Structure of Atom - NCERT Solutions (Part 5) | Class 11 Chemistry Chapter 2 | CBSE | 2024-25

[@motivationworld500]

sir question no. 43 me atomic no.17 chlorine hoga na

[@vaibhav-kh5zb]

Ha

[@BhauraoRathod-wo9ny]

Ha

[@vampirexsehaj3562]

43 question Mr Fluorine ki jaghe cl ana ha 😂😂😂😂

[@mariyakhan6251]

Tumko bahut aata hai itna hans rahi hi sir ji jagah tumhi padha diya karo

[@MaheshKumar-os4ck]

Sir accche hai

[@Sunita_Ranga-ja8tv]

sir bohot acche se padhate hai 😊😊

[@sumanmehta8064]

Yeah chlorine aeyga

[@snehamishra7427]

12:57 To the students studying: I WANT TO CORRECT SIR HERE, HE'S A GOOD TEACHER, AND IM SURE THAT HE GAVE FLOURINE AS THE ANSWER BY MISTAKE, THE REAL ANSWER IS CHLORINE.

[@princetanwar2009]

THANK YOU SIR FOR THIS EXPLANATION , I RECCOMENDED THIS CHANNEL TO MY FREINDS AND TUTORS ALSO . THANK YOU SIR

[@sonukumar-hd6vm]

Q 54 ka work function is 1.21763 into 10 to the power minus 15

[@MagnetBrainsEducation]

Thank you for your positive feedback! We're glad you're enjoying our educational videos and finding them helpful. Keep up the good work!

[@shalni2k23]

Thanku sir best teacher chemistry

[@MagnetBrainsEducation]

Thank you for your comment! We're glad you found our video helpful. As India's No.1 Online Education Platform, we offer the best quality educational videos to help students learn and succeed. We hope you continue to find our videos helpful in your studies.

[@vinodghlawat7082]

Q.54-7.6 into 10 ki power 3eV

[@MagnetBrainsEducation]

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[@Epicgaming-l6f]

Questions number 47 A- 4.8×10°14 B- 9×10°9 C- 3.2×10°-19 D- 6.25×10-20

[@sonukumar-hd6vm]

Q no 47 answer (a) ka 4.8 into 10 to the power 14 (c) ka 3.168 into 10 the power minus 19 (d) ka 6.3 into 10 to the power 18

[@MagnetBrainsEducation]

Thank you so much for your positive feedback! We're so happy to hear that you're enjoying our educational videos and that they're helping you in your studies. Keep up the good work and we hope you continue to find our videos helpful.

[@DAURAS2008]

Sir one shot of chapter 2 please sir 😊😊

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[@HarjinderK-c2f]

30:40 sir answer for question 47 :-😁😁😁😁 (a) 4.85x10^14 Hz (b) 9x10^9 meter (c) 3.2*10^15 (d) 6.25x10^-17

[@alsh9311]

12:37 sir its A chlorine

[@onmyway.ot7]

Sir please complete our syllabus fast , in my school thermodynamics have completed now

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[@rajatpal1258]

Freq1=509.3 into 10 ki power 12 Freq2= 508.8 into 10 ki power 12 Delta E = 3.3 into 10 ki power -22 Hemant sir zindabad 😄

[@pagoluganesh9374]

Wavelength of transition = 1285 nm = 1285 × 10-9 m (Given) v=3.29×1015(132−1n2)since,v=cλ =3.0×108ms−11285×10−9mNow,v=2.33×1014s−1 Substituting the value of ν in the given expression, 3.29×1015(19−1n2)=2.33×1014 19−1n2=2.33×10143.29×1015 19−0.7082×10−1=1n2 →1n2=1.1×10−1−0.7082×10−1 1n2=4.029×10−2 n=√14.029×10−2n=4.98n≈5 Hence, for the transition to be observed at 1285 nm, n = 5. The spectrum lies in the infra-red region.

