2.3 Solutions
107
mv
2
/r = kr
(2)
solving (1) and (2)
v
2
=
n
m
k
m
1
2
(3)
r
2
=
n
√
km
(4)
(c) E = U + T =
1
2
kr
2
+
1
2
mv
2
(5)
Substituting (3) and (4) on (5) and simplifying
E = n(k/m)
1/2
(d) ΔE = E n − E n−1 =
k
m
1
2 = 1.05 × 10
−34
1,180
3 × 10 −26
1
2
J
= 0.13 eV
λ =
1,241
0.13
= 9,546 nm
2.17 E n = −
13.6
n 2
ΔE = E n+1 − E n = 13.6
1
n 2 −
1
(n + 1) 2
=
13.6(2n + 1)
n 2 (n + 1) 2
In the limit n → ∞, ΔE ∝
n
n 4 =
1
n 3
2.18 The wavelength λ = 486.1 nm corresponds to the transition energy of E =
1241/486.1 = 2.55 eV Looking up Fig. 2.1, for the energy level diagram
for hydrogen atom, the transition n = 4 → 2 gives the energy difference
−0.85 − (−3.4) = 2.55 eV
The line belongs to the Balmer series.
Fig. 2.1 Energy level
diagram for hydrogen atom
2.19 Orbital velocity, v =
e
2
2nhε 0
, a 0 = n
2 h
2
ε 0 /π e
2 m
Orbital frequency f = v/2πa 0