text book exercise year n0tes chemistry new 4 (b) which ever gas is used in the discharge tube the...
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CHAPTER 5
ATOMIC STRUCTURE
TEXT BOOK EXERCISE
Q1. Select the most suitable answer for the given one. i. The nature of the positive rays depends on
(a) The nature of the electrode
(b) The nature of the discharge tube
(c) The nature of the residual gas
(d) all of the above
ii. The velocity of photon is
(a) Independent of its wavelength
(b) Depends on its wavelength
(c) Equal to square of its amplitude
(d) Depends on its source
iii. The wave number of the light emitted by a certain source is 2 x
106
m. the wavelength of this light will be
(a) 500nm (b) 500m
(c) 200nm (d) 5 x 107
m
iv. Rutherford’s model of an atom failed because
(a) The atom did not have a nucleus an electron
(b) It did not account for the attraction between protons and
neutrons
(c) It did not account for the stablility of the atom
(d) There is actually no space between the nucleus and the
electrons
v. Bohr model of atom is contradicted by
(a) Planck quantum theory
(b) Pauli’s exclusion principle
(c) Heisenberg’s uncertainty principle
(d) all of the above
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vi. Splitting of spectral lines when atoms are subjected to
strong electric field is called.
(a) Zeeman effect (b) Stark effect
(c) Photoelectric effect (d) Compton Effect
vii. In the ground state of an atom, the electron is present
(a) In the nucleus (b) in the second shell
(c) Nearest to the nucleus
(d) farthest from the nucleus
viii. Quantum number values for 2p orbitals are
(a) n=2, l=l (b) n=1, l=2
(c) n=1, l=0 (d) n=2, l=0
ix. Orbitals having same energy are called
(a) Hybrid orbitals (b) valence orbitals
(c) Degenerate orbitals (d) d-orbitals
x. when 6d orbitals is complete, the entering electron goes into
(a) 7f (b) 7s (c) 7p (d) 7d
Ans. (i)c (ii)a (iii)a (iv)c (v)c
(vi)b (vii)c (viii)a (ix)c
(x)c
Q.2 Fill in the blanks with suitable words (i) B-particles are nothing but _______moving with a very high
speed.
(ii) The charge on one mole of electrons is ________coulombs.
(iii) The mass of hydrogen atom is _________grams.
(iv) The mass of one mole of electron is _________.
(v) Energy is ________when electron jumps from higher to a
lower orbit.
(vi) The ionization energy of hydrogen atom can be given by
formula ________.
(vii) For d sub-shell, the azimuthal quantum number has a value of
________.
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(viii) The number of electrons in a given sub-shell is given by
formula ________.
(ix) The electronic configuration of H- is ________.
Ans. i)electrons ii)96500 iii)1066x10-24
iv)5.484x10-7kg v)emitted vi)
Bohr vii)2 viii)2(2l+1) ix) 1s2
Q.3 Indicate true or false as the case may be. (i) A neutron is slightly lighter particle than a proton.
(ii) A photon is the massless bundle of energy but has momentum.
(iii) The unit of Rydberg constant is the reciprocal of unit of
length.
(iv) The actual isotopic mass is a whole number.
(v) Heisenberg’s uncertainty principle is applicable to
macroscopic bodies.
(vi) The nodal plane in an orbital is the plane of zero electron
density.
(vii) The number of orbitals present in a sublevel is given by the
formula (2l_1)
(viii) Magnetic quantum number was introduced to explain Zeeman
and stark effects.
(ix) Spin quantum number tells us the direction of spin of
electron around the nucleus.
Ans. i)False ii)True iii)True
iv)False v)False vi)True vii)True
viii)True ix)Ture
Q.4 Keeping in mind the discharge tube experiment , answer the
following questions. (a) Why is it necessary to decrease the pressure in the discharge
tube to get the cathode rays?
Ans. There will be no flow of current through the gas when the
pressure in the discharge tube is high. In the presence of high
pressure, the cathode rays will not flow from the cathode surface.
