IISER Physics - Atoms, Molecules And Nuclei
Exam Duration: 45 Mins Total Questions : 30
In Rutherford's experiment, silver foil is replaced by a copper foil of the same thickness. The number of alpha particles scattered through the same angle per minute in copper foil is proportional to
- (a)
\(\frac { { Z }_{ Cu } }{ { Z }_{ Ag } } \)
- (b)
\({ \left( \frac { { Z }_{ Cu } }{ { Z }_{ Ag } } \right) }^{ 2 }\)
- (c)
\(\frac { { Z }_{ Ag } }{ { Z }_{ Cu } } \)
- (d)
\({ \left( \frac { { Z }_{ Ag } }{ { Z }_{ Cu } } \right) }^{ 2 }\)
Number of alpha particles scattered per minute through an angle \(\theta \) is
\(N\infty \frac { Z^{ 2 } }{ sin^{ 4 }\left( \frac { \theta }{ 2 } \right) } \)
\(N_{ Ag }\propto Z^{ 2 }_{ Ag }\)
\(N_{ cu }\propto Z^{ 2 }_{ Cu }\)
\(\frac { N_{ Cu } }{ N_{ Ag } } \propto \frac { Z^{ 2 }_{ cu } }{ Z^{ 2 }_{ Ag } } \)
In Rutherford's experiment on alpha particle scattering, the ratio of number of alpha particles through an angle of 60° and 90° is
- (a)
4 : 1
- (b)
\(\sqrt { 3 } :1\)
- (c)
3 : 1
- (d)
9 : 1
\(N\propto \frac { 1 }{ sin^{ 4 }\left( \frac { \theta }{ 2 } \right) } \)
\(\frac { N_{ 60^{ 0 } } }{ N_{ 90^{ 0 } } } =\left( \frac { sin45^{ 0 } }{ sin30^{ 0 } } \right) ^{ 4 }=\left( \frac { 1/\sqrt { 2 } }{ 1/\sqrt { 2 } } \right) ^{ 4 }=\left( \sqrt { 2 } \right) ^{ 4 }=\frac { 4 }{ 1 } \)
Rutherford's atomic model has the deficiencies, namely,
- (a)
since electron has centripetal acceleration, it must fall into nucleus along a spiral path
- (b)
there must be infinite number of orbits
- (c)
both (a) and (b)
- (d)
None of (a) and (b)
Choose CORRECT statement
- (a)
Rutherford's scattering experiment proves the existence of positively charged nucleus
- (b)
The trijectory of a scattered alpha-particle is a hyperpola
- (c)
According to Maxwell's classical theory of electro-magnetic theory, the Rutherford model of atom is unstable
- (d)
ALL OF THE ABOVE
Rutherford's alpha-particle scattering experiment gave experimental information about
- (a)
the charge on alpha-particle
- (b)
the size of atom
- (c)
the size of nucleus
- (d)
the force between alpha-particle and proton in the nucleus
Distance of closest approach of alpha-particle gives the size of the nucleus. This distance is
D= \(\frac { 1 }{ 4\pi \in _{ 0 } } \frac { 2Ze^{ 2 } }{ E_{ k } } \)
where Ek is the kinetic energy of alpha-particles approaching the target nucleus.
