104
2 Quantum Mechanics – I
2.9 In atomic physics the atomic units are as follows:
(i) (a) The Bohr radius
2
/m e e
2 is used as the unit of length. (b) The energy
is measured in multiples of the ionization energy of hydrogen m e e
2
/2
2
(c)
2
= 1 (d) e
2
= 2
(ii) In atomic units the Schrodinger equation
−
2
2m e
∇
2 u −
e
2 u
r
= Eu
would read as
−∇
2 u −
2u
r
= Eu
2.10 (a) Let the separation between the two nuclei each of charge q be 2d, then the
negative charge Q on the electrons is at a distance d from either nuclei
Total potential energy due to electrostatic interaction between three objects
is
q Q
d
+
q Q
d
+
q
2
2d
≤ 0
Taking the equality sign and cancelling q
Q = −
q
4
(b) Let T 0 be the initial kinetic energy and p 0 the momentum of the ion and T
the kinetic energy and p the momentum of the composite molecule and Q
the excitation energy.
T 0 = T + Q
Energy conservation
(1)
p 0 = p
Momentum conservation
(2)
∴ (2mT 0 )
1/2
= (2.2mT )
1/2
(3)
The mass of the composite being 2m as the excitation energy is expected
to be negligible in comparison with the mass of the molecule. From (3) we
get
T =
T 0
2
(4)
Using (4) in (1), we find Q =
T 0
2
=
10
−19
2
= 5 × 10
−20 J
2.11(a) Stationary orbits will be such that the circumference of a circular orbit is
equal to an integral number of deBroglie wavelength so that constructive
interference may take place i.e. 2πr = nλ
But λ = h/ p
∴ L = r p = nh/2π (Bohr’s quantization condition)
(1)
2 Quantum Mechanics – I
2.9 In atomic physics the atomic units are as follows:
(i) (a) The Bohr radius
2
/m e e
2 is used as the unit of length. (b) The energy
is measured in multiples of the ionization energy of hydrogen m e e
2
/2
2
(c)
2
= 1 (d) e
2
= 2
(ii) In atomic units the Schrodinger equation
−
2
2m e
∇
2 u −
e
2 u
r
= Eu
would read as
−∇
2 u −
2u
r
= Eu
2.10 (a) Let the separation between the two nuclei each of charge q be 2d, then the
negative charge Q on the electrons is at a distance d from either nuclei
Total potential energy due to electrostatic interaction between three objects
is
q Q
d
+
q Q
d
+
q
2
2d
≤ 0
Taking the equality sign and cancelling q
Q = −
q
4
(b) Let T 0 be the initial kinetic energy and p 0 the momentum of the ion and T
the kinetic energy and p the momentum of the composite molecule and Q
the excitation energy.
T 0 = T + Q
Energy conservation
(1)
p 0 = p
Momentum conservation
(2)
∴ (2mT 0 )
1/2
= (2.2mT )
1/2
(3)
The mass of the composite being 2m as the excitation energy is expected
to be negligible in comparison with the mass of the molecule. From (3) we
get
T =
T 0
2
(4)
Using (4) in (1), we find Q =
T 0
2
=
10
−19
2
= 5 × 10
−20 J
2.11(a) Stationary orbits will be such that the circumference of a circular orbit is
equal to an integral number of deBroglie wavelength so that constructive
interference may take place i.e. 2πr = nλ
But λ = h/ p
∴ L = r p = nh/2π (Bohr’s quantization condition)
(1)
