Numericals : 2 - Dual Nature of Radiation & Matter Video Lecture - Class 12

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assume that the molecule is moving with
the root-mean-square speed of molecules
at this temperature so now here we have
to calculate the de Broglie wavelength
of nitrogen molecule now for de Broglie
wavelength we need mass and velocity so
in this case it says that velocity is
nothing but root mean square speed
because the molecule is moving with root
mean square velocity right so what would
be the kinetic energy of the molecule it
will be nothing but half M V RMS square
right also we know that at a temperature
T the kinetic energy is related to
temperature by this relation T by 2 KT
where K is Boltzmann constant we have
studied all these things in our previous
lessons right in class 11th we have
studied all these relations so kinetic
energy is on one side is equal to half
MV square on the other hand it is also
equal to 3 by 2 K T where T is the
temperature right now let us calculate
the mass of nitrogen molecule because in
order to calculate de Broglie wavelength
we need mass as well as V RMS so for
mass we need to calculate the mass of
nitrogen molecule so this will be this M
is mass of n2 molecule now what is given
in the problem atomic mass of nitrogen
is given here now one molecule of
nitrogen consists of two atoms so one so
the mass of one nitrogen molecule is
equal to 2 into atomic mass because one
nitrogen molecule consists of two
nitrogen atoms so this will be 2 into 14
point 0 0 7 6 atomic mass unit so if
you want to convert it into kgs what is
one atomic mass unit it is nothing but
mass of one proton which is 1 point 6 7
into 10 to the power minus 27 kg s so
this is the mass of one nitrogen
molecule right now let us calculate so
we have calculated M now we have to
calculate V RMS right now from Equation
1 and equation 2 from 1 & 2 we can say
that half M V RMS square is equal to 3
by 2 KT so 2 2 will get cancelled from
this we can say V RMS square is equal to
3 KT divided by M or we can say V RMS is
equal to root over 3 KT divided by M so
we know K which is Boltzmann constant
that is 1 point 3 8 into 10 to the power
minus 23 T is given as 300 Kelvin and M
we have already calculated here so now
we can calculate lambda is equal to H
divided by M into V RMS so this is equal
to 8 divided by M into V RMS we can
write it as 2 over 3 KT divided by M so
this can be written as H divided by root
over 3 KT M now if we put all the values
here we get we find that lambda comes
out to be zero point zero two eight
nanometers so this comes out to be the
wavelength or the de Broglie wavelength
of a nitrogen molecule let us look at
problem 10 an x-ray tube produces a
continuous spectrum of radiation with
its short wavelength end at 0.45 and
strong so that means the wavelength is
given at as 0.45 and
from which is nothing but 0.45 will be
10 to the power minus 10 meters what is
the maximum energy of a photon in the
radiation so the wavelength of the
radiation is given so what would be the
energy of the photon energy of a photon
is given by H nu which can be written as
HC by lambda because frequency is
related to wavelength by this relation
nu is equal to C by lambda so this comes
out to be 6 point 6 7 into 10 to the
power minus 34 into 3 into 10 to the
power 8 divided by 0.45 into 10 to the
power minus 10 so this comes out to be
44 point 4 6 into 10 to the power minus
16 choose if you want to convert it into
electron volts which is divided by 1
point 6 into 10 to the power minus 19
joules which comes out to be 27 point 6
kilo electron volts so this would be the
maximum energy of a photon in the
radiation
here comes problem number 11 in an
accelerator experiment on high-energy
collisions of electrons with positrons a
certain event is interpreted as an
emulation of an electron-positron pair
of total energy 10 or 8 to be Evi into
two gamma rays of equal energy that
means it was talking about a process of
annihilation of an electron positron
rare that means an electron is combining
with a positron and this results in two
gamma rays of equal energy what is the
wavelength associated with each gamma
ray so we have to calculate the
wavelength associated with each gamma
ray it is given that the energy of each
gamma ray is the same so if this is e
this is also e now in this case what is
the thunder that we will apply here the
total energy remains conserved in case
of a collision right so what is the
total energy total energy should
be conserved right so let us calculate
the total energy before collision before
collision what is the total energy an
inhalation of electron-positron pair of
total energy 10.2 vev so the total
energy before collision is 10.2 de Vie
which comes out to be invaluable in
electron volt is 10 to the power 9
electron volts that is 10 point 2 into
10 to the power 9 electron volts this
comes out to be 10 point 2 into 10 to
the power 9 into 1 point 6 into 10 to
the power minus 19 joules so this is the
total energy before collision
now what happens after collision
now after collision total energy let us
call this energy as T 1 let us call this
total energy as T 2 so what will be the
total energy after collision after
collision what happens to gamma
particles of equal energies are formed
so let us suppose each of them has
energy D so what is the total energy
after collision 2e right now
according to conservation of energy T 1
should be equal to T 2 that is total
energy before collision should be equal
to total energy after collision so 10
point 2 into 10 to the power 9 into 1
point 6 into 10 to the power minus 19
should be equal to 2e this implies T
will be equal to 10 point 2 into 10 to
the power 9 into 1 point 6 into 10 to
the power minus 19 divided by 2 so this
e comes out to be eight point one six
into 10 to the power minus 10 joules so
this would be the energy of each
gamma-ray what is e e is the energy of
each gamma-ray we have to calculate the
wavelength associated with each gamma D
so how do we calculate the
wavelength goes to retail if this e is
the energy associated so this energy is
equal to H nu and this is equal to HC by
lambda so from this we can say that
wavelength lamda is equal to HC divided
by e so H is 6 point 6 7 into 10 to the
power minus 24 into 3 into 10 to the
power 8 this whole divided by e which is
a 20 26 into 10 to the power minus 10 so
what is the wavelength so we get that
lambda is equal to 2 point 4 3 6 into 10
to the power minus 16 meters right okay
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