An electron revolves round the nucleus of charge Ze. In order to excite the electron from the state n =2 to n =3, the energy required is 47.2 eV. The value of Z is
- (a)
3
- (b)
2
- (c)
5
- (d)
4
Energy of H-atom
\(=13.6\left( \frac { 1 }{ n^{ 2 }_{ 1 } } -\frac { 1 }{ n^{ 2 }_{ 2 } } \right) =13.6\left( \frac { 1 }{ \left( 2 \right) ^{ 2 } } -\frac { 1 }{ \left( 3 \right) ^{ 2 } } \right) \)
\(=13.6\times \frac { 5 }{ 36 } =1.89\quad e\quad V\)
Atomic number of H-atom ZH = 1
\(E_{ n }\propto Z^{ 2 }\)
\(\frac { E_{ 1 } }{ E_{ 2 } } =\frac { Z^{ 2 }_{ 1 } }{ Z^{ 2 }_{ 2 } } =\frac { 47.2 }{ 1.89 } =25\)
Putting Z2 = 1
\(Z_{ 1 }=\sqrt { 25\times 1 } =5\)
The ionisation energy of H2 is 13.6 eV. The ionisation energy of He atom would be
- (a)
13.6 eV
- (b)
27.2 eV
- (c)
6.8 eV
- (d)
54.4 eV
Required ionisation energy = z2 x 13.6 ev
= 4X13.6 eV
= 54.4 eV
Frequency of the series limit of Balmer series of hydrogen atom in terms of Rydberg constant R and velocity of light c is
- (a)
\({ R }_{ c }\)
- (b)
\(\frac { { R }_{ c } }{ 4 } \)
- (c)
\(4{ R }_{ c }\)
- (d)
\(\frac { 4 }{ { R }_{ c } } \)
Wavelength (\(\lambda \)) of the series limit in Balmer series is
\(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ 2^{ 2 } } -\frac { 1 }{ \infty } \right) =R\)
\(\lambda =\frac { 4 }{ R } \)
\(v=\frac { c }{ \lambda } =\frac { Rc }{ 4 } \)
The order of energies of energy levels A, B and C is \({ E }_{ A }<{ E }_{ B }<{ E }_{ C }\). If the wavelength corresponding to transitions \(C\longrightarrow B,\quad B\longrightarrow A\quad and\quad C\longrightarrow A\quad are\quad { \lambda }_{ 1 },\quad { \lambda }_{ 2 }\quad and\quad { \lambda }_{ 3 }\) respectively, then which of the following relations is correct?
- (a)
\({ \lambda }_{ 1 }+{ \lambda }_{ 2 }+{ \lambda }_{ 3 }=0\)
- (b)
\({ \lambda }_{ 3 }^{ 2 }={ \lambda }_{ 1 }^{ 2 }+{ \lambda }_{ 2 }^{ 2 }\)
- (c)
\({ \lambda }_{ 3 }={ \lambda }_{ 1 }+{ \lambda }_{ 2 }\)
- (d)
\({ \lambda }_{ 3 }=\frac { { \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } } \)
E is energy (Ec-EA) =(Ec-EB)+(EB-EA)
ECA= ECB+EBA
\(\frac { hc }{ \lambda _{ 3 } } =\frac { hc }{ \lambda _{ 1 } } +\frac { hc }{ \lambda _{ 2 } } \)
\(\frac { 1 }{ \lambda _{ 3 } } =\frac { 1 }{ \lambda _{ 1 } } +\frac { 1 }{ \lambda _{ 2 } } \)
\(\lambda _{ 3 }=\frac { \lambda _{ 1 }\lambda _{ 2 } }{ \lambda _{ 1 }+\lambda _{ 2 } } \)
Rydberg constant is
- (a)
a universal constant
- (b)
same for all elements
- (c)
different for lighter elements but same for heavier elements
- (d)
different for different elements
The maximum number of photons emitted when an electron jumps from an energy level n = 4 to n = 1 is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
For each .jump of electron from higher to lower energy level. one photon is emitted.
\(4\rightarrow 3;3\rightarrow 2;2\rightarrow 1\)
In an atom, two electrons move around the nucleus in circular orbits of radii 2R and 8R. The ratio of the time taken by them to complete one revolution is
- (a)
1 : 4
- (b)
4 : 1
- (c)
1 : 8
- (d)
8 : 1
Time period,T= \(\frac { 2\pi r }{ v } \)
\(r\propto n^{ 2 }\) ...(i)
\(v\propto \frac { 1 }{ v } \)
\(T\propto n\)
\(T\propto \left( \sqrt { r } \right) ^{ 3 }\) ....From equation (i)
or \(T\propto r^{ 2 }\)
