Atoms - Quiz 3

ATOMS QUIZ-3

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. Check the correctness of the following statements about Bohr model of hydrogen atom: (i) The acceleration of the electron in n=2 orbit is more than that in n=1 orbit (ii) The angular momentum of the electron in n=2 orbit is more than that in n=1 orbit (iii) The KE of the electron in n=2 orbit is less than that in n=1 orbit

•  All the statements are correct
•  Only (i) and (ii) are correct
•  Only (ii) and (iii) are correct
•  Only (iii) and (i) are correct

Solution

(c) Centripetal acceleration =mv^2/r Further, as n increase, r also increases. Therefore, centripetal acceleration for n=2 is less than that for n=1. So, statement (i) is wrong. Statement (ii) and (iii) are correct

Q2. The shortest wavelength of Lyman series of hydrogen is equal to the shortest wavelength of Balmer series of a hydrogen-like atom of atomic number Z. The value of Z is

•   4
•  2
•  3
•  6

Solution

(b) R[1/1^2 -1/∞^2 ]=RZ^2 [1/2^2 -1/∞^2 ] or Z=2

Q3. Figure shows five energy levels of an atom, one being much lower than the other four. Five transitions between the levels are indicated, each of which produces a photon of definite energy and frequency.Which one of the spectra below best corresponds to the set of transitions indicated?

•  
  
•  
  
•  
  
•  
  

Solution (d) The first three transitions from the left fall in the Lyman series of the hydrogen spectrum which corresponds to ultraviolet radiation The fourth transition falls in the Balmer series of the spectrum which corresponds to the visible light emission. The last transition falls in the Paschen series which corresponds to the infrared radiation Thus, frequencies of the last two transitions are closer to each other on the extreme left of the frequency spectrum whereas the frequencies of the first three transitions are closer to one another and fall on the right corner of the frequency spectrum The spectrum of the transitions is thus best represented in diagram(d)

Q4. The wavelength of radiation required to excite the electron from first orbit to third orbit in a doubly ionized lithium atom will be

•  134.25 Å
•  125.5 Å
•  113.7 Å
•  110 Å

Solution

Q5. A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively. The values of n and Z are, respectively,

•  6 and 6
•  3 and 3
•  6 and 3
•  3 and 6

Solution

Q6. Difference between n^th and (n+1)^th Bohr’s radius of hydrogen atom is equal to (n-1)^th Bohr’ radius. The value of n is

• 1
• 2
• 3
• 4


Solution (d) We know, r_n∝n^2 So, (n+1)²-n²=(n-1)²⇒n=4

Q7. When the hydrogen atom emits a photon in going from n=5 to n=1 state, its recoil speed is nearly
•  10^(-4) ms^(-1)
•  2×10^(-2) ms^(-1)
•  4 ms^(-1)
•  8×10² ms^(-1)
Solution





Q8. In a hydrogen atom following the Bohr’s postulates, the product of linear momentum and angular momentum is proportional to (n)^x where ‘n’ is the orbit number. Then ‘x’ is
•  0
•  2
•  -2
•  1
Solution

(a) Linear momentum ⇒mv∝1/n Angular momentum ⇒mvr∝n Therefore, product of linear momentum and angular momentum ∝n⁰

Q9. The angular momentum of an electron in a hydrogen atom is proportional to:
•  1/√r
•  1/r
•  √r
•  r²
Solution





Q10. When photons of wavelength λ_{1 are incident on an isolated sphere suspended by an insulated thread, the corresponding stopping potential is found to be V. When photons of wavelength λ2 are used, the corresponding stopping potential was thrice the above value. If light of wavelength λ3 is used, calculate the stopping potential for this case}

•  hc/e [1/λ₃ +1/(2λ₂ )-1/λ₁ ]
•  hc/e [1/λ₃+1/(2λ₂)-3/〖2λ〗₁]
•  hc/e [1/λ₃+1/λ₂-1/λ ₁ ]
• hc/e [1/λ₃-1/λ₂-1/λ₁]
Solution