 fine so now we will take atoms question this came in 1994 came in 1994 for 5 marks so hydrogen atom hydrogen like atom okay with atomic number z this is not hydrogen atom hydrogen like is in higher excited state of quantum number n quantum number n it is in excited state the excited atom can make a transition to first excited state this can go to what is the first excited state 2 n1 equal to 2 n1 equal to n from n it goes to n equal to 2 by emitting 2 photons of 10.2 electron volt and 17 electron volt respectively okay this is given understood right atomic number z x that is an excited state n goes to n equal to 2 by emitting 2 photons successively yes alternatively the atom from the same excited state can make a transition to second excited state as in from n to it goes to 3 by emitting 2 photons of 4.25 electron volt and 5.95 electron volts respectively okay determine the value of n and z just write down two equations you'll get what is given is ionization energy for hydrogen atom is 13.6 electron volt which i guess you know it already you know the formula energy in the nth orbit for hydrogen like atom is what r into z square n square this is 13.6 by n square n z square so definitely the value of n should be more than 3 right because it talks about jumping to n equal to 3 yes sir sir is n 5 there's n 5 question number 20 no n is not given as 5 you might have done some cellular okay should i do it very easily right yeah okay let me do it now so the atomic number is z right and let's say content number is n so energy in the nth state should be this and energy in the second state will be equal to what will be equal to 13.6 divided by 2 square which is 4 right this into z square now what is the difference in these two energy levels it will be some of these two yes or no yes sir right kondanya and ramcharan yes sir yes sir this oh you you got you you got six right yes sir and kondanya you got sir no sir okay so see i'm not converting electron volt into joules because you know i am finding the left hand side energy also in terms of electron volt this is your first equation and then you can write the second one also 13.6 divided by 3 square into z square minus 13.6 by n square z square is equal to 4.25 plus 5.95 right so this is your equation number 2 fine so if you subtract them you you'll get you know only z in the expression and you'll get z as 6 okay and then you substitute the value of z in the first equation you'll get the value of n are you guys clear yes sir yes sir okay now the natural tendency when you start solving this question is to try to find out that it emits two photons so what is the intermediate level right from n it goes to which level and then n equal to 2 it goes right so that that is what is our natural tendency because that that's how we are trained to solve every question we go step by step manner okay but then if you look at what is given and what you have to find okay so you'll get a direct relation between initial and final are you getting it yes yes okay let us take another question okay this came in 2002 for five marks when i wrote j and i don't even remember whether i got it correct or not okay so hydrogen like atom again it's a hydrogen like atom it's not a hydrogen atom is observed to emit six wavelengths okay it can emit six wavelengths fine so it can emit six different types of photons originating from all possible transitions between the group of levels so all possible transition is happening and you're getting six different wavelengths okay these levels have energies between minus 0.85 electron volt and minus of 0.5 double four electron volt fine so this is your initial energy and this is your final i mean between these two energy levels six wavelengths are containing you need to find out first the atomic number which is z z is equal to what and second at the first you find out atomic number then we'll talk about the second okay are you guys stuck or able to proceed stuck stuck so six six possible wavelengths right so which level that means from fourth fourth level fourth level right so if you see that number of wavelengths that are emitted when it is at nth state is nc2 right so n into n minus one into two this should be equal to six so n should be equal to four fine so this is clear cut that so this is when it's uh to the from some level to the ground state no all possible so if it is at n equal to four then if it is here then only it'll emit six different wavelengths otherwise six is not possible so this is from four to one it can also be from five to two like that right no you're not getting it what i'm trying to say here see if you are at n equal to four only six possible wavelengths are there if you are at n equal to five then how many possible wavelengths are there five into four divided by two ten wavelengths are there okay but the question says that it emits six wavelength originating from all possible transitions so only n equal to four has exactly six because you can't stop at n equal to two the second possible transition is from two to one but it can't go below one are you getting it ramcharan yes okay so we got that initial level is four okay now this is the energy in the fourth 0.85 electron volt fine oh sorry ramcharan you're correct you're correct it uh it says that all possible transition between