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(b) Which ever gas is used in the discharge tube the nature of the
cathode rays remains the same. Why?
Ans. Cathode rays are composed of negatively charged particles
(electrons). They are constituents of all gases. So, cathode rays are
independent of the nature of the gas in the discharge tube.
(c) Why e/m value of the cathode rays is equal to that of electron?
Ans. Cathode rays are composed of electrons, so their e/m value is just
equal to that of electron.
(d) How the bending of the cathode rays in the electric and magnetic
fields shows that they are negatively charged?
Ans. when cathode rays are passed through an electric field created by
two charged metal plates, they are deflected towards the positively
charged plate. This shows that they are negatively charged.
When cathode rays are passed between the poles of a
magnet, the magnet neither attracts nor repels but cause them to
move in a curved path perpendicular to the line drawn between the
poles of the magnet. This shows that they are negatively charged.
(e) Why positive rays are also called canal rays?
Ans. Positive rays pass through canals or holes in the cathode, so they
are called canal rays.
(f) The e/m value of positive rays for different gases are different but
those for cathode rays the e/m values is the same Justify it.
Ans. The e/m value for positive rays are different for different gases
because they differ in mass. the mass of the positive particles is the
same as that of the atom or molecule form which it is created.
Heavier the gas, smaller the e/m value.
The e/m value for cathode rays is the same because
cathode rays are composed of electrons which have constant mass.
(g) The e/m value for positive rays obtained from hydrogen gas
is 1836 time less than that of cathode rays. Justify it.
Ans. The e/m value for positive rays obtained form hydrogen gas is
1836 time less than that of cathode rays. This is because the mass
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of proton which is created from H-atom is 1836 time more than
that of an electron (cathode rays particle).
Q.5 (a) Explain Milliken’s oil drop experiment to determine the
charge of an electron.
(b) What is J.J Thomson’s experiment for determining e/m
value of electron?
(c) Evaluate mass of electron from the above two
experiments.
Q.6 (a) Discuss Chadwick’s experiment for the discovery of
neutrons. Compare the properties of electron, proton, and
neutron.
(b) Rutherford’s atomic model is based on the scattering of a-
particles from a thin gold foil. Discuss it and explain the
conclusions.
Q.7 (a) Give the postulates of Bohr’s atomic model. Which
postulate tell us that orbits are stationary and energy is
quantized?
(b) Derive the equation for the radius of nth orbit of hydrogen
atom using Bohr’s model.
(c) How does the above equation tell you that?
(i) Radius is directly proportional to the square of the number of
orbit.
(ii) Radius is inversely proportional to the number of proton in the
nucleus.
Ans. The equation for the radius of nth orbit is:
r=
Since, , h, ,m and e are constant , therefore , the factor is
constant.
Therefore, r=constant x
Or
r =
Hence, we can say:
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(i) Radius is directly proportional to the square of the number of
orbit.
(ii) Radius is inversely proportional to the number of protons in the
nucleus
(d) How do you come to know that the velocities of electrons in
higher orbits are less than those in lower orbits of hydrogen atom?
Ans. According to Bohr, since the electron keeps on revolving around
the nucleus:
Therefore, centrifugal force=Electrostatic force of attraction
=
or r=
For H-atom,Z=1
r=
Since, e, , and m are constant, therefore, the factor is constant.
Therefore, r=constant x
r=
Hence, the radius of a moving electron is inversely proportional
to the square of its velocity. The smaller the radius of the orbit, the
higher is the velocity of electron. Hence, the velocities of electrons in
higher orbits are less than those in lower orbits of hydrogen atom.
(e) Justify that the distance gaps between different orbits go on
increasing form the lower.
Ans. We know that: r=0.529 x[n2]
When n=1 r1=0.529
When n=2 r2=0.529 x4=2.11
When n=3 r3=0.529 x9=4.75
When n=4 r4=0.529 x16=8.4
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When n=5 r5=0.529 x25=13.22
Distance between orbits are:
r2 –r1=(2.11 – 0.529) =1.581
r3 –r2=(4.75 – 2.11) =2.64
r4 –r3=(8.4 – 4.75) =3.65
r5 –r4=(13.22 – 8.4) =4.82
From the data of radius difference, it is clear that the distance
gaps between different orbits go on increasing from the lower to
the higher orbits.
