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?

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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.