JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.
Q1. One of the lines in the emission spectrum of Li(2+) has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is
Q2. The wavelength of the first line of Balmer series is 6563 â„«. The Rydberg’s constant is
Q3. A hydrogen atom and a Li(++) ion are both in the second excited state. If lH and lLI are their respective electronic angular momenta, and EH and ELi their respective energies, then
Q4. Given: mass number of gold = 197, density of gold= 19.7 g cm(-3), Avogadro’s number =6×1023. The radius of the gold atom is approximately:
Q5. Magnetic field at the center (at nucleus) of the hydrogen-like atoms (atomic number =z) due to the motion of electron in n^th orbit is proportional to
Q6. The recoil speed of a hydrogen atom after it emits a photon in going from n=5 state to n=1 state is
Q7. The wavelength Kα of X-rays produced by the X-ray tube is 0.76 Å. The atomic number of the node material of the tube is
Q8. The force acting on the electron in a hydrogen atom depends on the principal quantum number as
Q9. As the quantum number increases, the difference of energy between conservative energy levels:
Q10. If 10,000 V are applied across an X-ray tube, find the ratio of wavelength of the incident electrons and shortest wavelength of X-ray coming out of the X-ray tube, given e/m, of electron =1.8×1011 C kg(-1)