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Frequency and Wavelength of emitted Radiation(In Hindi)
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In this lesson we will learn about basic concept of energy difference calculation during transition of electrons for IIT-JEE students

Niranjay Kumar Dwivedi
(N.K.D.Sir) is a Complete Chemistry faculty for IIT~JEE, and NEET examination. with more then 8 year of Teaching experience

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1. Problem & Concept OF Frequency and wave length of emitted radiation (Bohrs atomic model) By-N.K.D. Sir Please -Rate review and share this lesson

2. Frequency and wave length of emitted radiation Suppose an electronic transition takes place from orbit n2 to n , in hydrogen atom (Z-1) then the emitted energy is where EEnergy of electron in n2 orbit, E," = Energy of electron in n1 orbit (E,-> E,i) 2 -me Now 2 nzh n h 2 -me or

3. v is the frequency of emitted radiation Now So or where s the wave length of emitted radiation and 1/ ls wave number (wave per centimeter). Hence v wave number -R 22

4. where R is a constant known as Rydberg constant. For hydrogen like species, the wavelength of emitted radiation is given by the following relation where Z is the atomic numbeir O For an electronic transition between stationary states ni and n2, the value of increases decreases as Z o R- Rydberg constant for hydrogen is 2me 3 Unit of R is cm or meter For Hydrogen the value of R is 109678 cm

5. The wave number of the first Balmer line of L ion is 1,36,800 cm1 the wave number of the series first line of (in cm') (a)68,400 (c)76,000 Balmer of hydrogen atom IS -1 (b) 15,200 (d) 30,800. Solution: Atomic number of Li'is 3 1,36,800-Rx9 22 3 1 ,36,800 = 1 5,200 22 3

6. The of Ha line of Balmer series is 6500A. What is the of H line of Balmer series? Solution: For H, lines of Balmer series n 2, n2-3 For H, line of Balmer series n, -2, n, 4 22 3 and By Eqs. (1) and (2) 4 16 80 108 80