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choli [55]
3 years ago
13

The balmer series is formed by electron transitions in hydrogen that

Physics
1 answer:
RSB [31]3 years ago
4 0

Answer:

The Balmer series refers to the spectral lines of hydrogen, associated to the emission of photons when an electron in the hydrogen atom jumps from a level n \geq 3 to the level n=2.

The wavelength associated to each spectral line of the Balmer series is given by:

\frac{1}{\lambda}=R_H (\frac{1}{2^2}-\frac{1}{n^2})

where R_H is the Rydberg constant for hydrogen, and where n is the initial level of the electron that jumps to the level n = 2.

The first few spectral lines associated to this series are withing the visible part of the electromagnetic spectrum, and their wavelengths are:

656 nm (red, corresponding to the transition 3 \rightarrow 2)

486 nm (green, 4 \rightarrow 2)

434 nm (blue, 5 \rightarrow 2)

410 nm (violet, 6 \rightarrow 2)

All the following lines lie in the ultraviolet part of the spectrum. The limit of the Balmer series, corresponding to the transition \infty \rightarrow 2, is at 364.6 nm.

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How do surface currents affect climate
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3 years ago
Suppose a uniform electric field of 4 N/C is in the positive x direction. When a charge is placed at and fixed to the origin, th
Yuliya22 [10]

Answer:

E_total = 3 N / A

Explanation:

The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.

               E_total = E + E₁

the expression for the field of a point charge is

                E₁ = k q₁ / r²

for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative

                 E_total = E -k q₁ / r₂

we substitute

                   0 = E - k q₁ / r²

                   q₁ = \frac{E r^2}{k}

let's calculate

                   q₁ = \frac{4 \ 2^2}{9 \ 10^{-9}}

                   q₁ = 1.78 10⁻⁹ C

now we can calculate the field for position x = 4 m

                   E_total = 4 - 9 10⁹  1.78 10⁻⁹ / 4²2

                   E_total = 3 N / A

8 0
3 years ago
What is similar and different about ionic and covalent bonds?
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The similarity is that they both are types of bonds in molecules.
Ionic bonds are between a metal and a nonmetal.
Covalent bonds are between two nonmetals.
8 0
3 years ago
Help pleaseeeeeeeeeeeee
Tasya [4]
A it’s not that hard tbh
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3 years ago
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