Answer:
Explanation:
Utilizing Rydber's equation:
ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have
n=1 to n= infinity
ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)
n= 3 to n= infinity
ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J
b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .
1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹
1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m
λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm
Answer:
(a) n=1 → E= -1.96x10⁻¹⁷ J
n=3 → E= -2.18x10⁻¹⁸ J
(b) λ = 11.4 nm
Explanation:
(a) To find the energies needed to remove an electron from the n=1 state and n=3 state in the Li²⁺ ion, we need to use the Rydberg equation for energy change:
[tex] \Delta E = R \cdot Z^{2} \cdot h \cdot c (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}) [/tex]
where ΔE: is the energy change, R: is the Rydberg constant, Z: is the atomic number, h: is the Planck constant, c: is the speed of light, nf: is the final state of the electron and ni: is the initial state of the electron.
To remove an electron from the state n=1 means that ni=1 and nf=∞, so the energy is:
[tex] \Delta E = (1.097\cdto 10^{7} \frac{1}{m})(3^{2})(6.62\cdot 10^{-34} J.s)(3.00\cdot 10^{8}\frac{m}{s})(\frac{1}{\infty} - \frac{1}{1^{2}}) = -1.96 \cdot 10^{-17}J [/tex]
Similarly, to remove an electron from the state n=3 means that ni=3 and nf=∞, hence the energy is:
[tex] \Delta E = (1.097\cdto 10^{7} \frac{1}{m})(3^{2})(6.62\cdot 10^{-34} J.s)(3.00\cdot 10^{8}\frac{m}{s})(\frac{1}{\infty} - \frac{1}{3^{2}}) = -2.18 \cdot 10^{-18}J [/tex]
(b) To calculate the wavelength of the emitted photon in a transition from n=3 to n=1 we can use the Rydberg equation:
[tex] \frac{1}{\lambda} = R \cdot Z^{2} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}) [/tex]
The transition is ni=3 and nf=1, therefore the wavelength of the emitted photon is:
[tex] \frac{1}{\lambda} = (1.097 \cdot 10^{7} \frac{1}{m})(3^{2})(\frac{1}{1^{2}} - \frac{1}{3^{2}}) = 8.78 \cdot 10^{7} m^{-1} [/tex]
[tex] \lambda = 1.14 \cdot 10^{-8} m = 11.4 nm [/tex]
I hope it helps you!
