1.) Calculate the frquency of red light of wavelength 6.50 * 10² nm

2.) 6.42 * 10-19 J of energy is required to remove an electron from a silver atom. What is the maximum wavelength of light that can do this?

3.) Calculate the DE when the following transition occurs in a hydrogen atom:

n = 3 ---> n = 1

3b.) Calculate the wave length of light emitted as a hydrogen electron move from n = 1 to n = 3

4.) Give the quantum numbers l, m_l, and the number of orbitals for n = 4

5.) Give the electron configurations for:

a.) Sr   b.) Pt   c.) S

6.) Calculate the ionization energy of the Li²⁺ ion with the electron in its ground state.

7.) Find the wavelength emitted from an electron (m = 9.11 * 10^-31 kg) traveling at 2.0 * 10^-8 m/s

1.) v = c / l

c = 2.9979 * 10⁸ m/s

l = 6.50 * 10² nm * (1m/10⁹ nm) = 6.50 * 10^-9 m

(2.9979 * 10⁸ m/s) / ( 6.50 * 10^-9 m ) = 4.61 * 10¹⁴ Hz

2.) DE = hv     (h = 6.6326 * 10^-34 J * s)

v = c / l

Replace the v in the first equation with the lower equation:

DE = (hc) / l

Plugging in all the constants (remember, both h and c are constants), and plugging in for DE, the answer is: 3.09 * 10^-7, which converts to 309 nm.

3.) E = -2.178 * 10^-18 J (Z²/n²)

E1 ==> n = 3 ===> -2.178 * 10^-18 (1²/3²) = -2.42 * 10^-19

E2 ==> n = 1 ===> -2.178 * 10^-18 (1²/1²) = -2.18 * 10^-18

DE = E2 - E1 = -2.18 * 10^-18 + 2.42 * 10^-19 = -1.936 * 10^-18 J

3b.) The same process is done, except E1 will have n = 1 and E2 will have n = 3. The final step:

DE = E2 - E1 = -2.42 * 10^-19 + 2.18 * 10^-18= 1.94 * 10^-18 J

You then convert to wavelength using l = hc/DE

l = 103 nm

4.)

l = (n-1) = 4

m_l = -3, -2, -1, 0, 1, 2, 3

7 orbitals (count the m_l)

5.)

a.) [Kr]5s²

b.) [Xe] 6s²4f¹⁴5d⁸

c.) [Ne] 3s²3p⁴

6.) It is important to note that this is Li²⁺, not Li²⁺ has an electron configuration of 1s¹

E = -(Z² / n²) * 1310 kJ/mol = -( 3² / 1² ) * 1310 kJ/mol = 1.18 * 10² kJ

7.) l = h/mv (v is velocity, not frequency) (m in kg)

l = h / (9.11 * 10^-31)(2.0 * 10⁸)