Engineering Chemistry S Datta

ATOMS AND MOLECULES 7
∴ a sin kl + b cos kl = 0
It can only be possible when,
kl = nπ or k = n π
l
where x is called quantum number and is equal to 1, 2, 3 ... ∞.
Substituting this value in eqn. 4 and also putting V = 0; we get,
n
2 2
l
π
2
2
8π
m
= E
h
So, (E, the total energy of the electron) = K.E. of the electron.
2 2
∴ E n
= nh where n = 1, 2, 3 ... ∞.
2
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The above equation means that the particle in a box does not possess any arbitrary
amount of energy, rather it possesses discrete set of energy values i.e., its energy is quantised.
A few energy levels are given below:
E 1
=
2
2
h
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2
2
E 2
= 4 h
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E 3
= 9 h
2
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The reason why a particle in a box i.e., bound particle possesses
a quantised energy whereas a free particle does not, can easily be
deduced from the above equation.
2
Highlight: Calculation of minimum energy of a particle in a box from Heisenberg’s
uncertainty principle.
Here ∆ x = l, and the particle bounces back in the box. So, ∆p = 2p because the momentum
changes from + p to – p continuously. So we can write,
n=3
n=2
n=1
E 3
E 2
E 1
x=0 x=L 0
Fig. 1.5
Energy
∆ x × ∆p = l . (2p) = h or, p = h 2l
2 2
Now, E = 1 2 mv2 = ( mv)
p
=
2m
2m
∴ E =
2
h
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2
.
PROBABILITY DISTRIBUTION
The directional properties of an election in an orbital of the hydrogen atom cannot be represented
in one diagram. Two separate diagrams are required to meet the demand.