• Every matter has its existence from the
basic unit called atom
• All atoms of a particular substance are
alike
• Atoms of different elements are
different in size and electronic
configuration
g
• Atoms of one element combine with
atoms of other elements to form
molecules (and molecular compounds)
• JJ Thompson proposed plum pudding
model for an atom. Though he could
explain the stability of a large number of
elements he failed to explain the large
elements,
angle scattering of α (alpha) particles by
gold foil
• J. J. Thompson model also failed to
explain the cause for emission of
radiations by Hydrogen atoms
• Rutherford proposed the model based
on the experiment of α – scattering by
gold foil
• Rutherford model suggested a very
small space at the center of an atom in
which all the positive charges are
l t d Electrons
located.
El t
orbit
bit around
d the
th
nucleus in circular orbits
• Rutherford assumed that electrons
accelerate and radiate energy
• However,
Rutherford’s
model
favored
continuous
emission
spectrum
p
of hydrogen
y g
atoms.
Accelerating electrons accounts for
instability of atoms.
atoms
Hence
Hence,
Rutherford model failed to give a
stable
bl atom model
d l
• The quantization rule corresponding to the
g
momentum of electron and
angular
energy
transition/absorption
were
proposed by Neils Bohr. Non
Non‐radiating
radiating or
stationary orbit accounted for stability of
an atom.
atom Quantised orbit levels also
explained line emission spectrum of
Hydrogen atom.
atom Bohr model holds good
for Hydrogen and Hydrogen like atoms.
• Hyperfine splitting of energy levels with
the application of electric field and
magnetic field could not be explained by
Bohr’ss model.
Bohr
model
• Each principle quantum level generates
only a few transitions whereas a large
number
b off lines
li
were observed
b
d by
b using
i
powerful optical instruments.
• Sommerfield included additional
quantum
t
numbers
b
which
hi h accounted
t d
for elliptical orbits.
• Sommerfield model failed to explain
Zeeman effect and Stark effect and
transition of alkali metals.
• Vector models introduced the
concept of spatial quantization and
spin quantization of electron to
explain
l i multi‐electron
lti l t
system.
t
• De‐Broglie dual nature of matter
explains
p
Bohr’s second p
postulate of
quantization.
1. What is the radius of Iodine atom? (At no.53, mass no. 126)
(a) 2.5 X 10‐11m
(b) 2.5 X 10‐9m
(c) 7 X 10‐9m
(d) 7 X 10‐6m
2. Hydrogen atom emits blue light when it changes from n=4
it changes from n
4 energy level to n
energy level to n=2
2 energy level. Which color of light would the atom emit when it changes from n=5
the atom emit when it changes from n=5 level to n=2 level?
(a) Red
(c) Green
(b) Yellow
(d) Violet
3. The binding energy of the electron in
the lowest orbit of the Hydrogen
y g atom is
13.6eV. The energies required in eV to
remove an electron from the three
lowest orbits of the Hydrogen atom are
(a) 13.6, 6.8, 8.4
(c) 13.6, 27.2, 40.8
(b) 13.6, 10.2, 3.4
(d) 13.6, 3.4, 1.5
4. An electron jumps from the 4th orbit to
2nd orbit of hydrogen atom.
atom Given the
Rydberg’s
constant
R=105cm‐1,
the
frequency in hertz of the emitted radiation
will be
(a)3/16 X 105
( ) 9/16 X 1015
(c)
(b) 3/16 X 105
(d) 3/4 X 1015
5. In a hypothetical Bohr Hydrogen atom,
the mass of the electron is doubled. The
energy E0 and radius r0 of the first orbit
will be (a0 is the Bohr radius)
(a) E0=‐27.2
=‐27 2 eV; r0=a0/2
(b) E0=‐27.2 eV; r0=a0
(c) E0=‐13.6
= 13 6 eV; r0=a0/2
(d) E0=‐13.6 eV; r0=a0
7. If the difference between the two energy states in an atom is 2eV and h/e =
energy states in an atom is 2eV and h/e = 4 X 10‐15, then the wavelength of photon emitted as a result of the above itt d
lt f th b
transition will be
( )
(a) 1000Å
(c) 3000Å
( )