Q8. Derive the formula for calculating the energy of an electron in
nth orbit using Bohr’s model. Keeping in view this formula explain
the following:
(a) The potential energy of the bounded electron is negative.
Ans. According to Bohr, the energy of electron is calculated from the
equation:
En=- 2.178 x 10-18[]
When n= , then En=0
Consider that an electron is present at an infinite distance from
the nucleus, so there is no interaction between the two. The energy of
this electron is zero.
Now, suppose that the electron moves closer and closer to the
nucleus. Since electron is negatively charged and nucleus is positively
charged, no work needs to be done on the electron. The electron can
move towards the nucleus by itself due to electrical attraction. Thus,
work is done by the electron itself as it moves towards the nucleus. As a
result, the potential energy falls, i.e. it become less than zero. Any value
less than zero is negative. Hence, the potential energy of electron
becomes more and more negative as the electron moves closer and
closer towards the nucleus.
(b) Total energy of the bounded electron is negative.
Ans. When the electron is at infinite distance from the nucleus, there is
no electrostatic interaction between the two. Therefore, the energy of the
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system in this state is assumed to be zero. As the electron moves closer
to the nucleus, it does some work and energy is of the electron becomes
negative. The negative value of energy would keep increasing as the
electron moves to the energy levels nearer to the nucleus. the negative
value of total energy shows that electron is bound by the nucleus i.e,
electron is under the force of attraction of the nucleus.
(c) Energy of an electron is inversely proportional to n2
, but energy
of higher orbits are always greater than those of the lower orbits.
Ans. Energy of electron: The energy of electron in different orbits can be calculated by
using the following equation:
En = - kJmol-1
Energy of an electron is inversely proportional to n2.
When n =1 E1= =1312.36kJ mol-1
When n =2 E2= =328.09kJ mol
-1
When n =3 E3= =145.82kJ mol
-1
When n =4 E4= =82.023kJ mol
-1
When n =5 E1= =52.49kJ mol
-1
The energy of an electron is inversely proportional to n2 .
As the value of ‘n’ increases, the value of energy increases. The
energy of higher orbits are always greater than those of the lower
orbits.
E5 >E4 >E3> E2 >E1
(d) The energy difference between adjacent levels goes on
decreasing sharply.
Ans. The energy difference between adjacent levels can be found as:
E=E2 – E1 =(- 328.09) – (-1312.36) =984.27kJmol-1
E=E3 – E2 =(- 145.82) – (- 328.09) =182.27kJmol-1
E=E4 – E3 =(- 82.023) – (- 145.82) =63.797kJmol-1
E=E5 – E4 =(- 52.49) – (- 82.023) =29.533kJmol-1
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From the data of energy difference, it is found that the energy
difference between adjacent levels goes on decreasing sharply.
Q9. (a) Derive the following equation for hydrogen atom which
are related to:
i. Energy difference between two levels, n1 and n2 .
ii. Frequency of photon emitted when an electron jumps
from n2 to n1 .
iii. Wave number of the photon when the electron jumps
from n2 to n1 .
(b) Justify that Bohr’s equation for the wave number can
explain the spectral lines of Lyman, Blamer and paschen
series.
Q10. (a) What is spectrum? Differentiate between continuous
spectrum, and line spectrum.
(b) Comparison between line emission and line absorption
spectra.
Ans.
Line emission spectrum Line absorption
spectrum
1. “An atomic
spectrum which
consists of bright
lines against a dark
background is called
line emission
spectrum. “
2. it is produced when
radiations emitted
by an excited
substance are
analysed in a
spectroscope.
1. “An atomic
spectrum which
consists of bright
lines against a dark
background is called
line emission
spectrum. “
2. it is produced
when white is
passed through the
gaseous element
and the transmitted
rays are analysed in
a spectroscope.
