PHY 111 - Ch 2 -Atomic Structure - PDF

CHAPTER 2 Atomic Structure  Bohr in 1913, proposed, what is now called the Bohr model of the atom. The model success to explain the spectral emission lines of hydrogen atom. Bohr postulates for H- atom 1- The electron revolves in stable circular orbit around the (+ve) nucleus without ra...

CHAPTER 2 Atomic Structure  Bohr in 1913, proposed, what is now called the Bohr model of the atom. The model success to explain the spectral emission lines of hydrogen atom. Bohr postulates for H- atom 1- The electron revolves in stable circular orbit around the (+ve) nucleus without radiate energy 2- The angular momentum of electron ∴ 𝑳 = 𝒏 ሺ𝒉Τ𝟐𝝅ሻ n is a quantum number 3- Emission or absorption of radiation occurs only when an electron moves from one orbit to another 𝒉𝒄 𝟏𝟐𝟒𝟎 𝜟𝑬 = 𝑬𝒇 − 𝑬𝒊 = 𝒉𝒇 = = 𝒆𝑽 𝝀 𝝀 𝒏𝒎 Where (f) is final orbit and (i) initial orbit  1- An atom absorbs a photon of frequency 6.2×1014 Hz. By how much does the energy of the atom increase? Answer 𝑬 = 𝒉𝒇  − 𝟏𝟓 𝑬 = 𝟒. 𝟏𝟒 × 𝟏𝟎 𝒆𝑽 𝒔 ሺ𝟔. 𝟐 × 𝟏𝟎𝟏𝟒 𝑯𝒛ሻ = 𝟐. 𝟓𝟔𝟕 𝒆𝑽 2- An atom absorbs a photon having a wavelength of 375 nm and immediately emits another photon having a wavelength of 580 nm. How much net energy was absorbed by the atom in this process? Answer 𝒉𝑪 𝒉𝑪 ∵ 𝜟𝑬 = 𝑬𝒂𝒃𝒔𝒐𝒓𝒃𝒆𝒅 − 𝑬𝒆𝒎𝒊𝒕𝒕𝒆𝒅 = − 𝝀𝒂 𝝀𝒆 𝟏 𝟏 ∴ 𝜟𝑬 = 𝟏𝟐𝟒𝟎 − 𝒆𝒗 = 𝟏. 𝟏𝟕𝒆𝑽 𝟑𝟕𝟓 𝟓𝟖𝟎 3- A line in the x – ray spectrum of gold has a wavelength of 18.5 pm. The emitted x – ray photon correspond to a transition of the gold atom between two stationary states, the upper one of which has the energy – 13.7 keV. What is the energy of the lower stationary state? Answer 𝒉𝒄 ሺ𝟒. 