[@HarjinderK-c2f]

40:49 sir answer is 5x10^15 Hz for both the wavelengths

[@ArijitisStudying]

30:24 (d) 187 x 10^17 quanta

[@armanthakurvlb9837]

Sir question no 43 not F its Cl

[@ManishaKumari-d1u]

Sir 49 no.answer 8.25*10power -10 is this right answer sir plz reply me

[@GoluKumar-rm5on]

Sir questions no 54 ka answer nahi aa raha hai sir please explain 🙏

[@naincy5911]

Q47 : a) 4.8×10¹⁴ c) 3.21×10 ^ - 19 d)6.2 × 10¹⁸ Q49 : 8.25×10^ - 10

[@furyboy2666]

Can u tell me q 49

[@HarjinderK-c2f]

38:34 sir answer for 49 is 8.25x10^-9 😁😁😁😁

[@lunlelunle-r3w]

43 = cl❤

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[@SatyamSuman-w5u]

Sir.. When we complete syllabus of class 11th...

[@MagnetBrainsEducation]

We will be happy to know that you have completed the syllabus of class 11th. We appreciate your effort and look forward to your future success.

[@himanshurathorerathore4277]

Thank you sir 😊

[@MagnetBrainsEducation]

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[@chandujee]

1:13:05 55. n=5

[@parnikabijalwan570]

plz answer stopping potential ka formula tou k.e upon charge of electron hota h toh question 53 mai sir ne stopping potential to sidha k.e kai equal kese rkhdiaa

[@VikasVerma-bo5bl]

Sir please recheck question 43

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Thank you for letting us know. We will recheck the question and update the video accordingly. We appreciate your feedback!

[@ShaliniSharma-le8fr]

Answer 50 :- Frequency 1 = 5.093 × 10¹⁴ Hz Frequency 2 = 5.088 × 10¹⁴ Hz And ∆E = 3.45 × 10²² ✅

[@ShivaniKirar-pv9ff]

Kaise aaya

[@Spinogameing]

Q.48 is 9.54 photons 👍 ≈10

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Thank you for your positive feedback on our educational video! We're glad you found it helpful and we hope you continue to use our platform as a resource for your studies.

[@sunitakushwaha8979]

Thank u so so much sir you are very good teacher❤❤

[@OmRanjan-z9r]

Tukur tukur dekho😂😂😂😂...sahi h sir ji...best explanation ❤❤😅

[@mausamkesarwani493]

Sir 3 aur redox 0dha dijiye October se phle

[@MagnetBrainsEducation]

Thanks for writing in! We'll be sure to upload more videos on redox reactions in the coming weeks. Stay tuned and keep up the good work!

[@allncertmathsolution1134]

Question 51 frequency = 4.6×10^14

[@piyushraj167]

Q. 53 for ev=1240/wavelength(nm) simple.. Ans Sirf 1 min me nikal jayega

[@sonukumar-hd6vm]

Q no 51 velocity =5 into 10 to the power 5

[@MagnetBrainsEducation]

Your answer is correct! Well done! We're so happy to hear that you're finding our videos helpful. Keep up the good work!

[@tanzeemnaqvi178]

kaise aaya??

[@ananyasingh3969]

Sir, Can u plzz teach us little faster as our midterms are coming

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Hi! Thank you for watching our videos. We appreciate your feedback and will try to make our videos more concise in the future. Best of luck on your upcoming midterms! :)

[@motivationworld500]

Sir ji question no. 55, n=5 means Pfund series and it is belongs to infrared region

[@naincy5911]

Q55: n= 5 and lies in p fund series of infrared spectra region

[@Epicgaming-l6f]

Question no 49 Answer= 8.25×10°-10

[@furyboy2666]

Bro how ??

[@kapilbhagat9662]

Sir please make a one shot video of atomic structure chapter

[@MagnetBrainsEducation]

Thank you for your comment! We are India's No.1 Online Education Platform and we offer educational videos to school students. We will make a one shot video of atomic structure chapter for you.