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(c) What is the origin of the line spectrum?
Q11. (a) Hydrogen atom and He+
are monoelectronic system, but
the size of He+
is much smaller than H+
, why?
Ans. H-atom and He+
are monoelectronic system. It means both H-
atom and He+
have one electron in the valence shell. H-atom has
one proton in the nucleus whereas He+
has two proton in the
nucleus. So, the force of attraction between two protons and one
electron is greater than one proton and one electron. Hence, the
size of He+
is much smaller than H-atom.
Also, we know that: r=0.529
The size of H-atom: r=0.529 =0.2645
The size of He+
ion: r=0.529 =0.529 x =0.2645
Hence, the size of He+
is much smaller than H-atom. This is because the
nucleus of He+
has greater attraction for the electron as compared to H-
atom which contains one proton in the nucleus.
(b) Do you think that the size of li-2+
is even smaller than HE+
?
Justify with calculation.
Ans. The size of He+
ion: r=0.529 =0.2645
The size of li-2+
ion: r=0.529 =0.1763
The size of li-2+
ion is much smaller than the size of He+
ion:
Q12. (a) What are X-rays? What is their origin? How was the idea
of atomic number derived from the discovery of X-rays?
(b) How does the Bohr’s model Justify the Moseley‘s
equation?
Q13. Point out the defects of Bohr’s model. How these defects are
partially covered by dual nature of electron and Heisenberg’s
uncertainty principle?
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Q14. (a) Briefly discuss the wave mechanical model of atom.
How has it given the idea of orbital. Compare orbit and orbital.
Ans. Comparison between orbit and orbital:
Orbit Orbital
1. “A circular path around
the nucleus in which the
electron revolves is called
an orbit.”
2. It is circular in Shape.
3. It represents that an
electron moves around
the nucleus in one plane,
i.e., in a flat surface.
4. It is against Heisenberg’s
uncertainty principle.
5. The maximum number of
electrons in an orbit is
2n2, where ‘n’ is the
number of the orbit.
1. “The volume of space
within an atom in which
there is 95% chance of
finding an electron is
called orbital.”
2. It may be spherical,
dumbbell or double
dumbbell in shape.
3. It represent that an
electron can move
around the nucleus in
three dimensional space.
4. It is in accordance with
Heisenberg’s uncertainty
principle.
5. The maximum number of
electrons in an orbital is
two.
(b) What are quantum number? Discuss their significance.
(c) When azimuthal quantum number has a value 3, then
there are seven value of magnetic quantum number. Give reasons.
Q15. Discuss rules for the distribution of electrons in energy sub-
shells and in orbitals. (a) What is (n+ l ) rule. Arrange the
orbitals according to this rule. Do you think that this rule is
applicable to degenerate orbitals?
(b) Distribute electrons in orbitals of 57La, 29Cu, 79Au,
24Cr, 53I, 86Rn.
Ans. Electronic configurations:
57La=1s2 2s
2 2p
6 3p
6 4s
2 3d
10 4p
6 5s
2 4d
10 5p
6 6s
2 4f
4 (Actual configuration)
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=1s2 2s
2 2p
6 3p
6 4s
2 3d
10 4p
6 5s
2 4d
10 5s
2 5p
6 5d
1(Expected configuration)
29Cu =1s2 2s
2 2p
6 3d
9 4s
2 (Actual configuration)
=1s2 2s
2 2p
6 3p
6 4s
2 3d
10 4s
1 (Expected configuration)
79Au=1s2 2s
2 2p
6 3p
6 4s
2 3d
10 4p
6 5s
2 4d
10 4f
145s
2 5p
6 5d
10 gs
1
24Cr=1s
2 2s
2 2p
6 3s
2 3p
6 3d
4 4s
2 (Actual configuration)
=1s2 2s
2 2p
6 3s
2 3p
6 3d
4 4s
1 (Expected configuration)
53I=1s2 2s
2 2p
6 3s
2 3p
6 3d
10 4s
2 4p
6 4d
10 5s
2 5p
5
86Rn==1s
2 2s
2 2p
6 3s
2 3p
6 3d
4 4s
2 4p
6 4d
10 4f
14 5s
2 5p
6 5d
10 6s
2 6p
6
From the above configuration, it is important to note that there are three
irregularities in the general trend. The electronic configuration of Cr and
Cu show deviation from the expected configuration.
