(a) The H atom absorbs energy during the transition from n = 6 to n = 3.
The energy levels of an atom are quantized, which means that an atom can only occupy certain specific energy levels, or states. When an atom transitions from a higher energy level to a lower energy level, it must release the excess energy as electromagnetic radiation, such as light. This process is called emission.
On the other hand, when an atom absorbs energy and transitions from a lower energy level to a higher energy level, it must absorb energy from the environment. This process is called absorption.
In this case, the H atom is transitioning from a higher energy level (n = 6) to a lower energy level (n = 3), so it must release energy as electromagnetic radiation. This means that the H atom emits energy during the transition.
(b) The wavelength of the radiation associated with the spectral line can be calculated using the following formula:
wavelength = R * (1/nf^2 - 1/ni^2)
where R is the Rydberg constant (1.097 x 10^7 m^-1), nf is the final energy level (n = 3 in this case), and ni is the initial energy level (n = 6 in this case).
Plugging in the values, we get:
wavelength = (1.097 x 10^7 m^-1) * (1/3^2 - 1/6^2)
= 3.289 x 10^6 m^-1 * (-0.0555)
= -181.68 m^-1
Converting to nanometers (nm), we get:
wavelength = (-181.68 m^-1) / (1 x 10^-9 m/nm)
= -181.68 x 10^9 nm
= 181,680 nm
Thus, the wavelengths of the radiation associated with the spectral line is approximately 181,680 nm.