𝟏𝟒 × 𝟏𝟎−𝟏𝟓 ሻ × ሺ𝟑 × 𝟏𝟎𝟖 ሻ ∴ 𝜟𝑬 = = −𝟏𝟐 = 𝟔𝟕. 𝟏𝟑𝟓 𝒌𝒆𝑽 𝝀 𝟏𝟖. 𝟓𝒙𝟏𝟎 𝑬𝒊 = −𝟏𝟑. 𝟕 𝒌𝒆𝑽 ∵ 𝜟𝑬 = − 𝑬𝒇 − 𝑬𝒊 𝛌 = 𝟏𝟖. 𝟓 𝐩𝐦 𝑬𝒇 =? ? ∴ 𝜟𝑬 = 𝟔𝟕. 𝟏𝟑𝟓𝒌𝒆𝑽 = − 𝑬𝒇 − ሺ−𝟏𝟑. 𝟕𝒌𝒆𝑽ሻ Gold atom ∴ 𝑬𝒇 = 𝟖𝟎. 𝟕𝟏𝟕 𝒌𝒆𝑽 Help slide hydrogen atom The electron revolve in stable orbit under the effect of 𝒆𝟐 ❑ Coulomb force 𝑭𝒆 = 𝒌 𝟐 𝒓 𝒗𝟐 ❑ Centripetal force 𝑭𝑪 = 𝒎 𝒓 𝒆𝟐 𝒗𝟐 Fe=Fc ∴ 𝒌 𝟐 = 𝒎 𝒓 𝒓 Then, the kinetic energy is [times (1/2) in both sides in equation (1) 𝟐 𝟐 𝟐 𝟏 𝒆 𝟏 𝒆 𝒆 ∴ 𝑲. 𝑬 = 𝒎𝒗𝟐 = 𝒌 = = 𝟐 𝟐𝒓 𝟒𝝅𝜺𝒐 𝟐𝒓 𝟖𝝅𝜺𝒐 𝒓 Derive The radius of orbital (r n) of Hydrogen atom 𝒆𝟐 𝒗𝟐 𝒆 𝟐 Use (Fe=Fc ) ∴ 𝒌 𝟐 = 𝒎 ∴ 𝒌 = 𝒎𝒗𝟐 𝒓 𝒓 𝒓 times both sides by (r2 and m) to obtain the form of angular momentum (L= mvr) 𝒆𝟐 𝟐 𝟐 𝟐 𝟐 𝟐 𝟐 𝒉 ∴ 𝒌 𝒓 𝒎 = 𝒎 𝒗 𝒓 = 𝒎𝒗𝒓 = 𝑳 ∵𝑳=𝒏 𝒓 𝟐𝝅 𝟐 𝒉𝟐 𝟏 𝟐 𝟐 𝟐 𝒉 ∴ 𝒓 = 𝒏𝟐 × 𝟐 𝒉 𝟏 ∴ 𝒌𝒎𝒆 𝒓𝒏 = 𝒏 𝟐 𝒏 𝟒𝝅 𝟐 𝒌𝒎𝒆 𝟐 ∴ 𝒓𝒏 = 𝒏 × 𝟒𝝅 𝟒𝝅 𝟐 𝟏 𝒎𝒆𝟐 𝟐 𝟐 𝟒𝝅𝜺𝒐 𝟐 𝒉 𝟒𝝅𝜺𝒐 𝟐 𝒉 𝜺𝒐 ∴ 𝒓𝒏 = 𝒏 × ∴ 𝒓𝒏 = 𝒏 𝟒𝝅 𝟐 𝒎𝒆 𝟐 𝝅𝒎𝒆𝟐 e = 1.6x10-19 C 𝒉𝟐 𝜺𝒐 m =9.11x10-31 kg, ∴ 𝒓𝟏 = 𝟏 × = 𝟓𝟐. 𝟗𝟐 × 𝟏𝟎−𝟏𝟐 𝒎 ∴ 𝒓𝑩 = 𝟓𝟐. 𝟗𝟐 𝒑𝒎 𝝅𝒎𝒆 𝟐 o= 8.85x10-12 c2/Nm2 ∴ 𝒓𝒏 = 𝟓𝟐. 𝟗𝟐 × 𝒏𝟐 𝒑 𝒎 h = 6.63x10-34 J.s ∴ 𝒓𝒏 = 𝒓𝑩 𝒏𝟐 Derive The Energy of orbital (En) of Hydrogen Atom 𝟐 𝟐 𝟐 Use (Fe=Fc ) 𝒆𝟐 𝒗𝟐 𝟏 𝒆 𝟏 𝒆 𝒆 ∴ 𝒌 𝟐 𝒓 = 𝒎 𝒓 ∴ 𝑲. 𝑬 = 𝒎𝒗𝟐 = 𝒌 = = 𝟐 𝟐𝒓 𝟒𝝅𝜺𝒐 𝟐𝒓 𝟖𝝅𝜺𝒐 𝒓 𝟏 𝒆 −𝒆𝟐 𝑷. 𝑬 = 𝒒 × 𝑽 = ሺ−𝒆ሻ × 𝑽 = −𝒆 × = 𝟒𝝅𝜺𝒐 𝒓 𝟒𝝅𝜺𝒐 𝒓 𝟏 The total energy of electron is sum of kinetic energy (𝑲. 𝑬 = 𝒎𝝊𝟐) and potential energy(P.E) 𝟐 𝟐 𝟐 𝟐 𝒆 𝒆 −𝒆 𝑻. 𝑬. = 𝑲𝑬 + 𝑷𝑬 = − = 𝟖𝝅𝜺𝒐 𝒓 𝟒𝝅𝜺𝒐 𝒓 𝟖𝝅𝜺𝒐 𝒓 𝒉 𝟐𝜺 𝒐 Substituting by rn in E ∴ 𝒓𝒏 = 𝒏𝟐 𝝅𝒎𝒆𝟐 𝟐 −𝒆𝟐 −𝒆𝟐 −𝒆𝟒 𝒎 −𝒆𝟒 𝒎 𝟏 𝑬𝒏 = = 𝟐 = 𝟐 = 𝟐 𝟐 𝟐 = −𝟏𝟑. 