 Hello guys, welcome to the session. Today in the session, we are going to discuss some problems, subjective problems, based on atomic structure. So let's start with the session. So the first question here, you can see. If you want to try the question, you can pause the video and try. We'll start discussing the question. Excuse me. The question is naturally occurring boron consists of two isotopes. Consist of two isotopes whose atomic weight is given. The atomic weight of the natural boron is this, calculate the percentage of each isotope in natural boron. Is the question, everything is given, just you need to use the formula and get the answer. Suppose we have two isotope of boron. It is given in the question. So boron has two isotope. One has atomic mass 10.01. And the other one has 11.01. Percentage competition we need to find out. If it is x, then this we can assume it is 100 minus x. So the atomic weight of boron, atomic weight equals to 10.01 into x plus 100 minus x into 11.01 divided by the total composition we are assuming 100. And we have to solve this equation. So this left hand side, the value is 10.81, 10.01 x plus 11.01 minus 11.01 x divided by 100. So it is x left hand side equals to 11.01 minus 10.81. So it is 20. So x is, so this one is 20%. This one is 20%. And this one is rest of it. That is 80% present in the mixture, boron mixture. So answer is 20% and 80%. OK. The next one is 59. The energy of electron in the second and the third bores orbit of hydrogen atom is this erg. It is given in erg. Second and third bores orbit is given. Calculate the wavelength of the emitted radiation when the electron drops from third to second. OK. So the energy that is given in second, it is E2. 2 is, E2 is equals to, let's write all the data first, minus 5.42 into 10 to the power minus 12 erg. And third, E3 is equals to minus 2.41 into 10 to the power minus 12 erg. Sorry. So the electron jumps from second, it says, just from third to second orbit. Delta E is what? Delta E is E3 minus E2. That would be E3 minus E2, so 3.01. OK. When you solve this, you'll get 3.01 into 10 to the power minus 12 erg. Right? This is delta E. And this equals to we can write at C by lambda. OK. It is given in erg. So we can write down this as 3.01, or lambda is equals to because lambda we need to find out. So lambda is equals to 6.6 approximately, I'm taking. 10 to the power minus 27 erg into C. We should write that in CGS system. So the value is 3 into 10 to the power 10 centimeter per second divided by the energy 3.01 10 to the power minus 12. So if you take the approximation, then what we can say, this 3 and 3 will get cancelled. This becomes 10 to the power minus 15 and 10 to the power minus 5. So it is 6.6 into 10 to the power minus 5 we are getting. OK. That is centimeter because the CGS system is centimeter. Centimeter, if you convert this into angstrom, we know one angstrom is equals 10 to the power minus 10 meter or 10 to the power minus 8 centimeter. OK. So if you multiply this by 8 into 8, 6.6 into 10 to the power 3 angstrom. So this would be 6600 angstrom approximately. OK, approximate value is this. OK. The question was not difficult, but you have to take care of units here because R is given. Remember always, R is the CGS unit. OK. So everything I have taken in CGS here. Or you can do one more thing. You can convert this into joule here by 1 joule is equals to 10 to the power minus 7 R, 10 to the power 7 R. Right. So 1 R is equals to 10 to the minus 7 joule. That way also you can convert this into joule and then we can use this unit in angstrom or in SI system. Right. So this is the two subjective questions we have. You see the simple question? OK, this one. OK, the electron energy in hydrogen atom is given by this. OK, in hydrogen atom. OK, it's given by this. Calculate the energy required to remove an electron completely from N is equals to 2 orbit. N is equals to orbit. What is the longest wavelength of light that can be used to cause this transition? Fine. So similarly, we can find out delta E. Right. Delta E or simply E is equals to Hc by lambda we can use. OK. So in the second energy, second orbit, the energy is required to remove an electron is. The formula is given. Hence the question becomes so easier. Sometimes what happens nowadays, they won't give you the formula here. So that also you have to memorize. So formula is given E is equals to minus 21.7 10 to the power minus 12 by N square. Erg. Actually, the formula is Z square by N square. So Z is 1. Hence it is hydrogen atom. Hence the formula becomes this. OK. So for N is equals to 2, for N is equals to 2, the energy E is minus 21.7 into 10 to the power minus 12 divided by 4. So when you solve this, you'll get 5.4 approximately minus 5.42 into 10 to the power minus 12. So this is the energy in second orbit. E2 is this. Energy in the second orbit. E2 is this. OK. So this is the energy required to remove an electron because the energy that is there with the electron, the same energy you have to provide in the positive energy so that this negative positive becomes 0 and the electrons becomes free. OK. Now, the second question is what is the longest wavelength in centimeter of light that can be used to cause this transition? OK. So usually what happens is this transition happens when we provide the energy and this energy is what? This energy is this. The transition means what? We also calculate the energy required to remove an electron because you see at infinity, E infinity is 0. That is what the concept we have. And here the electron is free to move. So we have, suppose, the second orbit here and the infinite orbit, suppose we have here, second orbit and infinite orbit. So if you want to remove this electron, obviously you have to provide energy and that energy is the difference here. This energy difference, this delta E you have to provide. So usually they ask that what is the energy required to remove an electron? What we do if the electron present in nth shell, then it is minus 21.7. The formula is into 10 to the power minus 12. 