Determine the wavelenghts of photons

[soul5]

Homework Statement

Determine the wavelenghts of photons given off for the 3rd line in the Lyman series, the 2nd line in the Balmer series and the 1st line in the Paschen series.

The Attempt at a Solution

Lyman series: n=1
Balmer series: n=2
Paschen Series: n=3

What do I do?

 

[hage567]

If you look up the Bohr Model you should come across the equation(s) you need to solve this. See if you can find it.

 

[Moetasim]

To determine the wavelengths of photons for these series, we can use the Rydberg formula. This formula relates the wavelengths of photons emitted by hydrogen atoms to their energy levels. It is given by:

1/λ = R (1/n1^2 - 1/n2^2)

Where λ is the wavelength, R is the Rydberg constant (1.097x10^7 m^-1), and n1 and n2 are the energy levels of the electron.

For the Lyman series, n1=1 and n2=∞ (since the electron is transitioning from the highest energy level to n=1). Plugging these values into the formula, we get:

1/λ = R (1/1^2 - 1/∞^2)

1/λ = R (1 - 0)

1/λ = R

λ = 1/R = 91.2 nm

Therefore, the wavelength of photons emitted in the 3rd line of the Lyman series is 91.2 nm.

Similarly, for the Balmer series, n1=2 and n2=∞, so the formula becomes:

1/λ = R (1/2^2 - 1/∞^2)

1/λ = R (1/4 - 0)

1/λ = R/4

λ = 4/R = 364.8 nm

And for the Paschen series, n1=3 and n2=∞, giving us:

1/λ = R (1/3^2 - 1/∞^2)

1/λ = R (1/9 - 0)

1/λ = R/9

λ = 9/R = 1094.4 nm

In summary, the wavelengths of photons emitted in the 3rd line of the Lyman series, the 2nd line of the Balmer series, and the 1st line of the Paschen series are 91.2 nm, 364.8 nm, and 1094.4 nm, respectively.