𝟔 𝒆𝑽. 𝟖𝝅𝜺𝒐 𝒓𝒏 𝒉 𝜺𝒐 𝟐 𝒉 𝜺𝒐 𝟐 𝟖𝜺𝒐 𝒉 𝒏 𝒏 𝟖𝝅𝜺𝒐 𝒏 𝟖𝜺𝒐 𝒏 𝝅𝒎𝒆𝟐 𝟏 -19 e = 1.6x10 C 𝟐 m =9.11x10-31 kg, 𝟏 ∴ 𝑬𝒏 = −𝟏𝟑. 𝟔 𝒆. 𝑽. o= 8.85x10-12 c2/Nm2 𝒏 h = 6.63x10-34 J.s The Ionization Energy Its energy need to remove the electron from its orbit (energy level) 𝟏 𝟐 ∴ 𝑰𝒏 = −𝑬𝒏 = 𝟏𝟑. 𝟔 𝒆𝑽. 𝒏 Spectral series of Hydrogen (The Hydrogen Spectrum) ❑ The spectral lines of hydrogen correspond to jumps of the electron between energy levels. ❑ When an electron jumps from a higher energy to a lower, a photon of certain wavelength is emitted. 𝒉𝒄 𝟏𝟐𝟒𝟎 𝜟𝑬 = 𝑬𝒇 − 𝑬𝒊 = 𝒉𝒇 = = 𝒆𝑽 𝝀 𝝀 𝒏𝒎 Where (f) is final orbit and (i) initial orbit  How much work must be done by an external agent to pull apart a hydrogen atom if the electron initially is (a) in the ground state, and (b) in the state with 𝐧 = 𝟐. 𝟏𝟑. 𝟔 𝐈𝐧 = −𝐄𝐧 = 𝟐 𝐞𝐕 𝐧 (a) 𝟏𝟑. 𝟔 𝐈ሺ𝐧=𝟏ሻ = 𝟐 𝐞𝐕 = 𝟏𝟑. 𝟔 𝐞𝐕 𝟏 (b) 𝟏𝟑. 𝟔 𝐈ሺ𝐧=𝟐ሻ = 𝟐 𝐞𝐕 = 𝟑. 𝟒 𝐞𝐕 𝟐 Rydberg formula ❑ The energy differences between energy levels in the Bohr model, and hence the wavelengths of emitted or absorbed photons, is calculated by the Rydberg law 𝟏 𝟏 𝟏 =𝑹 𝟐− 𝟐 R = 1.0973732×107 m-1 𝝀 𝒇 𝒊 = 0.01097 (nm)-1 Derive the Rydberg formula 𝒉𝒄 𝟏𝟐𝟒𝟎 ∵ 𝜟𝑬 = 𝑬𝒇 − 𝑬𝒊 = 𝒉𝒇 = = 𝝀 𝝀 ሺ𝒏𝒎ሻ Use the energy formula of Bohr model 𝒉𝒄 𝒎𝒆𝟒 𝟏 𝟏 ∴ = 𝟐 𝟐 𝟐 − 𝟐 Where 𝑬𝒊 is the initial energy level, 𝑬𝒇 is 𝝀 𝟖𝜺𝟎 𝒉 𝒇 𝒊 the final energy level, and R is the 𝟏 𝒎𝒆𝟒 𝟏 𝟏 𝟏 𝟏 Rydberg constant. ∴ = 𝟐 𝟑 𝟐 − 𝟐 =𝑹× 𝟐− 𝟐 𝝀 𝟖𝜺𝟎 𝒉 𝒄 𝒇 𝒊 𝒇 𝒊 The spectral series Outer orbit 𝐄∞ = 𝟎. 𝟎 n=∞ n=6 Exited states 𝐄𝟓 = −𝟎. 𝟓𝟒 𝐞𝐕 n=5 Brackett series, IR 𝐄𝟒 = −𝟎. 