**Atomic Structure Practice Problems**:**1.)**
Calculate the frquency of red light of wavelength 6.50 * 10^{2} 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 Li^{2+} 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 * 10^{8} m/s

l = 6.50 * 10^{2} nm * (1m/10^{9} nm) = 6.50
* 10^{-9} m

(2.9979 * 10^{8} m/s) / ( 6.50 * 10^{-9} m ) = 4.61 * 10^{14}
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 (Z^{2}/n^{2})

E1 ==> n = 3 ===> -2.178 * 10^{-18} (1^{2}/3^{2}) = -2.42
* 10^{-19}

E2 ==> n = 1 ===> -2.178 * 10^{-18} (1^{2}/1^{2}) = -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]5s^{2}

**b.) **[Xe] 6s^{2}4f^{14}5d^{8}

**c.) **[Ne] 3s^{2}3p^{4}

**6.) **It is important to note that this is Li^{2+}, not Li^{2+}
has an electron configuration of 1s^{1}

E = -(Z^{2} / n^{2}) * 1310 kJ/mol = -( 3^{2} / 1^{2} )
* 1310 kJ/mol = 1.18 * 10^{2} kJ

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

l = *h* / (9.11 * 10^{-31})(2.0 * 10^{8})