Question

From the information, what is element X?

a.) The wavelength of the radio waves sent by an FM station broadcasting at 97.1 MHz is 30.0 million (3.00 x 10^7) times greater than the wavelength corresponding to the energy difference between a particular excite stateof the hydrogen atom and the ground state.

b.) Let V represent the principle quantum number for the valence shell of element X. If an electron in the hydrogen atom falls from shell V to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of 570 m/s.

c.) The number of unpaired electrons for element X in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations n=2, ml= -1, and ms= -1/2.

d.) Let A equal the principle quantum number for the lowest energy excited state for hydrogen. The value of A also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element X.

Answer 1

We know,  

Where   = wavelength, c =speed of light and = frequency

1. Wavelength of the element X = 3* 10^8/(30*10^7 *97.1*10^6)

                                                                                    =1.02 *10^-8 m

Let the transition be from n=1 to n=n, we know that

Putting the values we get,

(1/1.02 *10^-8)= 1.096*10⁷ (1/n²-1)

Solving for n we get n= 3. Therefore, the transition is from n=3 to n=1

1. From de Broglie’s equation, wavelength of an electron moving at a velocity v is

where h is Plank’s constant

Putting the values we get,

                                                      = 1.277 *10^-6 m

This transition is from shell V to 3

Therefore,

1/1.277 *10^-6 = 1.096*10⁷ (1/V²-1/3²)

Which gives V= 2

1. n=2, m_l= -1 and m_s=-1/2

m_l = -l, -(l-1),….

For l= 1, we get m_l= -1 which suggests p orbital. The principal quantum number is 2.

So, the electronic configuration is 1s² 2s² 2p¹

1. The lowest excited state of H is n=1 .Hence principal quantum number is 1 which is equal to the angular momentum quantum number for the subshell containing the unpaired electron(s).

Hence the element X is Boron (B)