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Questions : #14(JEE-2000) & #15(JEE-2002) (in Hindi)
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In this lesson solve questions #14 & #15 from MODERN PHYSICS for IIT JEE MAINS and ADVANCE

## Kumar Ketan is teaching live on Unacademy Plus

Kumar Ketan
#Specialisation in Electronics #6 years teaching experience #Physics Faculty in Kota #"magical education" (211k+ subs) #Writer

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thnks sir ...SWAYAM mai mphil nd B.ed hota hai sir
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3. #200 Most Important Questions of Modern Physics for IJT JEE MAINS and ADVANCE Electron Beam Dissociative Result Fiber 2e. Precursor Mirror Charged and Neutral Fragments Light source Non-dissociative Result ivot Parent Molecule Scale lonized Parent Molecule

4. #Q.14. : Imagine an atom made up of proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength (given in terms of the Rydberg constant R for the hydrogen atom) equal to (a) 9/5R (JEE 2000) (b) 36/5R (c) 18/5R (d) 4/R

5. Rhc In hydrogen atom, En Also, where, m is the mass of the electron. Here, the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in nth orbit will be given by 2Rhc The longest wavelength max (or minimum energy) photon will correspond to the transition of particle from n 3 to n-2. heE3 max This gives, max-18/5R :. The correct option is (c)

6. #Q.15. : A hydrogen like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between -0.85 eVand -0.544 eV (including both these values). (a) Find the atomic number of the atom (b) Calculate the smallest wavelength emitted in these transitions. (JEE 2002) (Take, he -1240 eV-nm, ground state energy of hydrogen atom -13.6 eV)

7. (a) Total 6 lines are emitted. Therefore, n (n -1) 2 =6 or n=4 So, transition is taking place between mth energy state and (m + 3) th energy state. E,--0.85 eV -1361-1=-0.85 or or = 0.25

8. Similarly, 0.544 eV or 13.6 --0.544 = (m + 3)2 or Solving Eqs. (i) and (ii) for Z and m, we get m 12 and Z 3 Ans. (b) Smallest wavelength corresponds to maximum difference of energies which is obviously Emax-0.544-(-0.85)-0.306 eV min 0.306 - 4052.3 nm Ans. max

9. #Q.16. : A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eVis emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is -13.6 eV. (JEE 2000)

10. Solution Let ground state energy (in eV) be E1. Then, from the given condition E2n E204 eV P - E, 204 eV 11-204 eV E2n En 408 eV 4-4 = 40.8 eV = 40.8 eV or 4n or and or 2 4n or From Eqs. (i) and (ii), we get 115 4n2 4n2 4n 4 - =1 or n-2 or