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)