1 by n square minus 1 by infinity square because this is the energy required to remove an electron. Because to remove this electron, you have to provide the energy difference between the orbit in which the electron is present and the energy at the infinite orbit, which is 0, we are 0. That is why it says that what is the longest wavelength of light that can be used to cause this transition from here to here? Because electron, the energy you have to provide in, you have to provide some energy in the form of light, radiation. So what wavelength we provide so that it gains this amount of energy and the electron becomes free from the nucleus. So this delta E we have already calculated. That is minus 5.42 into 10 to the power minus 12 is equals to Hc by lambda, 6.6 into 10 to the power minus 27. 3 into 10 to the power 10. And this divided by the lambda is a wavelength. So when you solve this for wavelength, you will get lambda is equals to 6.6 into 10 to the power minus 27 into 3 into 10 to the power 10 divided by 5.42 into 10 to the power minus 12. Now again, one thing you see here, the energy difference is what? The energy difference is positive always, because you always find out E greater minus, like no, we are providing this energy here. So this energy, the magnitude is this one. So we'll take the magnitude here, because the negative wavelength does not mean anything. So we'll take the positive energy here. That's why the magnitude I am taking here positive. So when you solve this, the approximate answer would be, I'll take it as 6.6 into 3 divided by 5.4 into 22 10 to the power minus 5. OK. Now when you solve this, you'll get 19.8 divided by 5.4. So 19.8 divided by 5.4 is 3. something yet, OK? So the approximate answer is 3.67 10 to the power minus 5 centimeter. This is the approximate answer for this, OK? So you can see these ones. Now the next question you see, question number 62 here, it says, give the reason why the ground is state of the outermost electronic configuration of silicon is this, OK? So this one is correct, and this one is wrong. The question is why this one is wrong. Obviously, the second one is wrong, because according to Hansl, we know that the unpaired electron present in an orbital of same substance of any orbital, unpaired electron present in any orbital must have the same configuration, right? The same spin electron must have. It can be either clockwise or anticlockwise, but it is not possible that one is clockwise and other one is anticlockwise. Either you can take all as anticlockwise, or you can take all as clockwise. So according to Hansl, this is not valid, hence the answer of this question is, OK? What is the maximum number of electrons that may be present in all the atomic orbitals with principle quantum number three and azimuthal quantum number two, right? OK, look at this question. Principle quantum number three means we can have 3S, we can have 3P, we can have 3D. These three subtleties we have. Azimuthal quantum number means L value is 2. So L value for S it is 0, 1, and 2. So L value is 2. It means it is 3D suction, right? And 3D suction we can have what? We can have five orbitals into this, and hence the number of electrons would be 5 into 2. That is 10 electrons. That is how you can do this question, OK? We also have the formula here, and the formula is the maximum number of electron, maximum number of electron, electron present in, present in a major shell, is 2N squared, is 2N squared, and a sub-shell if you have, if it is a sub-shell, 2 into 12 plus 1, depending upon the L value. So L value is given here, 2, right? So 2 into 2, 4 plus 1, 5 into 2, 10 electrons here, right? So with the formula also you can do it directly, or logically also if you have the concept, you can do it easily, OK? Easy one. Next question. According to the Bohr's theory, the electronic energy of hydrogen atom in Nth, Nth Bohr's orbit is given by this. It is given in Joule. Calculate the longest wavelength of light that will be needed to remove an electron from third Bohr's orbit of He plus, OK? Fine, so He plus we have, so E3 for He plus. What we can write? It is minus 21.76 into 10 to the power 19 ZZ squared by N squared, OK? So for helium, the atomic number is 2 minus 19 into 2 squared by N value is third orbit, so 3 squared. So 4 by 9, correct? So when you solve this, it would be 2.14 minus into 10 to the power 19 into 4. This is 4 minus 9.64 into 10 to the power minus 19 Joule. This is the energy we have, OK? Now for the longest wavelength, longest wavelength is, we know this energy E, the wavelength associated with Hc by lambda is equals to minus 9.64 into 10 to the power minus 19 Joule. So lambda is equals to 6.6 into 10 to the power minus 34. 