𝟖𝟓 𝐞𝐕 n=4 Paschen series,IR 2nd excited 𝐄𝟑 = −𝟏. 𝟓𝟏 𝐞𝐕 state n=3 Balmer series, visible light 1st excited 𝐄𝟐 = −𝟑. 𝟒𝟎 𝐞𝐕 n=2 state 𝐄𝟏 = −𝟏𝟑. 𝟔 𝐞𝐕 Lyman series, UV Ground state 𝐧=𝟏  Lyman series 1. Final (lower) energy level =1 2. Initial (upper) energy level =2, 3, 4….. 3. Their wavelengths find in ultraviolet region (UV) of EMW spectrum 4. Longest wavelength (i.e least energy) equivalent to transition between E2 to E1. 5. Shortest wavelength (i.e most energy) equivalent to transition between 𝑬∞ to E1. Balmer series  1. Final (lower) energy level =2 2. Initial (upper) energy level = 3, 4…. 3. Their wavelengths find in visible light region of EMW spectrum 4. Longest wavelength (i.e least energy) equivalent to transition between E3 to E2 5. Shortest wavelength (i.e most energy) equivalent to transition between 𝑬∞ to E2.   (a) What is the wavelength of the least energetic photon in the Balmer series? (b) what is the wavelength of the series limit for the Balmer series? Outer orbit E∞ = 0.0 n=∞ 𝟏 𝟏 𝟏 =𝐑 𝟐− 𝟐 n=6 𝛌 𝒇 𝒊 n=5 E4 = −0.85 eV n=4 𝟏 𝟏 𝟏 2nd excited 𝐚 = 𝟎. 𝟎𝟏𝟎𝟗𝟕 𝟐 − 𝟐 = 𝟔𝟓𝟔. 𝟑 𝐧𝐦 n=3 𝛌𝟑⟶𝟐 𝟐 𝟑 E3 = −1.51 eV state 1st excited E2 = −3.40 eV state n=2 𝟏 𝟏 𝟏 𝐛 = 𝟎. 𝟎𝟏𝟎𝟗𝟕 𝟐 − 𝟐 = 𝟑𝟔𝟒. 𝟓 𝐧𝐦 𝛌∞⟶𝟐 𝟐 ∞ Balmer series, visible light  From the energy level diagram for hydrogen, explain the observation that the frequency of the second Lyman series line is the sum of the frequencies of the first Lyman series line and the first Balmer series line. 