Calculate the wavelength for the emission transition if it starts from the orbit havingradius 1.3225 nm and ends at 211.6 pm. Name the series to which this transitionbelongs and the region of the spectrum

To solve this problem, we need to use the Rydberg formula for hydrogen, which is used to calculate the wavelength of light resulting from an electron moving between energy levels of a hydrogen atom.

The formula is:

1/λ = R * (1/n1² - 1/n2²)

Where:

• λ is the wavelength
• R is the Rydberg constant (approximately 1.097373 x 10^7 m^-1)
• n1 and n2 are the principal quantum numbers of the two energy levels (n1 < n2)

The radius of the nth orbit in a hydrogen atom is given by the formula:

r = 0.529 * n² (in Å)

We can use this formula to find the principal quantum numbers of the two energy levels.

For the initial level:

1.3225 nm = 1322.5 pm = 13.225 Å

So, n1² = 13.225 / 0.529 = 25

Therefore, n1 = sqrt(25) = 5

For the final level:

211.6 pm = 2.116 Å

So, n2² = 2.116 / 0.529 = 4

Therefore, n2 = sqrt(4) = 2

Now we can use the Rydberg formula to find the wavelength:

1/λ = R * (1/n1² - 1/n2²) = 1.097373 x 10^7 * (1/5² - 1/2²) = 1.644 x 10^6 m^-1

So, λ = 1 / (1.644 x 10^6) = 6.08 x 10^-7 m = 608 nm

This transition (n=5 to n=2) belongs to the Balmer series, which is in the visible region of the spectrum.