Expected Configurations: Cr=1s2 2s
2 2p
6 3s
2 3p
6 3d
4 4s
1
Cu=1s
2 2s
2 2p
6 3s
2 3p
6 3d
4 4s
1
Actual Configurations: Cr=1s2 2s
2 2p
6 3s
2 3p
6 3d
4 4s
2
Cu=1s
2 2s
2 2p
6 3s
2 3p
6 3d
4 4s
2
This is because the half filled and fully filled configurations
(i.e.,d5,d
10,f
7,f
14 )have lower energy ot more stability. Thus, in order to
become more stable, one of 4s electrons goes into 3d orbitals so that 3d
orbitals get half filled or fully filled configuration in Cr and Cu
respectively.
Reasons for stability of Half filled and fully filled orbitals: 1. Exchange energy; All the orbitals in a given sub-shell have
equal energies. The electrons present in different orbitals of the same
sub-shell can exchange their positions. Such an exchange of electrons
results in release of energy called exchange energy . Greater the
exchange energy, more is the stability associated with the orbitals. It has
been observed that in case of exactly half-filled and fully filled orbitals
the electrons can exchange their positions more readily as compared to
other configurations. As a result, half-filled and fully filled
configurations are more stable.
2. Symmetry: If removal or addition of an electron results in the
symmetrical distribution of electrons in an orbital, the electronic
configuration becomes more stable. Therefore, half-filled and fully filled
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configurations in which each orbital contains one and two electrons
respectively , are more stable.
Irregular Structure of La: The electronic configuration of La is irregular
and does no t follow the general trend. This is because strong nuclear
attraction and less shielding effect caused by d and f electrons. In this
case remember before adding any electron in the 4f orbital, a single
electron is added to a 5d orbital. The remaining nine electrons enter the
5d sub-shell after the 4f sub-shell has been completely filled with
fourteen electrons. Similarly, one electron enters the 6d sub-shell before
any electron enters the 5f sub-shell.
Q16. Draw the shapes of s, p and d-orbitals. Justify these by keeping
in view the azimuthal and magnetic quantum numbers.
Q17. A photon of light with energy 10-19
j is emitted by a source of
light/
(a) Convert this energy into the 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 x
10-34 js
Formula used: E=hv
V=
V=
V=1.51 x 1014 s-1
V=1.51 x 1014 Hz [CPS=s-1
=Hz]
Now, =
E=10-19
J=10-19
kg m2 s
-2 [1 J=1 kf m
2 s
-2 ]
h =6.625 x 10-34
kg m2 s
-2 s
=6.625 x 10-34
kg m2s
-1
=
=19.875x 10-7
m
=1.9875 x 10-6
m=1.9875 x 10-4
cm
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Now,
=0.51 x 106
m-1
=5.1 x 105
m-1 =5.1 x 103 cm
-1
(b) Convert this energy of the photon into ergs and calculate the
wave length in cm, frequency in Hz and wave number in cm-1
.
h=6.625x 10-34
js or 6.625 x 10-27
ergs. C=3 x 108
ms-1
or 3x 10 +10
cms-1
.
Solution: E= 10
-19 J =10
-19 x 10
7 erg [1 j =10
7 erg]
=10-12
erg
=1.98 x 10-6
m=1.98 x10-6
x 102 cm=1.98 x 10
-4 cm
=5x105 m
-1 =5 x 10
5 x 10
-2 cm
-1 =5x 10
3 cm
-1
Q18. The formula for calculating the energy of an electron in
hydrogen atom gives by Bohr, s model
En =
Calculate the energy of the electron in first orbit of hydrogen
atom. The values of various parameters are same as provided in
Q.19.
Solution: n = E=?