[@NityanandswarupBiswal-if2sl]

question no 43 is incorrect sir .please reply me sir .........please sir. correct answer is chlorine

[@monicamarwaha9189]

Sir it's not fluorine it's chlorine

[@AiemanVohra]

Sir 51 ka velocity puche alag aarha hai or calculator mai alag aaraha hai

[@KumariPrachiMahato]

Q49- ans=8.2× 10^-10J

[@armankhanp448]

47(b) 0.0048×10ki power 17

[@SanjayYadav-gh4lq]

First ques. Me chlorine ho ga

[@GaneshKumarDube-cj8uj]

Sir please request 55 do

[@dhirasingh2009]

49) 82.5*10^-11

[@saritadevi-cs2jl]

Mein toh khud pr doubt krne lgi thi 😂

[@godgaming1702]

sir why in question no. 50 you use the formula hc/lambda , why not used hv

[@Staleburger667]

Maybe because it's easier as we can take terms common in that way

[@ManviSingh-fi2tm]

Sir 47 ka answer 1.44 x 10*22 hona chahiye but book me to 6.2 x 10*18 hai 😢

[@tanzeemnaqvi178]

sr pls jaldi se 3rd chaptern ka ncert numericals laiye pls sr jaldi kriye

[@malharILAVIA]

43 question - ans cl

[@sonukumar-hd6vm]

Q no 50 ka frequency1 = 5.093 into 10 the power 14 frequency 2=5.113 into 10 to the power 14 delta E = 3.3 into 10 the power 14

[@MagnetBrainsEducation]

Thank you so much for your positive feedback! We're glad you're enjoying our videos and finding them helpful. Keep up the good work!

[@SaloniSaloni-no2nj]

Sir 17 element cl hota hai

[@MagnetBrainsEducation]

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[@selimlaskar1247]

thanks

[@MagnetBrainsEducation]

Welcome! Stay connected with Magnet Brains and Top your exams!

[@laxmanprashad7976]

Ham logo October se exam hai usme chemical bonding Tak AA Raha hai

[@MagnetBrainsEducation]

Thank you for letting us know that our videos have been helpful to you in preparing for your upcoming exams! We're glad to have been able to provide you with the resources you need to succeed. Best of luck on your exams!

[@sonukumar-hd6vm]

Q no 51 ka frequency note = 4.606 into 10 to the power 14

[@MagnetBrainsEducation]

Thanks for letting us know! We're so happy to hear that our videos are helping you out. Keep up the good work!

[@SherabGyatso-fm4xj]

sir math

[@MagnetBrainsEducation]

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[@vaibhav-kh5zb]

Sir 43 is wrong

[@Listening-to-Trend]

Sir i am from class6 I have a doubt in maths that how can we subtract integers

[@MagnetBrainsEducation]

Hi! Thanks for watching our videos and for posting a question. subtracting integers is a great question! Here's how we do it: when the signs are the same, we add the numbers and keep the same sign. So, for example, if we're subtracting -6 from -4, we first add the numbers: -6 + 4 = -2. Then, because the signs are the same, we keep the negative sign and the answer is -2.

[@crazyhackerz8327]

Sir chapter wise entrance k liye mind map ka video b upload kijiye na

[@Epicgaming-l6f]

Sir jab lamda se kamm to rha tha ro frequency kis liye nikal wai 😢

[@surendrasharma81]

49 que 8.25*10-10

[@VikasVerma-bo5bl]

Sir your question 43 is wrong

[@MagnetBrainsEducation]

Thank you for pointing out the error in our video. We appreciate your feedback and are constantly working to improve the quality of our content. Thank you for watching our videos and we hope you continue to learn with us.

