Can't get my head around reduced mass (particle physics)

[Flucky]

I know it's probably simple but I just don't understand reduced mass.

I am trying to work out the reduced mass of ⁴He⁺.

m = m_em_N / m_e + m_N

Can somebody please just explain step by step what I do. This is only a segment of a 1 mark question and I'm getting my knickers in a frustratingly twisty twist. I've looked at the other threads and websites but still no clue.

The answer needs to be 3.99... but I always end up with 0.99...

 

[Orodruin]

Normally you would need to expand more than that on your attempted solution, but I have a suspicion.

What are the units you are quoting your answers in? Atomic mass units? In that case, did you insert the mass of one nucleon instead of that of a helium nucleus?

 

[Flucky]

Hi Orodruin

I'll give you the full question to give a better understanding-

Calculate the wavelength of the n = 4 → 3 transition in ⁴He⁺ to an accuracy of 4 significant figures. (R∞=109 737 cm^-1.) (Fine structure effects can be neglected.)

Now the equation that I'd use for this is:

$\frac{1}{λ} = \frac{m}{m_e} R_∞ (\frac{1}{n_1^2} - \frac{1}{n_2^2})$

Where λ is wavelength, m is the reduced mass, and $R_∞$ is Rydberg constant.

So I know the answer for the wavelength is 468.7 nm (I looked), and working backwards to try make sense of reduced mass I got m = 3.99.

I tried it in SI units using the mass in kg of an electron + 2 protons + 2 neutrons but it still didn't help.

 

[Flucky]

I will create a new thread with proper formatting, and pose the question as a whole - so could you delete this thread please?

Edit: properly formatted thread at https://www.physicsforums.com/threads/calculate-the-wavelength-of-the-n-4-3-transition-in-he.791525/

 

[HalfLostAndSearching]

Reduced mass is a concept used in particle physics to simplify calculations involving two or more particles. It is defined as the effective mass of a system of two or more particles in motion relative to each other. In the case of 4He+, the reduced mass is used to simplify calculations involving the motion of the 4He+ ion and an electron.

To calculate the reduced mass, you will need to know the masses of the particles involved. In this case, we have the mass of the 4He+ ion (m) and the mass of the electron (me). The mass of the neutron (mN) is also involved, but since it is much larger than the electron mass, we can approximate it as infinity.

The formula for calculating the reduced mass is:

μ = memN / (me + mN)

Substituting the values we have, we get:

μ = (4.0026 x 1.6605 x 10^-27 kg) / (9.1094 x 10^-31 kg + infinity)

Since infinity is essentially a very large number, we can ignore it in this calculation. So, the reduced mass is approximately equal to:

μ = (4.0026 x 1.6605 x 10^-27 kg) / 9.1094 x 10^-31 kg

= 6.6419 x 10^-27 kg

This is the reduced mass of the system. To get the final answer of 3.99, you may need to round off the value to the appropriate number of significant figures. I hope this helps clarify the concept of reduced mass for you. If you have any further questions, please don't hesitate to ask.