3 into 10 to the power 8 divided by 9.64 into 10 to the power minus 19, OK? So answer would be in the order of 10 to the power minus 17, OK? So this would be, when you solve this, this thing will get cancelled. It will have your 10 to the power minus. So 19 we have, so it is 17 and then 10 to the power minus 7. We have this. And this would be 19.8 divided by 9.64 into 10 to the power minus 7. So if you solve this, you'll get 2 point something into 10 to the power minus 7 meter, which you can also write in terms of angstrom by multiplying by 1,000 in both numerator and denominator, OK? So answer would be 2 point something into 10 to the power minus 7 meter for this question. If you want to put the answer in terms of angstrom, which is generally we put, then we can convert this into angstrom, OK, easily. You know, 1 to the power minus 10, 1 angstrom is equal to 10 to the power minus 10 meter, so that way we can do this. OK, the next one you see. Estimate the difference in energy between first and second board orbit of hydrogen atom. And at what minimum atomic number, a transition from N2 to N1 energy level would result in the emission of x-ray with lambda this, which hydrogen atom, like species, does the atomic number correspond to, OK? So the question is, you see, the energy, estimate the energy difference between first and second board orbit of a hydrogen atom, right? So E2 minus E1 we need to find out, OK? We know the energy is the formula of energy E is equals to minus 21.76 into 10 to the power minus 19 Z square by N square joule per atom, OK? Then we can find out delta E is equals to E2 minus E1, which is the first part here. And that would be minus 21.76 10 to the power minus 19 right? It is 1 by E2, right? So minus 1 minus 1 by 4, 1 minus 1 by 4. This is what you need to solve. So delta E we can find out from this, OK? So this value, you see this question here. Estimate the estimate the energy difference between this and this, which is this is coming out to be 3 by 4. So 3 by 4 is minus 21.76 10 to the power minus 19 into 3 by 4. So it is 5 point something. So minus 5.44 into 3 into 10 to the power minus 19. So the answer is 16.32 into 10 to the power minus 19 joule. This is the energy we have. This is the energy between E2 minus E1. Now, at what minimum atomic number, a transition from N2 to N1, which is this, would result in the emission of x-rays with lambda this. So wavelength is given here, OK? Wavelength is given. So first of all, we'll find out what is the energy difference we have, which results in this particular wavelength. That is what the question is. OK, so say similarly, for any atom, we do not know what atom it is. For any atom, if you write down the expression for nth energy, it is equals to what? Minus 21.76 into 10 to the power minus 19. z squared by n squared, OK? So it says what atom we have. It says here, what atom we have. What hydrogen-like atom species does this atomic number corresponds to, OK? So delta E we know, that is E2 minus E1, according to the question, which is also equals to minus 16.32, 10 to the power minus 19. And this is equals to, this is equals to, this is equals to, we can write, at C by lambda, right? But we won't use this expression here, but we use 21. minus 21.76 into 10 to the power 19. z squared, because this is what we need to find out. And 1 by n squared is, we'll have here, when you solve this E2 minus E1, you get 3 by 4. This is equals to, at C by lambda, 6.6 into 10 to the power minus 34, 3 into 10 to the power 8, divided by lambda value, we have 3 into 10 to the power minus 8. So when you solve this expression for z, you'll get the atomic number. So basically, the question was what? That in hydrogen species, the energy difference we have between 1 and 2. For what other atoms, the energy difference would be same. And the wavelength that comes out is this. This is what we have. When you solve this, the z value will get z squared from this. It is approximately 4 you get. And hence, z value is 2 for this question. Second one, right? 6 to 6. What transition in the hydrogen spectrum would have the same wavelength as the Bamer transition for n is equals to 4 and n is equals to 2 for he plus. This one is also easy. We know the wavelength formula. 1 by lambda is equals to rh for hydrogen atom. Then z is square, 1 by 422, right? So 1 by nf square minus 1 by ni square, OK? So what transition in hydrogen spectrum would have the same wavelength as Bamer, right? So fine. So rh, you let it be as it is. Rh, since it is for helium, so z value is 2 square. 1 by nf square is 2 square. 1 by ni square is 4 square. So when you solve this, or if you compare this with hydrogen, which is 1 by lambda is equals to rh, 1 square, to 1 by nf square minus 1 by ni square. Always don't solve this expression for lambda in this kind of question, OK? Now if you see, if you multiply by 2 here, what you'll get over here? Rh into 1 square, multiply by 2 square insides, 1 by 1 square minus 2 square insides. So it is 4, and 4 will get cancelled. So 1 by 2 square will get here. So it is nf should be 1 for this, and ni should be 2, right? So 2 to 1 transition we have, which gives the same wavelength as we get the wavelength in case of He plus Bamer series transition, that is n4 to n2 transition, So this is the question we have discussed, OK? So I hope you understood all this question. We'll come back again with some more interesting questions. Till then, take care. Bye-bye.