CBSE Class 12 Physics 2020 Outside Delhi Set 1 Paper

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Question : 37
Total: 37
(a) Derive the law of radioactive decay N=Noeλt.
(b) The half life of 92238U undergoing α-decay is 4.5×109 years. Find its mean life.
(c) What fraction of the initial mass of a radioactive substance will decay in five half-life periods ?
OR
(a) State the postulates of Bohr's model of hydrogen atom and derive the expression for Bohr radius.
(b) Find the ratio of the longest and the shortest wavelengths amongst the spectral lines of Balmer series in the spectrum of hydrogen atom.
Solution:  
(a) Radioactive decay law : The rate of disintegration of a radioactive substance at an instant is directly proportional to the number of nuclei in the Radioactive substance at that time.
N=Noeλt , where symbols have their usual meanings
Let us consider a radioactive substance having No atoms initially at time (t=0) .
After time (t) , no. of atoms left undecayed be N.
If dN is the no. of atoms decayed in time dt , then according to radioactive decay law:
dN
dt
α
N

Or,
dN
dt
=λN
.......(i)
Here, λ is decay constant and negative sign indicates that a radioactive sample goes on decreasing with time.
Equation (i) can also be written as
dN
N
=λdt

Integrating
lnN=λt+K ........(ii)
Here, K is constant of integration
At t=0,N=N0
K=lnN0
Substituting K in equation (ii),
lnN=λt+lnN0
ln
N
No
=λt

Or,
N
No
=eλt

N=N0eλt
(b) Half-life =( Mean life )×ln2
Or, 4.5×109 years =( mean life )×ln2
Or, Mean life =(
4.5×109
ln2
)
years
Or, Mean life =(
4.5×109
0.693
)
years =6.5×109 years
(c) No. of half lives =5
So,
N
No
=(
1
2
)
5
=
1
32
=
fraction of mass of the radioactive substance left undecayed
So, fraction of mass decayed =1
1
32
=
31
32

OR
(a) Postulates of Bohr Model of Hydrogen atom:
Postulate - I: The electrons revolve in a circular orbit around the nucleus. The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons provide necessary centripetal force for circular motion.
Postulate - II: The electrons can revolve only in certain selected orbits in which angular momentum of electrons is equal to the integral multiple
h
2π
, where h is Planck's constant. These orbits are known as stationary or permissible orbits. The electrons do not radiate energy while revolving in theses orbits.
Postulate - III: When an electrons jumps from higher energy orbit to lower energy orbit, energy is radiated in the form of a quantum or photon of energy hv, which is equal to the difference of the energies of the electron in the two orbits.
Expression for Bohr radius :
Let us consider
m= mass of an electron
r= radius of the circular orbit in which the
electron is revolving
v= speed of electron
e= charge of electron
From 1st postulate
Centripetal force = Electrostatic force
mv2
r
=
1
4πε0
e2
r2

v2=
1
4πε0
e2
mr
.......(1)
From 2nd postulate
mvr=
nh
2π

Or, v=
nh
2πmr

Or, v2=
n2h2
4π2m2r2
.......(2)
Comparing eqns (1) and (2),
1
4πε0
e2
mr
=
n2h2
4π2m2r2

Or,
1
4πε0
e2
mr
=
n2h2
4π2m2r2

Bohr radius =r=
ε0n2h2
πme2

(b) Shortest wavelength in Balmer series:
1
λS
=R(
1
22
1
)

λS=
4
R

Longest wavelength in Balmer series:
1
λL
=R(
1
22
1
32
)

λL=
36
5R

So,
λL
λS
=
36
5R
4
R
=
9
5
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