TEXT BOOK EXERCISE year n0tes chemistry new 4 (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

1st year n0tes chemistry new

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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


-1


-1


-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


=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


=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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TEXT BOOK EXERCISE year n0tes chemistry new 4 (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