Formula: En =-2.178 x 10-18
[ ]J
En=-2.178 x 10-18
[ ]J
E=-2.178 x 10-18
J
E=-2.18 x 10-18
J
Q19. Bohr’s equation for the radius of nth orbit of electron in
hydrogen atom is
rn =
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(a) When the electron moves from n=1 to n=2, how much does the
radius of the orbit increases.
Solution:
=8.85 x10-12
C2 J
-1 m
-1 ; m=9.108 x 10
-31 kg
h=6.624 x 10-34 js ; e=1.602 x 10-19
C ;
=3.14
Form n=1 to n=2 : J=kg m2 s-2 : C=
s-1
Formula: rn =
r1=
r1=
r1=
r1= 5.29 x 10-11
jm-1
s2 kg
-1
r1= 5.29 x 10
-11 kg
m
2 s
2m
-1 s
2 kg
-1
r1= 5.29 x 10
-11 m
r1= 5.29 x 10-1
r1= 0.529
Also
r2= 0.529 [22]
R2= 0.529 x 4 =2.116
Increase in radius, (r2 - r1) =2.116 - 0.529
=1.587
(b) What is the distance traveled by th electron when it goes from
n=2 to n =3and n =9 to n=10?
=8.85x 10-12
C2 J
-1 m
-1 , h=6.24x10
-34 js, =3.14,
m=9.108 x10-31
kg , C=1.602 x 10-19
C
While doing calculations take care of units of energy
parameter.
J=kgm2 s-2
, c=kg m s-1
Solution: r=0.529
(n2)
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For n=2 r2 =0.529 (22) =2.116
n =3 r3=0.529 (32) =4.761
Distance traveled by the electron when it goes from n=9 to n
=3
r3 – r2 =4.761 - 2.116 =Answer
For n=9 r9 =0529(92)=42.849
n =10 r10=0.529 (102)=52.9
Distance traveled by the electron when it goes grom n=9 to n=10
r10 – r9 =52.9 - 42.849 =10.051 Answer
Q20. Answer the following questions, by performing the calculation
s.
(a) Calculate the energy of first five orbits of hydrogen atom an
determine the energy differences between them.
(b) Justify that energy difference between second and third orbits
is approximately five times smaller than that between 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 group of the spectral lines of He+
are at
different places than those for hydrogen atom? Give reasons.
Q21. Calculate the value of principal quantum number if an electron in
hydrogen atom revolves in an orbit of energy - 0.242 x 10-18
j.
Solution: E= - 242 x 10
-18 j ; n=?
Formula: E=- 2.178 x 10-18
n2= - 2.178 x 10
-18 x
n2=- 2.178 x 10-
18x
n2=
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n2=9
n= 3 Answer
Q22. Bohr’s formula for the energy levels of hydrogen atom for any
system say H, He+,
Li-2+
, etc is
En=
Or
En= - k [ ]
For hydrogen Z=1 and for He+
, Z=2
(a) Draw an energy level diagram for hydrogen atom and He+
.
(b) Thinking that k = 2.18 x 10-18
j, calculate the energy needed
to remove the electron from hydrogen atom and from He+
.
Solution: For H; Z=1 ; n=1 ; k=2.18 x 10
-18 j
Formula En =-k[ ]
E1 = -2.18 x 10-18
[ ]
E1=-2.18 x 10-18
j
=-2.18 x 10-18 [ ] J
=-2.18 x 10-18
0j=0
Now energy required to remove an electron from the orbit:
= - E1
=0 – (-8.72 x10-18
J)
=8.72 x 10-18
J Answer
(c) How do you justify that the energies calculated in (b) are the
ionization energies of H and He+
?
Ans. The energy difference between first and infinite levels of energy
for H atom is 2.18x 10-21
kJ and for He+
ion is 8.72x10-21
kJ are the
ionization energies of H and He+
respectively. These values are the
same as determined experimentally.
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(c) Use Avogadro’s number to convert ionization energy values
in kJ mol-1
or H and He+
.
