Question

Answer these and explain your reasoning.

1. How much energy would it take to remove the electron from unexcited hydrogen completely so that it is no longer bound to its proton?

2. Suppose a hydrogen atom in the n=3 state emitted light by decaying into the n=2 state. What color would that light be seen as? What about one starting at n=4 and ending in n=2? Can a hydrogen atom emit infrared light and end up in the n=2 state?

3.   If you can see light with wavelengths as short as 385 nm, how many lines of the Balmer series might you be able to see by eye with a bright light source and a spectrometer to make a spectrum?

4. Suppose that you could measure accurately the wavelengths of the first two lines of the Balmer series. How could you use those measurements to predict the wavelength of the first line of the Paschen series?

This is the method by which we can find the energy levels of atoms experimentally.

Answer 1

1.) To remove the electron from the atom completely is referred to as moving the electron from the ground state to infinity.

where R is the Rydberg constant( 1.097 x 10⁷ m^-1 )

n₁ = 1 and n₂ = infinity

So,

   (Since 1/infinity = 0)

  

Frequency,

  

Energy,

  

where h is the Planks constant( 6.626 x 10^-34 Js)

= 2.18 x 10^-18 J = 13.6 eV

This is the required energy to remove an electron from an hydrogen atom.

2.) In the excited state,

where, n₁= 2 and n₂ = 3

m = 655.9 nm

According to the visible light spectrum range, this wavelength corresponds to red colour so one sees red colour.

When n₂ = 4 and n₁ = 1 :

48.587 x 10^-8 m = 485.87 nm

This wavelength corresponds to green colour.

The hydrogen atom cannot emit infrared light and end up in n=2 state as the maximum wavelength it can emit is 655.9 nm while range of infrared spectrum is from 700 nm to 1 micro meter.

3.) The minimum wavelength to see the spectrum is 400nm corresponding to violet colour.If the wavelength is below 400nm it falls to the ultraviolet region and no spectrum can be seen with the eye.

4.) For the first line of Balmer series,

   n₁ = 2 and n₂ = 3

     

For the second line for Balmer series,

n1 = 2 and n₃ = 4

For the first line of Paschen series,

n₁ = 3 and n₂ = 4

Therefore the wavelength of first line of paschen series can be found out by subtracting the wavelength of first line of Balmer series from the wavelength of second line of Balmer series.