Atoms - Quiz 6

ATOMS QUIZ-6

Dear Readers,

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. In order to break a chemical bond in the molecules of human skin, causing sunburn, a photon energy of about 3.5 eV is required. This corresponds to wavelength in the

•  Infrared region
•  X-ray region
•  Visible region
•  Ultraviolet region

Solution

Q2. Which of the following is true when Bohr gave his model for hydrogen atom?

•  It was not known that hydrogen lines could be explained as differences of terms like R/n² with R being a constant and n an integer
•  It was not known that positive charge is concentrated in a nucleus of small size
•  It was not known that radiant energy occurred in energy bundles defined by hv with h being a constant and v a frequency
•  Bohr knew terms like R/n² and in the process of choosing the allowed orbits to fit them he got “angular momentum =n_i/2π as a deduction

Solution

(d) a. No, since Balmer formula was known b. No, since Rutherford scattering experiment was known c. No, since Einstein’s photon theory was known d. Bohr’s chose ‘allowed’ energy levels ∝1/n² and these led to angular momentum quantized as a derivation

Q3. Let v₁ be the frequency of series limit of Lyman series, v₂ the frequency of the first line of Lyman series, and v₃ the frequency of series limit of Balmer series. Then which of the following is correct?

•   v₁-v₂=v₃
•  v₂-v₁=v₃
•  v₃=1/2 (v₁+v₂ )
•  v₂+v₁=v₃

Solution

Q4. The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Å. The wavelength of the first line is

•  27/20×4861 Å
•  20/27×4861 Å
•  20×4861 Å
•  4861 Å

Solution
(c)

Q5. The orbiting speed v_n of e^- in the n^th orbit in the case of positronium is x-fold compared to that in n^th orbit in a hydrogen atom, where x has the value

•  1
•  √2
•  1/√2
•  2

Solution

Q6. The frequency of revolution of an electron in n^th orbit is f_n. If the electron makes a transition from n^th orbit to (n-1)^th orbit, then the relation between the frequency (v) of emitted photon and f_n will be

•  v=f_n²
•  v=√(f_n)
•  v=1/f_n
•  v=f_n

Solution

Q7. Electron with energy 80 keV are incident on the tungsten target of an X-ray tube. K shell electrons of tungsten have -72.5 keV energy. X-rays emitted by the tube contain only

•  A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of -0.155 Å
•  A continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
•  A continuous X-ray spectrum of tungsten
•  A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155 Å and the characteristic X-ray spectrum of tungsten

Solution

Q8. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

•  1215 Å
•  1640 Å
•  2430 Å
•  4687 Å

Solution

Q9. An X-ray tube is operating at 150 kV and 10 mA. If only 1% of the electric power is converted into X-rays, the rate at which the target is heated, in cal s^(-1), is

•  3.57
•  35.7
•  4.57
•  15

Solution

Q10. K_α wavelength emitted by an atom of atomic number Z=11 is λ. The atomic number for an atom that emits K_α radiation with wavelength 4λ is

•  6
•  4
•  11
• 44

Solution