(b) 6000Å
(d) 9000Å
8. The minimum excitation potential of 8.
The minimum excitation potential of
Bohr’s first orbit in hydrogen atom is (a) 13.6 V
(c) 10 2 V
(c) 10.2 V
(b) 3.4 V
(d) 3 6 V
(d) 3.6 V
E2 2 ‐ E1 1 = ‐3.4 – ((‐13.6) = 10.2 eV )
and E.P = 10.2V
9. The ratio of minimum to maximum wavelength in Balmer
l
th i B l
series is
i i
((a)5:9
)
(c) 1:4
(b) 5:36
(b)
(d) 3:4
10. In the Bohr model of Hydrogen atom, let R, V & E represent the radius of the orbit, sped of the electron & total energy of the electron respectively. Which of the
of the electron respectively. Which of the following quantities is proportional to the quantum number n
quantum number n
(a)R/E
(c) RE
(b) E/V
(b)
E/V
(d) VR
11. An α‐particle of energy 5MeV is scattered through 1800 by a fixed Uranium scattered through 180
by a fixed Uranium
nucleus. The distance of the closest approach is of the order of
h i f th
d
f
((a)1Å
)1Å
(c) 10‐12cm
10cm
(b) 10
(b)
10‐10
(d) 10‐15cm
12. The ratio of the frequencies of the long wavelength limits of the Lyman to Balmer
wavelength limits of the Lyman to Balmer
series of Hydrogen is
(a)27:5
(c) 4:1
(c) 4:1
(b) 5:27
(d) 1:4
(d) 1:4
13. The ratio of the shortest wavelength observed in Lyman series to Pfund series of emission spectrum of hydrogen is
(a)4/3
(c) 1/25
(c) 1/25
(2) 525/376
(4) 960/11
(4) 960/11
14. Ionization potential of Hydrogen atom is 13 6Ev The least energy of photon of
is 13.6Ev. The least energy of photon of Balmer series is (a)3.4 eV
( ) 10 2 V
(c) 10.2 eV
(b) 1.89 eV
(d) 8 5 V
(d) 8.5 eV
For Balmer series, n
, 1=2, n
, 2=3. The least energy corresponds to the transition from )
3→2 i.e. E3 3 – E2 = ‐1.511 – ((‐3.4) = 1.89 eV
15. If the series limit og Balmer series is 6400Å th th
6400Å, then the series limit of Paschen
i li it f P h
series will be
(a)6400Å
(c) 14400Å
Å
(b) 18680Å
(d) 2400Å
Å
16. In a Hydrogen atom, electron makes
transition from an energy level with
quantum number n to another with quantum
number (n‐1).
(n 1) If n>1,
n>1 the frequency of
radiation emitted is proportional to
(a)1/n
( ) 1/n
(c)
1/ 3/2
(b) 1/n2
(d) 1/n
1/ 3
17. When an electron in hydrogen atom in the ground state absorbs a photon of energy 12.1 eV, its angular momentum
(a) Decreases by 2.11 X 10‐34Js
(b) Increases by 1.055 X 10‐34Js
(c) Decreases by 1.055 X 10‐34Js
(d) Increases by 2.11 X 10‐34Js
18. Let v1 be the frequency of the series limit of the Lyman series v2 be the limit of the Lyman series, v
be the
frequency of the first line of Lyman series and v3 be the frequency limit of Balmer
and v
be the frequency limit of Balmer
series. Then
(a)v1 + v2 = v3
( ) 1 – v2 = v3
(c) v
(b) v2 – v1 = v3
(d) 3 = ½ (v
(d) v
½ ( 1 + v2)
19. When a Hydrogen atom emits a photon in going from n=5 to n=1 its recoil speed is
in going from n=5 to n=1, its recoil speed is almost
(a)10‐4 m/s
(c) 2 X 102 2 m/s
(c) 2 X 10
(b) 8 X 102 m/s
(d) 4 m/s
(d) 4 m/s
20. Which energy state of the triply ionized beryllium has the same electron orbital
beryllium has the same electron orbital radius as the of ground state hydrogen? Given Z for Be = 4
Given Z for Be = 4
(a)n = 4
(a)n
4
(c) n = 2
(b) n = 3
(b)
n 3
(d) n = 1 21. In a Hydrogen atom, the electron is in the nth excited state. It comes down to the first excited state. It comes down to the first
excited state by emitting ten different wavelengths. The value of n is
wavelengths. The value of n is
(a) 9
(c) 7
(b) 8
(b)
8
(d) 6
22. The binding energy of an electron in the ground state of He is equal to 24 6 eV
the ground state of He is equal to 24.6 eV. The energy required to remove both the electrons is
l t
i
((a)49.2 eV
)49 2 V
(c) 38.2 eV
(b) 24.6 eV
(b)
24 6 V
(d) 79.0 eV
E1 = ‐ z2 (13.6) = ‐ 55.4 eV
Therefore, I.E = 54.4 + 24.6 = 79.0 eV
23. According to Bhor’s theory of hydrogen p
g
gy
atom, the product of the binding energy of electron in nth orbit of its radius in the nth orbit
(a)Is proportional to n
(a)Is
proportional to n2
(b)Has a constant value 7.2 eV‐Å
(c) Has a constant vale 10.2 eVÅ
Has a constant vale 10 2 eVÅ
(d)Is inversely proportional to n3
24. The ionization energy of 10 times ionized sodium atom is
ionized sodium atom is
(a)13.6/11 eV
(c) 13 6 X (11)2 eV
(c) 13.6 X (11)
(b) 13.6/112 eV
(d) 13 6 eV
(d) 13.6 eV
IE = +Z2EH ; Z for Na is 11
27. If the atom 100Fm257 follows the Bohr
model
d l and
d the
h radius
di off 100Fm
F 257 is
i n times
i
the
h
Bohr’s radius, find n.
