Bohr model numericals - Structure of Atom Video Lecture - Class 11 | Best Video for Class 11

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atom part 23 is brought you by exam fear
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watching this video please make sure
that you have watched part 1 to part
went to so the question is find the
frequency in wavelength of a photon
emitted during transition from n is
going to 5 to n is going to that is one
polymer in hydrogen so I know that e is
nothing but 2 point 1 8 into 10 to the
power minus 18 into when my n initial
minus n fine both spot this is the
energy right you put the values here so
this become 2 point 1 8 into 10 to the
power minus 18 n initial was 5 so v
square minus n final verse 2 2 square so
if you solve this you get minus 4 point
5 8 into 10 to the power minus 9 so this
much energy is emitted when the
electrons jump from 5 to 2 if you see 5
to 2 energy has to be emitted and this
much energy is emitted when the
electrons jump from 5 to also- science
miss the mission is negative that means
it has emitted it has lost more actually
also naívi got the energy we can find
the frequency is equal to H mu very
easily we know the value of H you know e
you can find the frequency also we
define wavelength this is nothing but HC
by lambda so we know H you know C you
can find lambda so this we can do on you
you define again the wavelength of light
emitted with the electron of the
hydrogen atom move from 4 to 2 similar
version e is equal to 0.18 18 1 by n
initial square minus n by n final square
so any nature was 4 finalists to 2.18
1824 -
correct so if you solve this you get
minus 4 point 0 8 7 5 into 10 to power
minus 19 Joule this is what you get this
is negative that means it is imagine
here
you can use the same formula e is equal
to HC by lambda so lambda you will get e
by HC I have got the value of e this
value H sorry lambda is HC by lambda is
HC by H I know clients constant C I know
speed of light e I know I just form put
the values you get for 686 nanometers as
the answer how much energy is required
to ionize hydrogen if the electron of a
bias n is equal to 5
orbit compare your answer with the
ionization enthalpy of hydrogen energy
required to remove electron from n is
equal to so you are saying that in one
case you have to remove hydrogen from n
is equal to 5 second years you are asked
to remove hydrogen from n is going to
one orbit let's see the difference so
the formula says e
Delta e to be specific running back to
point 1 8 into 10 to power minus 18 1
minus initial minus n final square N
final is my infinity here always so in
case 1 and initial was 5 so this becomes
2 point 1 8 into 10 to power minus
getting 1 by v square minus 1 by
infinity square this becomes 0 if you
solve this you get eight point 17 into
10 to power minus 2 n DSU in second case
same formula 2 point 1 a into a row
minus 18 here n initial is 1 1 by 1
square n final is infinity in the square
so you saw Lizzie gate to point 1 8 into
10 to power minus 80 zoo
make everything 10 to the power 18 only
right so this becomes point zero eight
seven so you see this guy is less and
this is more does you can see that to
remove the electron from fifth orbit it
is easy to remove electron from first
orbit is defeated sorry what is the
maximum number of invest in line when
they like excited electron of hydrogen
is n is equal to six drops to Knowles
tape so if you see from five it can go
to 4 5 2 3 5 2 to 5 to 1 right or sorry
it goes from six six two five six two
four six two three six two two six two
one a so there are five possible option
here from fifth it can go possible
option from for can go to three or two
to four to one key possible option from
third it can go to to third it can go to
one two possible options from second it
can go to one only one possible version
one possible option you add all five
plus four plus three plus two plus one
you get okay so if you see here from six
it can go to 5 it can go to a fool it
can go to 3 it can go to do it can go to
one find out from fifth it can go to 4
it can go to 3 it can go to 2 equal to 1
so possible fourth it can go to 3 it can
go to 2 it can be the one he bustled
option from 3 it can go to 2 and you do
one two possible option from - it can go
to only one so there are 15 possible
options right
so you see this is also equal to n into
n minus 1 right you'll see this if you
put the value of n will 6 this becomes 6
into I invite to you this way so you can
write as n formula for the maximum
number of MS and lines for n is equal to
2 or for the highest value of n to the N
you get n into n minus 1 by 2 maximum
deviation that's correct because it's AP
actually if you see right 1 plus 2 Plus
3 plus 4 plus 1 with a difference of 1
this becomes n into n minus 1 by 2 so we
have seen that in this case the energy
associated with the first orbit is this
guy we have to find the energy so it
with the 5th offered C energy associated
with any orbit something but some
constant by n Square where n is my orbit
number so you can see that energy should
be the 5th orbit is nothing but energy
by one or by 5 square because energy of
1 is nothing but 1 here correct energy
certainly v is nothing but energy it
will be the first orbit by PI square
this is nothing but minus 2 point 1 8
into 10 to power minus 18 by PI square
is 25 we saw this is equal to minus 8
point 7 1 minus 20 G's correct and if
you talk about radius radius and the
Corbitt is nothing but point 0 5 to mine
nanometer
into n square so radius of v orbit is
nothing but point zero five two nine
nanometer into pi square so we solve
this this becomes 1 point 3 2 pi and
that's where it is of 5th orbit easy to
find
find the wave number of the longest
wavelength transition the bomber series
of hydrogen atom so in bomber series is
my n final is always do you notice
bombers it is n final is always - I have
to find the longest wavelength so energy
you know this guy for a jump is nothing
but 2 point 1 8 into 10 to power minus
18 1 minus n initial square minus n
final squares are making 2 right and
this is also equal to energy is also
equal to e in terms of wavelength HC by
lamda and we see everything is constant
here H is constant C is constant with a
is constant to his concert correct so
for max lamda the maximum and this has
to be mean you write this can be minimum
when this guy is very close to 2 so the
minimum value of initial can be see the
ni n I can be more than two a three four
five six seven like this I am saying I
want to maximize this lambda so minimize
this to minimize this I have to give
this guy's very even value correct so
the minimum value can be three and that
is my ID to find the wave number also
find out calculate the wave number
actually so for n is equal to n I is
going to create you find the wave number
so let us do that and I use the formula
V number is one point zero nine into
candy bar seven into and initial is 3 3
square 9 - finals
- two squares for this is what I get and
there is only one 1 by 5 - 3 6 into 10
to the power 6 per meter or what again
do is you can use this formula only and
you find the value of one malama there
is a wave number correct so the question
is where the energy in joules
please note energy in joules required to
shoot electron from n is equal to 1 to n
is equal to 5 I have this like this so n
is equal to 1 is here 1 2 3 4 5 from
here the electron jumps to here I had to
find the energy required also this
electron again jumps back to 1 and it
emits of light I have to find a
wavelength of the light which is the
first part Delta e when it jumps from 1
to 5 nothing bar 2.1 18 build a power
minus 18 1 by any nation is 1 1 square
minus n final is 5 5 squared 25 solve
this we get to the point 0 9 2 8 into 10
to the power minus 18 jus this is the
energy is required to move electrons
from n is equal to 1 to 5 and the same
energy will be released is the electrons
jump from 5 to 1 right so this is
nothing but e will be equal to energy
release will be equal to HC by lambda I
have to find lambda and energy is
nothing but the same value right so I
can say lambda is equal to Delta e by C
if you understand the same energy
release when electron jumps back from PI
to 1 Delta is lambda is HC by Delta
itself but the value of H
six point six to 6 into 10 to the power
minus 34 joule-seconds C is 3 into 10 to
the power speed of light winter per
second and Delta e we have in zoom that
is 2 point 0 ni into 8 into 10 to power
minus 8 region so we solve this you get
lambda is 9 point 4 9 8 to 10 to power
minus 8 min this is the wavelength of
the light emitted when the electron
jumps back from here to here
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