\(\frac { T_{ 1 } }{ T_{ 2 } } =\left( \frac { 2R }{ 8R } \right) ^{ 3/2 }=\left( \sqrt { \frac { 1 }{ 4 } } \right) ^{ 2 }=\frac { 1 }{ 8 } \)
Density of nuclear matter varies with nucleon number A as
- (a)
A
- (b)
A0
- (c)
A3
- (d)
A2
The density of a nucleus is independent of mass number. Radius of a nucleus, R=R0(A)1/3 , where R0 = 1.1X10-10 m
Density of nucleus = \(\frac { mass }{ volume } \)
\(=\frac { A }{ \frac { 4\pi }{ 3 } R^{ 3 }_{ 0 }A } =\frac { 3 }{ 4\pi R^{ 3 }_{ 0 } } \) = constant for all nuclei
If N1 = N0 \({ e }^{ -\lambda { t }_{ 1 } }\), then the number of atoms decayed during time interval from t1 and t2 (t2>t1) will be
- (a)
\({ N }_{ { t }_{ 1 } }-{ N }_{ { t }_{ 2 } }={ N }_{ 0 }[????{ e }^{ -\lambda { t }_{ 1 } }-{ e }^{ -\lambda { t }_{ 2 } }]\)
- (b)
\({ N }_{ { t }_{ 2 } }-{ N }_{ { t }_{ 1 } }={ N }_{ 0 }[{ e }^{ -\lambda { t }_{ 2 } }-{ e }^{ -\lambda { t }_{ 1 } }]\)
- (c)
\({ N }_{ { t }_{ 2 } }-{ N }_{ { t }_{ 1 } }={ N }_{ 0 }[{ e }^{ \lambda { t }_{ 2 } }-{ e }^{ -\lambda { t }_{ 1 } }]\)
- (d)
NONE OF THE ABOVE
Nt1 = N0et1
Nt1 = N0et2
Number of atoms decaying during time interval t1 and t2 is
Nt1- Nt2 = N0 [ et1 - et2]
Complete the equation for the following fission process :
\({ _{ 92 }U }^{ 235 }+{ _{ 0 }n }^{ 1 }\longrightarrow ...+{ _{ 38 }Kr }^{ 90 }+...\)
- (a)
\({ _{ 54 }Xe }^{ 143 }+3{ _{ 0 }n }^{ 1 }\)
- (b)
\({ _{ 54 }Xe }^{ 145 }\)
- (c)
\({ _{ 57 }Xe }^{ 142 }\)
- (d)
\({ _{ 54 }Xe }^{ 142 }+{ _{ 0 }n }^{ 1 }\)
\(_{ 92 }U^{ 235 }+_{ 0 }n^{ 1 }\rightarrow _{ 54 }x^{ 143 }+_{ 18 }k^{ 81 }+3_{ 0 }n^{ 1 }\)
Apply law of conservation of mases number and atomic number They should be same on L.H.S and R.H.S
Mass number : L.H.S 235+1 =236
R.H.S 143 +90+3=236
Atomic number ; L.H.S 92+0=92
R.H.S = 54+38+0=92
Hence alternative (a) is correct.
The energy released in nuclear fission mostly appears as
- (a)
heat
- (b)
kinetic energy of fission fragments
- (c)
radio-active radiations
- (d)
chemical energy
The fusion of hydrogen into helium is more likely to take place
- (a)
at high temperature and high pressure
- (b)
at high temperature and low pressure
- (c)
at low temperature and low pressure
- (d)
at low temperature and high pressure
The half life of radio-active substance is 40 days. The substance will disintegrate completely in
- (a)
40 days
- (b)
400 days
- (c)
4000 days
- (d)
infinite
From the following equations pick out the possible nuclear fusion reaction
- (a)
\({ _{ 6 }C }^{ 12 }+{ _{ 1 }H }^{ 1 }\longrightarrow _{ 7 }{ N }^{ 13 }+2\quad MeV\)
- (b)
\(4({ _{ 1 }H }^{ 1 })\longrightarrow _{ 2 }{ He }^{ 4 }+2{ e }^{ + }+26\quad MeV\)
- (c)
\({ _{ 1 }H }^{ 2 }+{ _{ 1 }H }^{ 2 }\longrightarrow { _{ 2 }He }^{ 4 }+Q\)
- (d)
ALL OF THE ABOVE
Nuclear forces are not
- (a)
always attractive
- (b)
stronger than gravitational force of attraction
- (c)
short range
- (d)
charge dependent
Nuclear forces are stronger than any other torce like gravitational (electrostatic and Magnetic) forces [or electromagnetic forces].
Nuclear forces are short range, always attractive and charge independent.
A nuclear holocaust of fusion bomb will make weather forever
- (a)
rainy
- (b)
hot
- (c)
cold
- (d)
changing rapidly
Nuclear waste after explosion hangs like a cloud in the earth atmosphere and absorbs the sun's radiation. This leads to a long winter and hence in turn, cold.
