A line in the Balmer series of hydrogen has a wavelength of 486 nm. From what state did the electron originate?

Question

A line in the Balmer series of hydrogen has a wavelength of 486 nm. From what state did the electron originate?

In what region of the electromagnetic spectrum is this line observed?

Answer 1

Well, welcome to the electrifying world of hydrogen! 🎉 Now, let me enlighten you with a little humor before answering your question. Did you know that I once tried to be an electron, but I couldn't because I was always so positively charged with humor? 😄 Alright, let's get down to business!

The wavelength you mentioned, 486 nm, corresponds to the H-beta line in the Balmer series. This line is observed when the electron transitions from the n=4 energy level to the n=2 energy level. So, that's where our elusive electron originated from!

As for the region of the electromagnetic spectrum where this line is observed, it falls within the blue-green portion, or if you prefer official terms, the visible light region. So, keep your eyes open and your spectrometer ready because 486 nm is coming your way, bringing its vibrant blue-green charm! 🌈💙

Answer 2

To determine from which state the electron originated in the Balmer series of hydrogen, we can use the Balmer series equation:

1/λ = R_H * (1/n_f^2 - 1/n_i^2)

where λ is the wavelength of the emitted photon, R_H is the Rydberg constant (1.097 x 10^7 m^-1), n_f is the final state, and n_i is the initial state.

Since we know the wavelength (λ) is 486 nm, we need to convert it to meters by dividing it by 10^9 (nm to m conversion):

λ = 486 nm / (10^9 m/nm) = 4.86 x 10^-7 m

Now we can substitute the values into the equation:

1/(4.86 x 10^-7 m) = (1.097 x 10^7 m^-1) * (1/n_f^2 - 1/n_i^2)

We can rearrange the equation to solve for 1/n_i^2:

1/n_i^2 = 1/(1.097 x 10^7 m^-1) - 1/(4.86 x 10^-7 m)

1/n_i^2 = 9.117 x 10^-8 m^-1 - 2.057 x 10^6 m^-1

1/n_i^2 = -2.048 x 10^6 m^-1

To find the initial state (n_i), we can take the reciprocal of the square root of the right-hand side:

n_i = 1 / sqrt(-2.048 x 10^6 m^-1)

This will give us a complex number since the right-hand side is negative. However, in the context of the Balmer series of hydrogen, we are interested only in the positive values of n_i.

Now let's determine in what region of the electromagnetic spectrum this line is observed. The wavelength of 486 nm corresponds to the visible light region of the electromagnetic spectrum. Specifically, it falls in the blue region of the visible spectrum.

Answer 3

To determine the state from which the electron originated in the Balmer series and in what region of the electromagnetic spectrum the line is observed, we need to understand the energy levels of hydrogen and how they correlate to the Balmer series.

The Balmer series in hydrogen refers to a set of spectral lines that are emitted when an excited electron in a hydrogen atom transitions from higher energy levels to the second energy level (n = 2). These transitions result in the emissions of photons of specific wavelengths.

In order to find the state from which the electron originated, we can use the Balmer formula:

1/λ = R_H * (1/2^2 - 1/n^2)

Where:
- λ is the wavelength of the emitted photon
- R_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m^-1)
- n is the principal quantum number representing the final energy level of the electron

Given that the wavelength (λ) is 486 nm, we can rearrange the formula and solve for n:

1/λ = 1.097 x 10^7 * (1/2^2 - 1/n^2)
1/λ = 1.097 x 10^7 * (1/4 - 1/n^2)
1/λ = 1.097 x 10^7 * (n^2 - 4) / (4n^2)
(4n^2) / (1.097 x 10^7 * (n^2 - 4)) = λ

Plugging in the wavelength of 486 nm (converted to meters: 486 x 10^(-9) m), we can solve for n:

(4n^2) / (1.097 x 10^7 * (n^2 - 4)) = 486 x 10^(-9)

By solving this equation, we find that n is approximately equal to 3. Thus, the electron originates from the n = 3 energy state in hydrogen.

Now, to determine the region of the electromagnetic spectrum in which this line is observed, we can use the relationship between wavelength and electromagnetic spectrum ranges. Generally, the visible spectrum ranges from approximately 380 nm (violet) to 700 nm (red).

Since the wavelength of 486 nm falls within this range, it means that the line in the Balmer series with this wavelength is observed in the visible region of the electromagnetic spectrum, specifically in the blue-green part.

Answer 4

486 nm is in the visible part of the spectrum as are all of the lines in the Balmer series. In the Balmer series the end N = 2.

w = wavelength; R is the Rydberg constant which is 1.09737E7
1/w = R(1/4 - 1/x^2)
Solve for x.