# Calculating Velocity & Field Strength of Gas: Help Needed!

• Barbequeman
In summary: For the first calculation, you would use 676.813 nm, not 676.816, 676.819, or 676.822. You would also need to calculate the velocity in meters per second, not centimeters per second.
Barbequeman
Homework Statement
From near to the centre of the solar disc, observations of Ni I line emission (which has rest wavelength of 676.800 nm) are analysed at two points, X and Y. At point X the observations show a triple-peaked spectral line with components at 676.813, 676.820 and 676.827 nm. The central component is not as bright as those at the longer and shorter wavelengths.

Measurements at a point Y around 10,000 km away from X on the Sun’s surface show three components at 676.816, 676.819 and 676.822. In this case, the central component is brighter than the components at longer and shorter wavelengths.

a. For each point X and Y determine the velocity of the gas containing the nickel relative to the observer.
b. Calculate the magnetic field strength at each point X and Y.
Relevant Equations
v=c (λ-λ0)/λ0

B=v/(c*λ)
For the first calculation of the velocity of the gas I use the first equation and this converted in meter would be look like this (first value as an example)

v=299792458 m/s * (6.76813x10^-7-6.768x10^-7)/6.768x10^-7 =5836.03m/s or 0.0019c
this was the velocity of the gas for the first spectral line at 676.813nm

the next I´m not sure if it is correct, to calculate the field strength in Tesla with the second formula

B=5836.03m/s / 299792458m/s * 6.67813x10^-7 m = 28.76 Tesla

With my formula the calculated field strength is much to high ... I´m not sure where the error lies

I would be glad to get some help in this case, or a response if I did it correctly

Last edited:
Barbequeman said:
Homework Statement:: From near to the centre of the solar disc, observations of Ni I line emission (which has rest wavelength of 676.800 nm) are analysed at two points, X and Y. At point X the observations show a triple-peaked spectral line with components at 676.813, 676.820 and 676.827 nm. The central component is not as bright as those at the longer and shorter wavelengths.

Measurements at a point Y around 10,000 km away from X on the Sun’s surface show three components at 676.816, 676.819 and 676.822. In this case, the central component is brighter than the components at longer and shorter wavelengths.

a. For each point X and Y determine the velocity of the gas containing the nickel relative to the observer.
b. Calculate the magnetic field strength at each point X and Y.
Relevant Equations:: v=c (λ-λ0)/λ0

B=v/(c*λ)

For the first calculation of the velocity of the gas I use the first equation and this converted in meter would be look like this (first value as an example)

v=299792458 m/s * (6.76813x10^-7-6.768x10^-7)/6.768x10^-7 =5836.03m/s or 0.0019c
this was the velocity of the gas for the first spectral line at 676.813nm

the next I´m not sure if it is correct, to calculate the field strength in Tesla with the second formula

B=5836.03m/s / 299792458m/s * 6.67813x10^-7 m = 28.76 Tesla

With my formula the calculated field strength is much to high ... I´m not sure where the error lies

I would be glad to get some help in this case, or a response if I did it correctly
May I inquire where you obtained your relevant equations?

Your first equation, v=c (λ-λ0)/λ0, while it might function as a rough approximation for relatively slow velocities, it is not exact. It's not precise enough for the number of significant figures that you used.

You might wish to use more precise equations. I would go here for a starting point:
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/reldop2.html

With some algebra, you can solve the velocity by manipulating:
$Z = \sqrt{\frac{1 + \beta}{1 - \beta}} - 1,$
where,
$Z = \frac{\lambda - \lambda_0}{\lambda_0},$
and
$\beta = \frac{v}{c}$

Your second equation, B=v/(c*λ), is obviously incorrect. It doesn't have the right dimensions. According to the equation, the dimensions of magnetic field are $[\mathrm{length^{-1}}]$ which is incorrect.

For that, you might want to start researching the "Zeeman effect."
https://en.wikipedia.org/wiki/Zeeman_effect
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html

I don't know which formula to use, but should relate to the Zeeman effect some way or another.

----------

Regarding your attempted solution for the Doppler effect, you should be using the center wavelength, not one of the splits.

Last edited:
BvU

## 1. How do you calculate velocity of a gas?

The velocity of a gas can be calculated using the formula v = √(3kT/m), where v is the velocity, k is the Boltzmann constant, T is the temperature in Kelvin, and m is the mass of the gas particles.

## 2. What is the relationship between velocity and temperature in a gas?

According to the kinetic theory of gases, the velocity of gas particles is directly proportional to the square root of the temperature. This means that as the temperature increases, the velocity of gas particles also increases.

## 3. How do you calculate the field strength of a gas?

The field strength of a gas can be calculated using the formula E = F/q, where E is the field strength, F is the force acting on the gas particles, and q is the charge of the gas particles.

## 4. What factors affect the velocity and field strength of a gas?

The velocity of a gas is affected by temperature, mass of gas particles, and the size of the container. The field strength of a gas is affected by the force acting on the gas particles and the charge of the particles.

## 5. How can the velocity and field strength of a gas be measured?

The velocity of a gas can be measured using various techniques such as the time-of-flight method or the effusion method. The field strength of a gas can be measured using an electric field meter or by observing the motion of charged particles in the gas.

Replies
49
Views
4K
Replies
23
Views
2K
Replies
4
Views
2K
Replies
2
Views
6K
Replies
9
Views
4K
Replies
17
Views
2K
Replies
8
Views
1K
Replies
1
Views
2K
Replies
17
Views
2K
Replies
4
Views
895