Recent content by Adams2020

Doesn't anyone exist?:smile:

I think an error could be to read the diffraction angle. Because people may read different angles, albeit with slight differences. And because the diffraction angle is related to the wavelength, can affect it.

Yes, I am familiar with diffraction patterns and its theory. No, this is not a school work question, it is a personal question, otherwise I would have sent the question in the relevant section.

In a diffraction grating experiment, what can be the sources of error and also what effect do these sources have on the unknown quantity (wavelength) in the experiment?

Table: To solve, using the conservation of energy: E0=(m(deutron)+m(16O)-m(17O)-m(p))c^2 so: E0=(2.014+15.994-16.999-1.008)931.5=0.93 MeV. so using the conservation of energy: 14.95+0.93=16.62+0+E' E'=- 0.74 MeV But the energy sign has become negative. I also calculated for the first excited...

no

I don't know. This is the question our professor asked us without explanation to think. What is the reason that it is a vector?

Why is the action a scalar? Please explain.

ok thanks

A similar formula was stated by our nuclear professor, but with the difference that there was b instead of c. So I thought in this case b = c

My data is as follows:

My final result is as follows: 4*pi*R^2 (1+ε+ (7/5)ε ^2+ (36/35)ε ^3 +...) Which has a bad difference with the result of the book.:oldconfused:

In "An Introduction to Nuclear Physics by W. N. Cottingham, D. A. Greenwood" for the surface area of an oblate ellipsoid, the following equation is written for small values of ε : The book has said this without proof. I found the following formula for the desired shape: No matter how hard I...

I understand now by your good example. Thankful.

I'm still confused. For example, in 'Introduction to Elementary Particles by Griffith', for relativistic collisions, the center of momentum frame is introduced to solve problems. But isn't the center of mass frame appropriate in relativistic collisions?