Chemistry (F.Sc.1) Chapter # 05 Numericals- Malik Xufyan

1. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
NUMERICLALS
Q 17. (a) A photon of light with energy 10-19
J is emmited by a source of light.
(b) Convert this energy into wavelength, frequency and wave number of the
Photon in terms of meters, hertz and m-1
respectively.
Solution:
E = 10-19
J
h =6.625 × 10-34
Js
c = 3 × 108
m/s
v = ?
λ = ?
ṽ = ?
Since E = h v
Or v = =
. ×
= 1.509 × 1014
s-1
Since λ = c v =
×
. ×
= 1.988 × 10-6
m
Since ṽ =
λ
=
. ×
= 5.030 × 105
m-1
(b) Convert this energy of photon into ergs and calculate the wavelength in cm,
frequency in Hz and wave number in cm-1
.
h = 6.625 × 10-34
Js c = 3 × 108
m/s
E = 10-19
J = 10-19
× 107
= 10-12
erg
h = 6.625 × 10-34
Js = 6.625 × 10-34
× 107
ergs = 6.625 × 10-27
ergs
c = 3 × 108
m/s = 3 × 1010
cm/s
v = ?
λ = ?
ṽ = ?
v = =
. ×
= 1.509 × 1014
s-1
λ = =
×
. ×
= 1.988 × 10-4
cm
ṽ =
λ
=
. ×
= 5.030 × 103
cm-1
1J = 107
erg
1m = 100 cm
Erg is the unit
Of energy in
c.g.s. system.

2. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
Q 18. The formula for calculating the energy of an electron in hydrogen atom given by
Bohr’s model.
En =
ε
Calculate the energy of electron in first orbit of hydrogen atom.
Solution
ɛ0 = 8.85 × 10-12
C2
J-1
m-1
h = 6.625 × 10 -34
Js
m = 9.1 × 10-31
Kg
e = 1.6022 × 10-19
C
E1 = ?
n = ?
energy in the first orbit is given by
E1 = -
ε
E1 = -
. × . ×
× ( . × ) × × ( . × )
E1 = -2.18 × 10 -18
J
Q 19. Bohr’s equation for the radius of nht orbit of electron in hydrogen atom is
rn =
While doing calculations take care of units of energy parameter.
(a) When an electron moves from n = 1 to n = 2, how much does the radius of the orbit
increases.
Soulution:
ɛ0 = 8.85 × 10-12
C2
J-1
m-1
h = 6.625 × 10 -34
Js
m = 9.1 × 10-31
Kg
e = 1.6022 × 10-19
C
Radius of nth orbit is given by
rn =
rn =
. × ( . × )
. ×( . × )( . × )
× n2
rn = 0.529 × 10-10
m × n2
= 0.529 A0
× n2
thus, for n = 1
r1 = 0.529 × 12
= 0.529 A0

3. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
Thus for n = 2
r2 = 0.529 × 22
= 2.11 A0
Hence increase in radius = r2 – r1 = 2.11 A0
– 0.529 A0
= 1.581 A0
(b) What is the distance travelled by the electron when it goes from n=2 to n=3 and n=9
to n=10?
Solution:
Since rn= 0.529 × 10-10
m × n2
= 0.529 A0
× n2
Thus for n = 2
r2= 0.529 × 22
= 2.11 A0
Thus for n = 3
r3= 0.529 × 32
= 4.75 A0
Hence distance travelled = r3- r2 = 4.75 A0
- 2.11 A0
= 2.65 A0
Also for n = 9
r9 = 0.529 × 92
= 42.849 A0
And for n = 10
r10 = 0.529 × 102
Hence distance travelled = r10 – r9 = 52.9 A0
– 42.849 A0
= 10.05 A0
Q 20. Answer the following questions, by performing the calculations.
(a) Calculate the energy of the first five orbits of hydrogen atom and determine the
energy difference b/w them.
(b) Justify that energy difference b/w second and third orbits is approximately five
times smaller that b/w first and second orbits.
(c) Calculate the energy of electron in He+
in first five orbits and justify that the energy
differences are different from those of hydrogen atom.
(d) Do you think that groups of the spectral lines of He+
are at different places than
those for hydrogen atom? Give reasons.
Solution:
(a) Energy of electron in nth orbit is given by
En = -2.18 × 10-18
×
Where Z = atomic number n = number of orbit
For hydrogen atom Z = 1
Therefore
En = -2.18 × 10-18
×
For n = 1, E1 = -2.18 × 10-18
× = -2.18 × 10-18
J
10-10
m = 1 A0

4. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
For n = 2, E2 = -2.18 × 10-18
× = -5.45 × 10-19
J = 0.545 × 10-18
J
Similarly we found the energy of other orbitals in this way.
Such as;
For n = 3, E3 = -2.18 × 10-18
× = -2.42 × 10-19
J = -0.242 × 10-18
J
For n = 4, E4 = -2.18 × 10-18
× = -1.36 × 10-19
J = -0.136 × 10-18
J
For n = 5, E5 = -2.18 × 10-18
× J = -8.72 × 10-20
J = -0.0872 × 10-18
J
Energy differences will be
E2-E1 = (-0.545 × 10-18
) – (-2.18 × 10-18
) = 1.635 × 10-18
J
E3-E2 = (-0.242 × 10-18
) – (-0.545 × 10-18
) = 0.303 × 10-18
J
E4-E3 = (-0.136 × 10-18
) – (-0.242 × 10-18
) = 0.106 × 10-18
J
E5-E4 = (-0.0872 × 10-18
) – (-0.136 × 10-18
) = 0.0488 × 10-18
J
(b) Justify that energy difference b/w second and third orbits is approximately five
times smaller that b/w first and second orbits.
Energy difference b/w E2-E1 and E3-E2 is given by the ratio of energy difference.
=
. ×
. ×
≈ 5
= (E2-E1) ≈ (E3-E2)
Hence, energy difference b/w E3-E2 is approximately five times smaller than E2-E1.
(c) Calculate the energy of electron in He+
in first five orbits and justify that the energy
differences are different from those of hydrogen atom.
For He+
ion z = 2
Therefore
En = -2.18 × 10-18
× = -2.18 × 10-18
×
Thus
For n = 1, E1 = -2.18 × 10-18
× = -8.72 × 10-18
J
For n = 2, E2 = -2.18 × 10-18
× = -2.18 × 10-18
J
For n = 3, E3 = -2.18 × 10-18
× = -9.68 × 10-19
J = -0.968 × 10-18
J
For n = 4, E4 = -2.18 × 10-18
× = -5.45 × 10-19
J = -0.545 × 10-18
J

5. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
For n = 5, E5 = -2.18 × 10-18
× = -3.488 × 10-19
J = -0.3488 × 10-18
J
Energy differences will be
E2-E1 = (-2.18 × 10-18
) – (-8.72 × 10-18
) = 6.45 × 10-18
J
E3-E2 = (-0.968 × 10-18
) – (-2.18 × 10-18
) = 1.21 × 10-18
J
E4-E3 = (-0.545 × 10-18
) – (-0.968 × 10-18
) = 0.423 × 10-18
J
E5-E4 = (-0.3488 × 10-18
) – (-0.545 × 10-18
) = 0.196 × 10-18
J
is different from hydrogen.
+
Hence difference of energy b/w the energy levels of He
(d) Do you think that groups of the spectral lines of He+
are at different places than
those for hydrogen atom? Give reasons.
Since energy difference b/w energy levels in He+
ion is different from hydrogen, therefore
spectral lines in He+
ions spectrum will be different from that of hydrogen spectrum.
(a) Q 21. Calculate the value of principal quantum number if an electron in hydrogen
atom revolves in an orbit of energy -0.242 × 10-18
J.
Solution:
E = -0.242 × 10-18
J
Z = 1
n = ?
Energy of electron nth orbit is given by
En = -2.18 × 10-18
×
Hence -0.242 × 10-18
= -2.18 × 10-18
×
Or n2
=
. ×
. ×
= 9
n = √9 = 3
Q 22. Bohr’s formula for the energy levels of hydrogen atom for any system say H, He+
,
Li2+
etc is En = -
En = -K
For hydrogen Z = 1and for He+
, Z = 2.
(a) Draw an energy level diagram for hydrogen atom and He+
.
(b) Thinking that K = 2.18 × 10-18
J, calculate the energy needed to remove the electron
from hydrogen atom and from He+
.
Solution:
K = 2.18 × 10-18
J
For hydrogen Z = 1
Energy in nth is given by

6. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
En = -K
For n = 1, En = -2.18 × 10-18
= -2.18 × 10-18
J
For n = ∞, E = -2.18 × 10-18
= 0 J
Hence to move an electron from hydrogen’s first orbit to an infinite distance, the energy
required will be
E − E = 0 – (-2.18 × 10-18
) = 2.18 × 10-18
J
This is the ionization energy of hydrogen atom.
For He+
ion Z = 2
Energy in nth orbit is given by
En = -K
For n = 1, En = -2.18 × 10-18
= -8.72 × 10-18
J
For n = ∞, E = -2.18 × 10-18
= 0 J
Hence to remove an electron from He+
ion’s first orbit to an infinite distance, the energy
required will be
E − E = 0 – (-8.72 × 10-18
) = 8.72 × 10-18
J
This is the ionization energy of He+
ion.
(c) How do you justify that the energies calculated in (b) are the ionization energies of
H and He+
?
The amount of energy required to remove an electron from an atom or ion to an infinite
distance is called ionization energy.
Hence, 2.18 × 10-18
J and 8.72 × 10-18
J are the ionization energies of H atom and He+
ion
respectively.
(d) Use Avogadro’s number to convert ionization energy values in kJmol-1
or H and
He+
.
The ionization energy of H-atom in kJmol-1
is given as
E = 2.18 × 10-18
×
. ×
= 1312.36 kJ/mol
The ionization energy of He+
-ion in kJ/mol is given as
E = 8.72 × 10-18
×
. ×
= 5249.4 kJ/mol
The experimental values of ionization energy of H and He+
are 1331 kJ/mol and
5250 kJ/mol respectively. How do you compare your values with experimental
values?

7. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
The calculated values of ionization energies of H-atom and He+
ion using Bohr’s
theory are 1312.36 kJ/mol and 5249.4 kJ/mol respectively.
These results agree well with the experimental results i.e. 1331 kJ/mol for H-atom and
5250 kJ/mol for He+
ion.
Both H-atom and He+
ion consists of one electron each and the above results clearly
shows that Bohr’s theory is perfectly applicable to one electron syste
Q 23. Calculate the wave number of the photon when the electron jumps from
(i) n = 5 to n = 2
(ii) n = 5 to n = 1
In which series of spectral lines these photons will appear.
Solution:
Ryberge constant = R = 1.097 × 107
m-1
.
When electron jumps from n = 5 to n = 2. The wave number of photon is given by the eq.
ṽ = R − = 1.097 × 107
−
ṽ = 1.097 × 107
− = 1.097 × 107
× = 2.30 × 106
m-1
This spectral line is present in visible region (Balmer series)
When electron jumps from n = 5 to n = 1.
The wave number of photon is given by the eq.
ṽ = R − = 1.097 × 107
−
ṽ = 1.097 × 107
− = 1.097 × 107
× = 1.05 × 107
m-1
This spectral lines is present in UV region (Lyman series)
Q 24. A photon of a wave number 102.70 × 105
m-1
jumps from higher to n = 1.
(a) Determine the number of that orbit from where the electron falls.
Solution:
Ryberge constant = R = 1.097 × 107
m-1
.
n1 = ?
n2 = ?
The wave number of photonis given by the eq.
ṽ = R −
102.7 × 105
= 1.097 × 107
× −
. ×
. ×
= 1 -
0.93637 = 1 -

8. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
= 1- 0.93637 = 0.0636
Or =
.
n2 = √
.
= 3.96 ≈ 4
(b) Indicate indicate the name of the series to which this photon belongs.
This spectral line is present in Lyman series.
(c) If the electron will fall from higher orbit to n = 2, then calculate the wave number of
the photon emitted. Why this energy difference is so small as compare to above
calculations?
When electron jumps from n = 4 to n = 2. The wave number of photon is given by the eq.
ṽ = R − = 1.097 × 107
−
ṽ = 1.097 × 107
−
ṽ = 1.097 × 107
× = 2.05646 × 106
m-1
Energy difference for n = 4 to n = 1 can be calculated by the eq.
∆E = 2.18 × 10-18
−
Thus ∆E = 2.18 × 10-18
−
∆E = 2.18 × 10-18
−
∆E = 2.18 × 10-18
× = 2.04 × 10-18
J
And the energy difference for n = 4 to n = 2 can be calculated by the eq.
Thus ∆E = 2.18 × 10-18
−
∆E = 2.18 × 10-18
−
∆E = 2.18 × 10-18
× = 4 × 10-19
= 0.4 × 10-18
J
The energy difference in second case is small.
It is because electron travel more distance from n = 4 to n = 1 than n = 4 to n = 2. And
since energy is directly related to the distance of the electron, hence energy difference in
second case is smaller than first case.
Q 25. (a) What is the de Brogile’s wavelength of an electron travelling at half a speed of
light?

9. Malik Xufyan F.Sc.1 (Chemistry)
0313-7355727 Chapter # 05
www.onlychemistrydiscussion.blogspot.com
Solution:
Mass of electron = m = 9.1 × 10-31
kg
Velocity of light = c =3 × 108
m/s
Velocity of electron = v = =
×
= 1.5 × 108
m/s
h = 6.625 × 10-34
Js
λ = ?
Wavelength of the electron is given by
λ = =
. ×
. × × . ×
= 4.85 × 10-12
m = 4.85 pm
(b) Convert the mass of the electron into grams and velocity of light into cms-1
.
Calculate the wavelength of an electron in cm.
Solution:
m = 9.1 × 10-31
kg = 9.1 × 10-31
× 1000g = 9.1 × 10-28
g
velocity of the light = c = 3 × 108
m/s = 3 × 1010
cm/s
velocity of the electron = v = =
×
= 1.5 × 108
m/s
Plank’s constant = h = 6.625 × 10-34
Js = 6.625 × 10-34
× 107
ergs = 6.625 × 10-27
ergs
λ = ?
wavelength of the electron is given by
λ = =
. ×
. × × . ×
= 4.85 × 10-10
cm
(c) Converts the wavelegth of electron from meters to
(i) nm (ii) A0
(iii) pm
Solution:
(i) 1m = 109
nm
Therefore
λ = 4.85 × 10-12
× 109
nm = 4.85 × 10-3
nm
(ii) 1m = 1010
A0
Therefore
λ = 4.85 × 10-12
× 1010
A0
= 4.85 × 10-2
A0
= 0.0485 A0
(iii) 1m = 1012
pm
Therefore
λ = 4.85 × 10-12
× 1012
pm = 4.85 pm
10-12
m = 1pm
1J = 107
erg
1m = 100 cm