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Atomic Structure Practice Problems:

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, ml, 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

ml = -3, -2, -1, 0, 1, 2, 3

7 orbitals (count the ml)

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)

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