## ATOMS QUIZ-1

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. The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to (where e,h, and c have their usual meanings in cgs system)
•  2Ï€hc/e2
•  er2h/2Ï€c
•  e2c/2Ï€h
•  2Ï€e2/hc
Solution

Q2. In Fig, E1 to E6 represent some of the energy levels of an electron in the hydrogen atom.Which one of the following transitions produces a photon of wavelength in the ultraviolet region of the electromagnetic spectrum?
•   E2-E1
•  E3-E2
•  E4-E3
•  E6-E4
Solution
(a) The wavelengths of the hydrogen spectrum could be arranged in a formula or series named after its discoverer. For ultraviolet spectrum the series is called Lyman series, for visible spectrum the Balmer series, and for infrared region we have the Paschen series The ultraviolet series is obtained when the energy of the atom falls from higher states to the energy level corresponding to n=1. Thus, ultraviolet radiation can only be possible with transition from E2 to E1 out of the given transitions

Q3. If elements of quantum number greater than n were not allowed, the number of possible elements in nature would have been
•   1/2 n(n+1)
•  {n(n+1)/2}2
•  1/6 n(n+1)(2n+1)
•  1/3 n(n+1)(2n+1)
Solution

Q4. The ratio between total acceleration of the electron in singly ionized helium atom and hydrogen atom (both in ground state) is
•  1
•  8
•  4
•  16
Solution

Q5. Consider a spectral line resulting from the transition n=5 to n=1, in the atoms and ions given below. The shortest wavelength is produced by
•  Helium atom
•  Deuterium atom
•  Singly ionized helium
•  Ten times ionized sodium atom
Solution

Q6. A hydrogen atom in around state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
•  1.05×10(-34) Js
•  2.11×10(-34) Js
•  3.16×10(-34) Js
•  4.22×10(-34) Js
Solution

(b)
Q7. In the Bohr model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass, and e is the charge on the electron and Îµ0 is the permittivity of vacuum, then the speed of the electron is:
•  0
•  e/√(Îµoaom)
•  e/√(4Ï€Îµoaom)
•  √((4Ï€Îµoaom)/e)
Solution

Q8. As the electron in Bohr orbit of hydrogen atom passes from state n=2 to n=1, the KE (K) and PE (U) change as
•  K two-fold, U also two-fold
•  K four-fold, U also four-fold
•  K four-fold, U two-fold
•  K two-fold, U four-fold
Solution
(b) KE∝1/n^2 and PE∝1/n^2
Q9. An electron is in an excited state in a hydrogen like atom. It has a total energy =-3.4 eV. The kinetic energy of electron is E and its de Broglie wavelength is Î»
•  E=6.8 eV; Î»=6.6×10(-10) m
•  E=3.4 eV; Î»=6.6×10(-10) m
•  E=3.4 eV; Î»=6.6×10(-11) m
•  E=6.8 eV; Î»=6.6×10(-11) m
Solution

Q10. When the voltage applied to an X-ray tube increases from V1=10 kV to V2=20 kV, the wavelength interval between KÎ± line and cut-off wavelength of continuous spectrum increase by a factor of 3. Atomic number of the metallic target is
•  28
•  29
•  65
• 66
Solution

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET & IIT JEE COACHING: Atoms - Quiz 1
Atoms - Quiz 1