these two energy levels oh okay so uh it it's not uh all possible uh transition till the ground level it need not be till the ground level okay but one thing is clear that whatever is happening there's if this is n1 what should be this n1 plus 3 are you clear about it yes right so if n1 correspond to energy let us say this is uh this will have lower energy right so energy corresponding to n1 should be equal to minus of 0.85 minus of 0.85 electron volt and energy corresponding to n1 plus 3 should be equal to minus of 0.544 electron volt okay we'll get two equation minus 13.6 divided by n square z square this will be equal to minus of 0.85 okay this is equation number one and you will be getting minus 13.6 by n plus 3 whole square into z square this is equal to what minus of 0.544 you got these two equations right yes yes yes so if you if you divide them you will get rid of z basically you'll get n plus 3 divided by n this will come out to be square root of 0.85 divided by 0.544 it is little calculation intensive but uh are you able to do it anyways once you solve this you will get answer 33 so n equal to 12 you will be getting okay you'll get n equal to 12 you might have to use log and anti-log to solve this particular question fine yes sir okay let us move to next question see you might be appreciating the questions which are actually coming in the exam they are different from the routine question which you guys might be practicing okay so uh I mean make sure you uh you solve the past year j questions okay I mean I'm not saying that you should do it right away or finish it off very quickly but keep in your mind that before taking the j exam or let's say j advance or take some I mean have some target in mind suppose by February or end of February you should be done with past year j question these questions are must before you take uh j exam okay let's do this question this came in 2006 for six months all right so in a hydrogen like atom the atomic number is 11 all right the nth line nth line of limon you know right limon series yes yes so what what is what is the lower content number in limon series and one is equal to one one good so nth line of limon series has a wavelength lambda okay and it is given that d brugli wavelength d brugli wavelength of the electron in the level okay in the level from which it originated let me read this statement again the deep rule is wavelength of the electron in the level from which it originated is also lambda getting the question yes sir you have to find the value of n what is this principal content number let me know if you have not understood the question i'll read it again so could you read it once more okay in a hydrogen like atom where z is equal to 11 nth line of limon series has wavelength lambda okay nth line of limon series has wavelength lambda the d brugli's wavelength of electron in which d brugli wavelength of electron in the level from which it originated is also lambda find the value of n i hope you have understood the question nth line of limon series means what is the content number is it n no content number is one no the second content number is one it should land up in n equal to one what's from where okay yeah from the nth all right no it is nth line it is not nth content number so yeah n plus one n plus one should i do it so no so calculate by the way the Bohr's model the second postulate which says that m v r is equal to n h by 2 pi this is valid for hydrogen like atoms also there is no atomic number though in this okay and you know if you combine d brugli and the Bohr's model you should also get 2 pi r should be equal to n times lambda where n is a content number okay so these are the hints now can you try it out where r is equal to 0.5 i think i said 53 amstrong is the Bohr's radius the first orbit that's not correct actually it is 0.52 amstrong okay 0.5729 into n square by z this is the radius of the nth orbit for a Bohr's so sorry for for a hydrogen like atom if if the content number is n plus one then you can just substitute the value of n here as n plus one okay so you'll be getting here as 2 pi r is equal to n plus one times lambda right and r is what r is 0.529 into n square now here n is n plus one this is n plus one square by z this is equal to n plus one times lambda okay so you will be able to get the value of lambda from here lambda will be equal to 2 pi 0.529 divide by z is given 11 fine this into n plus one this should be the value of lambda okay fine and it is given this is using this d brugli condition this is the wavelength of the electron at n plus 1th level which is nth line of lineman series okay this is your first equation and this wavelength should also be the wavelength of the photon that is emitted from the nth line which is what which is given by this equation 1 by lambda should be equal to rickbook constant into z square also right rickbook constant z square multiplied by 1 divided by n plus 1 whole square sorry the first level I should write it as 1 lineman series 1 square minus 1 divided by n plus 1 whole square okay so if you use these two equations okay you can write down the value of n plus 1 from here and substitute it here you get the value of lambda getting it so these are the two equation lambda and n they are the two variables and you get the answer okay okay