Question

Be sure to answer all parts. Solving the Rydberg equation for energy change gives ΔE = R[infinity]hc [ 1 n12 − 1 n22 ] where the Rydberg constant R[infinity] for hydrogen-like atoms is 1.097 × 107 m−1 Z2, and Z is the atomic number.(a) Calculate the energies needed to remove an electron from the n = 1 state and the n = 3 state in the Li2+ ion. n = 1 × 10 J n = 3 × 10 J (Enter your answer in scientific notation.(b) What is the wavelength (in nm) of the emitted photon in a transition from n = 3 to n = 1?

Answer 1

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 2

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:  

$\Delta E = R \cdot Z^{2} \cdot h \cdot c (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$

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:

$\Delta E = (1.097\cdto 10^{7} \frac{1}{m})(3^{2})(6.62\cdot 10^{-34} J.s)(3.00\cdot 10^{8}\frac{m}{s})(\frac{1}{\infty} - \frac{1}{1^{2}}) = -1.96 \cdot 10^{-17}J$

Similarly, to remove an electron from the state n=3 means that ni=3 and nf=∞, hence the energy is:

$\Delta E = (1.097\cdto 10^{7} \frac{1}{m})(3^{2})(6.62\cdot 10^{-34} J.s)(3.00\cdot 10^{8}\frac{m}{s})(\frac{1}{\infty} - \frac{1}{3^{2}}) = -2.18 \cdot 10^{-18}J$

(b) To calculate the wavelength of the emitted photon in a transition from n=3 to n=1 we can use the Rydberg equation:

$\frac{1}{\lambda} = R \cdot Z^{2} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$

The transition is ni=3 and nf=1, therefore the wavelength of the emitted photon is:

$\frac{1}{\lambda} = (1.097 \cdot 10^{7} \frac{1}{m})(3^{2})(\frac{1}{1^{2}} - \frac{1}{3^{2}}) = 8.78 \cdot 10^{7} m^{-1}$

$\lambda = 1.14 \cdot 10^{-8} m = 11.4 nm$

I hope it helps you!