Solution: For H: =2.18 x 10
-18 J
The value of energy obtained for the electron is in J/atom. If this
quantity is multiplied by Avogadro’s number and divided by 1000,
the value of E will become.
=
=13.1236 x 102 kJ mol
-1
=1312.36 kJ mol-1
Answer For He+
;
=8.72 x 10-18
J/atom
=8.72 x 10-18
x
=52.4944 x 102 kJ mol
-1
=5249.44 kJ mol-1
Answer
(e) The experimental values of ionization energy of H and He+
are
1331 kJ mol-1
and 5250 kJ mol-1
respectively. How do you compare
your values with experimental values ?
Q23. 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 and spectral regions these
photons will appear.
Solution: (i) n=5 to n =1 . =?
Formula : =1.09678 x 107 [ - ]m
-1
=1.09678x 107[ - ]m
-1
=1.09678x 107[ ]m
-1
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=1.09678x 107[ ]m
-1
=1.09678x 10
7m
-1 =2.3x10
6 m
-1 Answer
The photon will appear in the Balmer series
ii) n =5 to n= 1
Solution: n = 5 to n=1 ; =?
Formula: =1.09678 x 107 [ - ]m
-1
=1.09678 x 10
7 [ - ]m
-1
=1.09678 x 10
7 [ - ]m
-1
=1.09678 x 10
7 [ ]m
-1
=1.09678 x 10
7 m
-1 Answer
The photon will appear in the Lyman series.
Q24. A photon of a wave number 102.70 x 105
m-1
is emitted when
electron jumps form higher to n=1.
(a) Determine the number of that orbit from where the electron falls.
Solution: =102.70 x 10
5 ; n
2=?
Formula used: =1.09678 x 107 [ - ]m
-1
102.70x 10
5 m
-1 =1.09678 x 10
7 [ - ]m
-1
=[ - ]
0.936=1-
=1-0.936
=0.064
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=
=16
n2 = =4 Answer
(b) Indicate the name of the series to which this photon belongs.
Ans. Since the electron falls from n=4 to n=1, therefore, the name of
the series is Lyman series
(d) 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 compared to that in part (a)?
Solution: n1 =2 ; n2=4
=?
Formula: =1.09678 x 107 [ - ]m
-1
=1.09678 x 10
7 [ - ]m
-1
=1.09678 x 10
7 [ - ]m
-1
=1.09678 x 10
7 [ ]m
-1
=0.20565x 10
7 m
-1
=0.20565x 10
5m
-1 Answer
Q25. (a) What is de Broglie’s wavelength of an electron in meters
traveling at half a speed of light ?
m =9.109 x 10-31
kg , c=3x 108 ms
-1
Solution:
M=9.109 x 10-31 kg : v= x3x 108
ms-1
=1.5x 108
ms-1
h =6.624 x 10-34
js =6.624 x 10-34
kg m2 s
-1
Formula: =
=
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=0.485 x 10-11
m
=0.485 x 10-12
m
=0.485 x 10-12
x 1010
=0.485 x 10-2
=0.0485 Answer
(b) Convert the mass of electron into grams and velocity of light
into cms-1
, and then calculate the wavelength of an electron in cm.
Solution: m=9.109 x 10
-31 kg =9.109 x 10
-31 x 103 g =9.109 x 10
-28 g
c =3 x 10-8
ms-1
=3 x 10-8
x 102 cms
-1 =3 x 10
8 cms
-1
=4.85 x 10-12
m=4.85 x 10-12
x102 cm =4.85 x 10
-10 cm
=0.048 x 10-8
cm Answer
(c) Convert the wavelength of electron from meters to
i) nm ii) iii) pm
Solution: i) =4.85 x 10
-12 m=4.85 x 10
-12 x10
9 cm =4.85 x 10
-3 nm=0.048nm
ii) =4.85 x 10-12
m=4.85 x 10-12
x1010
=4.85 x 10-2
=0.0485 Ans.
iii) =4.85 x 10-12
m=4.85 x 10-12
x10-12
pm =4.485 x 10-24
pm Ans.
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