(a) 100
( )4
(c)
(b) 200
(d) 1/4
28. The atomic weight of Boron is 10.81 and it has two isotopes 510B and has two isotopes B and 511B. The ratio of B. The ratio of
the isotopes would be (1) 19:81
(3) 15:16
(3) 15:16
(2) 10:11
(4) 81:12
(4) 81:12
29. The density of the hydrogen nucleus with ZZ=1,
1, is 2.29 X 10
is 2.29 X 1017kg/m3. The density of gold . The density of gold
nucleus with Z=79 would be (a) 2.29/79 X 1017 kg/m3
(b) 2.29 X 79 X 1017 kg/m3
(b) 2.29 X 79 X 10
(c) 2.29 X 1017 kg/m3
(d) 2.29/√79 X 1017 kg/m3
(d) 2.29/√79 X 10
The density of the nucleus is the same for all the atomic nuclei.
the atomic nuclei. 30. The ratio between total acceleration of the electron in a singly ionised helium atom and electron in a singly ionised
helium atom and
hydrogen atom (both in ground state) is
(a) 4
(c) 16
(c) 16
(b) 8
(d) 1
(d) 1
31. The energy required to excite an electron from n 2 to n 3 energy level is 47 2eV The
from n=2 to n=3 energy level is 47.2eV. The charge number of the nucleus around which the electron is revolving will be
the electron is revolving will be
(a) 5
(c) 15
(b) 10
(b)
10
(d) 20
32. The hydrogen like ion having wavelength difference between first line of Balmer and difference between first line of Balmer
and
Lyman series is 593Å. What is the atomic number of the ion?
number of the ion?
(a) 4
(c) 2
(b) 3
(b)
3
(d) 1
33. The largest wavelength in the ultra violet region in the hydrogen spectrum is 122nm
region in the hydrogen spectrum is 122nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the
region of the hydrogen spectrum (to the nearest integer) is
(a)823nm
( ) 1882
(c) 1882nm
(b) 802nm
(d) 1648
(d) 1648nm
34. The electron in an hydrogen atom makes a transition from n n1 to n=n
a transition from n=n
to n n2 state. The time state The time
period of the electron in the initial state (n1) is 8 times that in final state (n2). The possible is 8 times that in final state (n
) The possible
vales of n1 and n2 are
(a) n1 = 4, n2 = 2
(b) n1 = 2, n
(b) n
= 2 n2 = 4
=4
(b) n1 = 8, n2 = 1
(c) n1 = 1, n
(c) n
= 1 n2 = 8 =8
35. As the electron in Bohr orbit of Hydrogen atom passes from state n=2
atom passes from state n
2 to n
to n=1,
1, the kinetic the kinetic
energy K and the potential energy U change as
(a)
(b)
(c)
(d)
K two‐fold, U four‐fold
K four‐fold,
K four
fold, U two
U two‐fold
fold
K four‐fold, U four‐fold
K two‐fold,
K two
fold, U also two
U also two‐fold
fold
36. The ionization P.E. of H2 atom in its ground
state is 13.6
13 6 V.
V The ionization P.E.
P E of H2 atom in
its first excited is
(a) 13.6 V
(c) 1.51
1 51 V
(b) 3.4 V
(d) 0.85
0 85 V
37. The total energy of electron in its first orbit of H2 atom is orbit of H
atom is ‐13.6eV.
13.6eV. The kinetic The kinetic
energy and potential energy of the electron in the first orbit are
in the first orbit are
(a) 13.6eV, 13.6eV, ‐27.2eV
27.2eV
(c) 13.6eV, 27.2eV
(b) ‐13.6eV,
(b)
13.6eV, ‐27.2eV
27.2eV
(d) ‐13.6eV, 27.2eV
38. The ratio of ground state energy of H2
atom is singly ionized He atom is
atom is singly ionized He atom is
(a)
(b)
(c)
(d)
4:1
1:4
2:1
1:2