1.
2. For hydrogen atom transitions where both quantum number states are real, calculate E, v, ?;
Initial Final Exists? Yes E (kJ) or No V (Hz) A (m)
Solution
a) This does not exist, because the only 2 allowed states of m s are +1/2 and -1/2. But this has a
value of -3/2 and hence does not exist.
b) n=4, l=0, m l = 0, ms = 1/2 .
n=4 is the energy level, l=0 and m l = 0 corresponds to s orbital and ms = 1/2 indicates a positive
spin.
This indicates that the electron is 4s 1
with a positive spin.
An electron from 4s 1
can go from n=4 to n=3 level, we can calculate the value of ?, the
wavelength using the following formula:
1/? =R?(1/ n f
2
?1/n i
2
),
Where,
? is the wavelength of the emitted photon;
R - Rydberg's constant = 1.0974 x 10 7
m ?1
;
n f - the final energy level - in your case equal to 3;
n i - the initial energy level - in your case equal to 4.
1/? = 1.0974 x 10 7
(1/ 3 2
?1/4 2
)
1/? = 1.0974 x 10 7
(1/ 9?1/16)
1/? = 1.0974 x 10 7
(7/144) = 5.334 x 10 5
m -1
=
? = 1/5.334 x 10 5
m -1
? = 1.87 x 10 -6 m

2.
Since E=hc/?, to calculate for the energy of this transition we have to multiply Rydberg's
equation by hc,
where,
h = 6.626 x 10 -34
Joule?s (planks constant)
c - the speed of light - 2.99 x 10 8
m/s
E = hc/?
E = (6.626 x 10 -34
Joule?s x 2.99 x 10 8
m/s)/ 1.87 x 10 -6
m = 10568737.8 x 10 -26
= 1.06 x 10 -19
J = 1.06 x 10 -19
x 10 -3
KJ
E = 1.06 x 10 -22 KJ
v= c/? = 2.99 x 10 8
ms -1
/ 1.87 x 10 -6
m = 1.60 x 10 14
s -1
v = 1.60 x 10 14 Hz