𝑪 𝟏 𝟏 Answer ∵ 𝒇 = = 𝑪𝑹 𝟐 − 𝟐 𝝀 𝒇 𝒊 ∴ 𝒇𝟏𝒔𝒕 𝒍𝒚𝒎𝒂𝒏 + 𝒇𝟏𝒔 𝒕𝑩𝒂𝒍𝒎𝒆𝒓 = 𝒇 𝟐𝒏𝒅 𝑳𝒚𝒎𝒂𝒏 𝟏 𝟏 𝟑 𝒇𝟏𝒔𝒕 𝒍𝒚𝒎𝒂𝒏 = 𝑪𝑹 𝟐 − 𝟐 = 𝑪𝑹 Outer orbit 𝟏 𝟐 𝟒 n=∞ n=7 𝟏 𝟏 𝟖 n=6 𝒇 𝟐𝒏𝒅 𝒍𝒚𝒎𝒂𝒏 = 𝑪𝑹 𝟐 − 𝟐 = 𝑪𝑹 n=4 𝟏 𝟑 𝟗 𝐧=𝟑 𝟏 𝟏 𝟓 Balmer series 𝒇𝟏𝒔𝒕 𝑩𝒂𝒍𝒎𝒆𝒓 = 𝑪𝑹 𝟐 − 𝟐 = 𝑪𝑹 𝐧=𝟐 𝟐 𝟑 𝟑𝟔 𝟑 𝟓 𝟖 Lyman series ∴ 𝒇𝟏𝒔𝒕 𝒍𝒚𝒎𝒂𝒏 + 𝒇𝟏𝒔 𝒕𝑩𝒂𝒍𝒎𝒆𝒓 = 𝑪𝑹 + = 𝑪𝑹 𝐧=𝟏 𝟒 𝟑𝟔 𝟗  5- What are (a) the energy, (b) momentum, and (c) the wavelength of the photon that is emitted when a hydrogen atom undergoes a transition from the state n = 3 to n = 1? Answer 𝟏𝟑. 𝟔 n=∞ ∵ 𝑬 𝒏 = − 𝟐 𝒆𝑽 n=7 𝒏 n=6 n=4 ∵ 𝛥𝐸 = ℎ𝑓 = 𝐸𝑓 − 𝐸𝑖 = 𝐸1 − 𝐸3 𝐧=𝟑 𝐧=𝟐 𝟏𝟑. 𝟔 𝟏𝟑. 𝟔 𝛥𝐸 = − 𝟐 − − 𝟐 𝒆𝑽 = 𝟏𝟐. 𝟎𝟗 𝒆𝑽 Lyman series ሺ𝑎ሻ 𝟏 𝟑 𝐧=𝟏 𝜟𝑬 𝟏𝟐. 𝟎𝟗 ሺ𝒃ሻ ∵ 𝑷 = = = 𝟒. 𝟎𝟑 × 𝟏𝟎 −𝟖 𝒆𝑽. 𝒔/ 𝒎 𝑪 𝟖 𝟑 × 𝟏𝟎 𝒎/𝒔 ሺ𝒄ሻ 𝒉𝑪 𝟏𝟐𝟒𝟎 ∵𝝀= = 𝒏𝒎 = 𝟏𝟎𝟐. 𝟓𝟔𝒏𝒎 𝜟𝑬 𝟏𝟐. 𝟎𝟗  6- Using Bohr’s formula to calculate the three longest wavelengths of Balmer series. Answer 𝟏𝟑. 𝟔 Outer orbit E∞ = 0.0 n=∞ ∵ 𝑬 𝒏 = − 𝟐 𝒆𝑽 𝒏 n=6 𝒉𝑪 𝟏𝟐𝟒𝟎 n=5 ∵ 𝜟𝑬 = = 𝒆𝑽 E4 = −0.85 eV n=4 𝝀 𝝀 2nd excited E3 = −1.51 eV state n=3 𝒉𝑪 𝟏𝟐𝟒𝟎 𝝀𝟏 = = 𝒏𝒎 = 𝟔𝟓𝟔. 𝟎𝟖𝒏𝒎 𝑬𝟑 − 𝑬𝟐 −𝟏. 𝟓𝟏 + 𝟑. 𝟒 1st excited E2 = −3.40 eV state n=2 𝒉𝑪 𝟏𝟐𝟒𝟎 Balmer series, visible light 𝝀𝟐 = = 𝒏𝒎 = 𝟒𝟖𝟔. 𝟐𝟕𝒏𝒎 𝑬𝟒 − 𝑬𝟐 −𝟎. 𝟖𝟓 + 𝟑. 𝟒 𝒉𝑪 𝟏𝟐𝟒𝟎 𝝀𝟑 = = 𝒏𝒎 = 𝟒𝟑𝟒. 𝟏𝟕𝒏𝒎 𝑬𝟓 − 𝑬𝟐 −𝟎. 𝟓𝟒𝟒 + 𝟑. 𝟒  7- A hydrogen atom is excited from a state with n = 1 to one with n = 4. (a) Calculate the energy that must be absorbed by the atom, (b) Calculate and display on an energy - level diagram the different photon energies that may be emitted if the atom returns to the n = 1 state. Answer 𝟏𝟑. 𝟔 ∵ 𝑬 𝒏 = − 𝟐 𝒆𝑽 𝜟𝑬 = 𝑬𝒇 − 𝑬𝒊 𝒏 a) 𝜟𝑬𝟒𝟏 = 𝑬𝟒 − 𝑬𝟏 = −𝟎. 𝟖𝟓𝒆𝒗 + 𝟏𝟑. 𝟔𝒆𝒗 = 𝟏𝟐. 𝟕𝟓 𝒆𝑽 n=∞ b) n=7 n=6 𝜟𝑬 = 𝑬𝟒 − 𝑬𝟏 = ሺ−𝟎. 𝟖𝟓ሻ𝒆𝒗 − ሺ−𝟏𝟑. 𝟔ሻ𝒆𝒗 = 𝟏𝟐. 