[@allncertmathsolution1134]

43me clorine hoga

[@Simu__virk]

Sir 52 sum ki calculation ka video dalo plz

[@jatinsingh5069]

Sir q; 44 me x+3 kaise hoga

[@ShivamRaj-fy1cb]

51 ko karna chahiye tha

[@keshav_nagar]

Aree sir exam aa gye or aap abhi 2nd chapter pe he ho

[@Itxmayank718]

Yes bro kya kare ab

[@SgsjskvSbsbsnsn-lq4ud]

Using bots for comments

[@Amitgoliyan-s9m]

Q. 49 8.28 × 10^-10

[@shalni2k23]

Sir questions no 54 ka answer nahi aa raha hai

[@shambhavi7850]

7.6×10³eV

[@furyboy2666]

Can anyone explain me q. 51 ?? Plzzz

[@Kimtaegua07]

You're given the work function of caesium which is 1.9eV . In the first option A. you have to find threshold frequency and work function is always equal to hVo (h - constant, Vo - threshol frequency) so simply Wo= hVo Now send Vo on the left side and Wo on the right side so it will become Vo=Wo/h now put the value , you will have the answer and yeah also to transform 1.9eV into J multiply 1.9 from 1.602 ×10-¹⁹ B. As we know ㅅ(lambda) = c/v change the v to Vo. put the value and question will be solved C. In this part ㅅ is 500 nm which will become 500 × 10-⁹m . Now the formula of K.E = h(V-Vo) Put the value of h and transform V into 1/ㅅ and Vo into 1/ㅅo You are already given the value of ㅅ which is 500 × 10-⁹m and ㅅo will be 654 ×10-⁹ which you had found in the part B of this question and after solving you'll get the K.E And as we know K.E is equal to 1/2mv² just put all of the value except v because that's what we have to find you get the answer Hope it will help

[@ashutoshshukla1571]

Q 49=6.6*10*-10

[@changdevghodke6641]

1+ like for Chemistry is easy 😅😅

[@laxmanprashad7976]

Sir syllabus piche chal Raha hai jaldi jaldi padhao

[@MagnetBrainsEducation]

Thank you for your valuable feedback. We are India's No.1 Online Education Platform and we offer educational videos to school students. We are constantly improving our videos and content to ensure that our students are getting the best education possible. Thank you for your support!

[@monicamarwaha9189]

Karado 55 bhi😅

[@karan_wagh]

sir solve kara karo calculation hamara galat aa raha hao

[@TopokidokDigbak-jg9oq]

Khon last july se start kar re

[@Janhavi-we5me]

Sir 43 question me flourine bhai sahab ki jagah chlorine behen aayegi 😂😂 Chlorine be like -------gaddari karbe😂😂

[@vineetasingh1776]

Magnet brains barbaad hai 😂

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[@KrishnaKumar-el2ms]

Toom bhi barbad ho

[@asifgamingnews53]

Q2.54 work function is 12.362×10 ki power -16 joule

[@MagnetBrainsEducation]

Thank you so much for your positive feedback! We're so glad you're enjoying our educational videos and finding them helpful. Keep up the great work and we'll keep making videos that you love!

[@HarjinderK-c2f]

44:44 sir answer for q50 for energy is 1.98x10^-16

[@sonukumar-hd6vm]

Q no 50 ka frequency1 = 5.093 into 10 the power 14 frequency 2=5.113 into 10 to the power 14 delta E = 3.3 into 10 the power 14

[@MagnetBrainsEducation]

Thank you for your comment! We're so happy to hear that you're enjoying our educational videos. We offer a variety of videos to school students and we're glad that you're finding them helpful. Keep up the good work and we hope you continue to enjoy our videos.

[@ShivaniKirar-pv9ff]

Kaise kiya?

[@parnikabijalwan570]

plz answer stopping potential ka formula tou k.e upon charge of electron hota h toh question 53 mai sir ne stopping potential to sidha k.e kai equal kese rkhdiaa

[@armankhanp448]

47(c) 0.026×10ki power -17

[@parnikabijalwan570]

plz answer stopping potential ka formula tou k.e upon charge of electron hota h toh question 53 mai sir ne stopping potential to sidha k.e kai equal kese rkhdiaa