From the information below, identify element \(\mathrm{X}\). a. The wavelength of the radio waves sent by an FM station broadcasting at \(97.1 \mathrm{MHz}\) is \(30.0\) million \(\left(3.00 \times 10^{7}\right)\) 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 \(570 . \mathrm{m} / \mathrm{s}\) c. The number of unpaired electrons for element \(\mathrm{X}\) in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations \(n=2\), \(m_{\ell}=-1\), and \(m_{s}=-\frac{1}{2}\) d. Let \(A\) equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of \(A\) also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element \(\mathrm{X}\).

Step by step solution

01

Solving for part a#

We know that the wavelength of the radio waves sent by the FM station broadcasting at \(97.1\, \mathrm{MHz}\) is \(30.0 \times 10^7\) times greater than the wavelength of the energy difference of the excited state of the hydrogen atom and the ground state. The frequency \(f\) is given as: \[f = 97.1 \times 10^6\, \mathrm{Hz}\] To get the wavelength \(\lambda_1\), we can use the speed of light, \(c\): \[c = 3.00 \times 10^8\, \mathrm{m/s}\] and the formula, \[c = \lambda_1 \times f\] Solving for wavelength \(\lambda_1\): \[\lambda_1 = \frac{c}{f}\] Now, \[\lambda_2 = 30.0 \times 10^7 \times \lambda_1\] Where \(\lambda_2\) corresponds to the energy difference between the excited state of the hydrogen atom and the ground state.

02

Solving for part b#

We are given that if an electron falls from shell V to the inner shell of excited state, the wavelength of emitted light is the same as the wavelength of an electron moving at a speed of \(570\, \mathrm{m/s}\), which is \(\lambda_2\). Using the de Broglie wavelength formula, \[\lambda_2 = \frac{h}{mv}\] Where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(v\) is its speed. Now we have: \[V = \frac{n_1^2 + n_2^2}{n_1^2 - n_2^2}\] Where \(n_1\) and \(n_2\) are principal quantum numbers, and \(V\) represents the principal quantum number for the valence shell of element X.

03

Solving for part c#

Now we need to find the number of unpaired electrons for element X, which is the same as the maximum number of electrons in an atom that can have the quantum number \(n=2, m_\ell=-1\), and \(m_s=-\frac{1}{2}\). Since there is only one electron that can have these designations, the element X must have one unpaired electron.

04

Solving for part d#

Lastly, we need to find the charge A of the stable ion that would form when the unknown element 120 forms ionic compounds. This charge A also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element X. Angular momentum quantum number \(l\) is given by \(A=l\): Since \(A=1\), the subshell containing the unpaired electron(s) would have the quantum number representation of: \[n=2, l=1, m_\ell=-1\] #Conclusion#: After analyzing all the given information, we can deduce that element X has unpaired electrons in the second shell's \(p\) subshell and one unpaired electron. Therefore, element X is most likely Boron.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!