𝟕𝟓𝒆𝒗 n=4 𝐧=𝟑 𝜟𝑬𝟑𝟏 = 𝑬𝟑 − 𝑬𝟏 = −𝟏. 𝟓𝟏𝒆𝒗 + 𝟏𝟑. 𝟔𝒆𝒗 = 𝟏𝟐. 𝟎𝟗 𝒆𝑽 𝐧=𝟐 𝜟𝑬𝟐𝟏 = 𝑬𝟐 − 𝑬𝟏 = −𝟑. 𝟒𝒆𝒗 + 𝟏𝟑. 𝟔𝒆𝒗 = 𝟏𝟎. 𝟐𝒆𝑽 𝐧=𝟏 Lyman series 9- Light of wavelength 486.1 nm is emitted by a hydrogen atom (a) what transition of the atom is responsible for this radiation? (b) To what series does this radiation belong? Answer 𝟏𝟑. 𝟔 ∵ 𝑬 𝒏 = − 𝟐 𝒆𝑽 𝒏 𝟏𝟑. 𝟔 𝟏𝟑. 𝟔 ∵ 𝑬 𝟐 = − 𝟐 = −𝟑. 𝟒 𝒆𝑽 ∵ 𝑬 𝟑 = − 𝟐 = −𝟏. 𝟓𝟏 𝒆𝑽 𝟐 𝟑 𝟏𝟑. 𝟔 ∵ 𝑬 𝟒 = − 𝟐 = −𝟎. 𝟖𝟓 𝒆𝑽 𝟒 𝟏𝟐𝟒𝟎 𝟏𝟐𝟒𝟎 𝜟𝑬 = = = 𝟐. 𝟓𝟓𝒆𝑽 𝝀 𝟒𝟖𝟔. 𝟏 𝜟𝑬𝟒→𝟐 = 𝑬𝟒 − 𝑬𝟐 = −𝟎. 𝟖𝟓 − ሺ−𝟑. 𝟒ሻ 𝜟𝑬𝟒→𝟐 = 𝟐. 𝟓𝟓𝒆𝑽 i= 4 and f=2 (Balmer series)  𝑯𝟏 Hydrogenic Atom An atom lost all electrons by ionization except one electron ❑ These atoms behave like hydrogen except that the nucleus has a positive charge of Ze, where Z is the atomic number of the atom ❑ Examples of hydrogenic atoms are Z=1 𝑯𝒆+𝟏 𝟐 𝑳𝒊𝟑+𝟐 Z=2 Z=3 ❑ The nucleus have a charge Q = Ze where Z is the atomic number.  Rules of Hydrogen atom Rules of Hydrogenic atom atomic number (Z) of hydrogen By replacing each (e2) in hydrogen atom is one (Z=1) atom formula by (Ze2) 1- Energy of orbits (n) 1- Energy of orbits (n) −𝒆𝟒 𝒎 𝟏 𝟐 𝟏 𝟐 ∴ 𝑬𝒏 = = −𝟏𝟑. 𝟔 𝒆𝑽 −𝒆𝟒 𝒎 𝒁𝟐 𝟏 𝟐 𝒁 𝟐 𝟖𝜺𝟐𝒐 𝒉𝟐 𝒏 𝒏 ∴ 𝑬𝒏 = = −𝟏𝟑. 𝟔 𝒆𝑽 𝟖𝜺𝟐𝒐 𝒉𝟐 𝒏 𝒏 2- Radius of orbits 2- Radius of orbits 𝒉𝟐 𝜺𝒐 𝟐 ∴ 𝒓𝒏 = 𝒏 = 𝒓𝑩 × 𝒏𝟐 𝒉𝟐 𝜺𝒐 𝒏𝟐 𝒏𝟐 𝝅𝒎𝒆𝟐 ∴ 𝒓𝒏 = 𝟐 × = 𝒓𝑩 × 3- The Rydberg formula 𝝅𝒎𝒆 𝒁 𝒁 3- The Rydberg formula 𝟏 𝒎𝒆𝟒 𝟏 𝟏 𝟏 𝒁𝟐 𝒎𝒆𝟒 𝟏 𝟏 = 𝟐 𝟑 𝟐 − 𝟐 = 𝟐 𝟑 − 𝟐 𝝀 𝟖𝜺 𝒉 𝒄 𝒇 𝒊 𝝀 𝟖𝜺 𝒉 𝒄 𝒇 𝟐 𝒊 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 =𝑹 𝟐− 𝟐 = 𝑹𝒁 𝟐 − 𝟐 𝝀 𝒇 𝒊 𝝀 𝒇 𝟐 𝒊 Problem Calculate the radius and energy of the 2nd orbit of 3L2+ ? What is the longest wavelength in Balmer series of this ion? N= 2 and Z= 3 Solution 𝒏𝟐 𝟐𝟐 𝟒 ∴ 𝒓𝒏 = 𝒓𝑩 × ∴ 𝒓𝟐 = 𝒓𝑩 × = 𝟓𝟐. 