
Any idea what the shape of graph is ?
0
reply
Callicious
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 9 months ago
#3
For the de Broglie, remember that p=\frac{h}{\lambda}. Come up with a proportionality for that and see how it changes when you say, half p or double \lambda. Use a proportionality to get your solution.

For the radioactive isotope decay, you know that A(t)=A_0\exp\left(-\lambda{t}\right), where \lambda is some decay constant. A describes the number of decays per second of the radioactive isotope to the stable product, in other words,





A(t) = -\frac{dN_{\textrm{unstable}}}{dt}




where I've thrown N_\textrm{unstable} in self-explanatorily. The negative sign should be obvious- you're losing radioactive isotope. Anyhow, you know that for every radioactive decay, you gain one stable product- can you think of a way to relate dN_\textrm{stable} to dN_\textrm{unstable}? Try to put that into the equation above and get an integral going to get N_\textrm{stable}(t).
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest

How are you feeling about your results?

They're better than I expected (116)
41.43%
They're exactly what I expected (66)
23.57%
They're lower than I expected (98)
35%

Watched Threads

View All