Question

From the information below, identify element X. a. The wavelength of the radio waves sent by an FM station broadcasting at 97.1 MHz is 30.0 million (3.00×107) times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state. b. Let V represent the principal 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 principal quantum number for the lowest energy excited state for hydrogen. This value of A also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element X.

Answer 1

Ans) The wavelength corresponding to 97.1MHz is hc/$\lambda$ = h x 9.71 x 10⁷ => $\lambda$ = c/(9.71 x 10⁷) = 3.08m

The wavelength corresponding to the energy level difference is 3.08/(3 x 10⁷ ) = 1.02 x 10^-07 m

Let the excited state be n. By Rydberg's formulae we know that 1/$\lambda$ = 1/(102 x 10^-09) = 1.097 x 10⁷(1- (1/n)²)

=> solving this we get n = 3

Now the wavelength associated with the electron moving with a speed of v=570m/s be $\lambda$₂ = h/mv. { By de broglie equation}

$\lambda$₂ = (6.626 x 10^-34)/{(9.1 x 10^-31)(570)}. => $\lambda$₂ = 127.7 x 10^-08m Now let the valence shell be V. The transition is from V to 3rd shell. Again by Rydberg's formulae we get 1.097 x 10⁷(1/9 - (1/V)²) = 1/(127.7 x 10^-08)

Solving this we get V= 5. So the principal quantum number is 5.

The set of quantum numbers in the question denote a 2p_x¹ configuration so it can have total of 5 electrons. So the number of unpaired electrons is 5

Principal quantum number of lowest energy state of hydrogen is 2 = A which designates d orbital

Therefore the unpaired electron configuration is d5.

Therefore the element X has 5s² 4d⁵. So the element is molybdenum atomic number 43