𝟗𝟐 × = 𝒑𝒎 𝒁 𝟑 𝟑 𝒁 𝟐 𝟑 𝟐 ∴ 𝑬𝒏 = −𝟏𝟑. 𝟔 𝒆𝑽 ∴ 𝑬𝟐 = −𝟏𝟑. 𝟔 = 𝒆𝑽 𝒏 𝟐 The longest wavelength in Balmer series produce due to transition from n=3 to n=2 𝟏 𝟐 𝟏 𝟏 𝟏 𝟏 𝟏 ∴ = 𝑹𝒁 𝟐 − 𝟐 ∴ = 0.01097 × 𝟑𝟐 − 𝟐 𝒏𝒎 −𝟏 𝝀 𝒇 𝒊 𝝀 𝟐 𝟐 𝟑 𝝀= 𝒏𝒎 Problem (i) Write the equation of radius of hydrogenic atom 2He+ (ii) Calculate the energy of the least energetic photon in the Lyman series of this ion? N= 2 and Z= 3 Solution 𝒏𝟐 𝒏𝟐 ∴ 𝒓𝒏 = 𝒓𝑩 × ∴ 𝒓𝟐 = 𝒓𝑩 × 𝒁 𝟐 𝒁 𝟐 ∴ 𝑬𝒏 = −𝟏𝟑. 𝟔 𝒆𝑽 𝒏 The least energetic photon in the Lyman series corresponding to transition between n= 2 and n= 1 𝟐 𝟐 𝟐 𝟐 ∴ 𝑬𝟏 = −𝟏𝟑. 𝟔 = −𝟓𝟒. 𝟒 𝒆𝑽 ∴ 𝑬𝟐 = −𝟏𝟑. 𝟔 = −𝟏𝟑. 𝟔 𝒆𝑽 𝟏 𝟐 𝜟𝑬𝟐→𝟏 = 𝑬𝟏 − 𝑬𝟐 = −𝟓𝟒. 𝟒 − −𝟏𝟑. 𝟔 = − 40.8 eV The negative sign means this energy lost from 2He+ ion Essay Questions 1- State Bohr postulates for the Hydrogen atom. 2- Using the semi – classical derivation to calculate: (i) the radii of the allowed orbits of the hydrogen atom, and (ii) the energies of the stationary states of the hydrogen atom. 3- Define the ionization energy and calculate it for the hydrogen atom. 4- Derive a relation for the wavelength of the emitted or absorbed radiation during a transition from one level to another in the hydrogen atom. 5- Draw the energy level diagram of the hydrogen atom. 6- State the spectral series of hydrogen atom and specify the quantum numbers of the upper and the lower energy states for each series. 7- Compare between hydrogen atom and hydrogenic atoms.