The radius of Bohr's first orbit is a0. The electron in nth orbit has a radius
- (a)
nao
- (b)
\(a_o\over n\)
- (c)
n2ao
- (d)
\(a_o\over n^2\)
\(r_n={r_on^2\over Z}, for \ H_2; Z=1\)
\(r_n=n^2a_0\)
Radius of the second Bohr orbit of singly ionised helium atom is
- (a)
0.53Ao
- (b)
1.06Ao
- (c)
0.265Ao
- (d)
0.132Ao
\(r_n=0.53{n^2\over Z}={0.52X2^2\over 2}=1.06A^o\)
An electron in Bohr's theory of hydrogen atom has an energy of -3.4eV. The angular momentum of the electron is
- (a)
\(h\over \pi\)
- (b)
\(h\over 2\pi\)
- (c)
\(nh\over 2\pi\)
- (d)
\(2h\over \pi\)
Given, \(E_n=-3.4 eV\)
n=2
\(L={nh\over 2\pi}\)= \(h\over\pi\)
Monochromatic radiation of wavelength \(\lambda\) is incident on a hydrogen sample which is in ground state. Hydrogen atoms absorb the light and subsequently emit radiations of ten different wavelengths. The value of \(\lambda\) is
- (a)
95 nm
- (b)
103 nm
- (c)
73 nm
- (d)
88 nm
Number of spectral lines obtained = \({n(n-1)\over 2}=10\)
n=5
So, energy required =E5-E1=13.06eV
and \(\lambda\) =photon wavelength = \(hc\over \Delta E\)=\(1240eV-nm\over13.06eV\)=95nm
A set of atoms in an excited state decays
- (a)
in general to any of the states with lower energy
- (b)
into a lower state only when excited by an external electric field
- (c)
all together simultaneously into a lower state
- (d)
to emit photons only when they collide
De-excitation is simulated and spontaneous
An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently, the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum KE of colliding electron
- (a)
10.2 eV
- (b)
1.9 eV
- (c)
12.1 eV
- (d)
13.6 eV
Energy of electron = excitation energy
= energy of photon emitted
=energy corresponding to 1st transition in Balmer series
=E3-E2=-1.51-(-3.40)=1.9eV
The radius of the shortest orbit in a one-electron system is 18pm. It may be
- (a)
hydrogen
- (b)
deuterium
- (c)
He+
- (d)
Li++
As, \(r_n={0.53{n^2\over Z}}={0.53D\over Z}=18pm(n=1)\)
\(\Rightarrow {0.53X10^{-10}\over18X10^{-12}}=Z\)
\(\Rightarrow Z={53\over 18}=2.944=3\)
This corresponds to Li2+
Hydrogen (1H1), Deuterium (1H2), singly ionised Helium (2He4)+ and doubly ionised lithium (2Li6)++ all have one electron around the nucleus. Consider an electron transition from n=2 to n=1. If the wavelengths of emitted radiation are \(\lambda_1,\lambda_2,\lambda_3 \ and \ \lambda_4\) respectively then approximately which one of the following is correct?
- (a)
\(\lambda_1=\lambda_2=4\lambda_3=\lambda_4\)
- (b)
\(\lambda_1=2\lambda_2=3\lambda_3=4\lambda_4\)
- (c)
\(4\lambda_1=2\lambda_2=2\lambda_3=\lambda_4\)
- (d)
\(\lambda_1=2\lambda_2=2\lambda_3=\lambda_4\)
As, \(\frac { 1 }{ \lambda } =RZ^{ 2 }\left( \frac { 1 }{ 1^{ 2 } } -\frac { 1 }{ 2^{ 3 } } \right) \)
\(\therefore \lambda =\frac { 4 }{ 3RZ^{ 2 } } \Rightarrow \lambda _{ 1 }=\frac { 4 }{ 3R } \)
\(\lambda _{ 2 }=\frac { 4 }{ 3R } \)
\(\Rightarrow \lambda _{ 1 }=\frac { 4 }{ 12R } \)
\(\Rightarrow \lambda _{ 4 }=\frac { 4 }{ 27R } \)
\(\Rightarrow \lambda _{ 1 }=\lambda _{ 2 }=4\lambda _{ 3 }-9\lambda _{ 4 }\)
Which of the following transitions in hydrogen atoms emit photons of highest frequency?
- (a)
n=2 to n=6
- (b)
n=6 to n=2
- (c)
n=2 to n=1
- (d)
n=1 to n=2
Emission spectrum would rises when electron makes a jump from higher energy level to lower energy level.
Frequency of emitted photon is proportional to change in energy of two energy levels, i.e., \(f\propto \) charge in energy
\(f=RcZ^{ 2 }\left( \frac { 1 }{ n^{ 2 }_{ 1 } } -\frac { 1 }{ n^{ 2 }_{ 2 } } \right) \)
highest frequency corresponds to n = 2 to n = 1