ATOMIC STRUCTURE QUIZ-7
Dear Readers,As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
Q1. The velocity of an electron must possess to acquire a momentum equal to the photon of wavelength 5200 A^°, will be
Solution
λ=h/mv ∴mv=(6.626×10^(-34))/(5200×10^(-10) )=1.274×10^(-27) For electron, m=9.1×10^(-31)kg 9.1×10^(-31)×v=1.274×10^(-27) v=1400 m/s
λ=h/mv ∴mv=(6.626×10^(-34))/(5200×10^(-10) )=1.274×10^(-27) For electron, m=9.1×10^(-31)kg 9.1×10^(-31)×v=1.274×10^(-27) v=1400 m/s
Q2.If the electron in the hydrogen atom is excited to n=5, the number of different frequencies of radiations which may be emitted is:
Solution
Frequencies emitted =∑▒〖(n-1)=∑▒〖(5-1)=∑▒4〗〗 =1+2+3+4=10
Frequencies emitted =∑▒〖(n-1)=∑▒〖(5-1)=∑▒4〗〗 =1+2+3+4=10
Q3. 3 p-orbital has :
Solution
Radial node =n-l-1; Angular node =l.
Radial node =n-l-1; Angular node =l.
Q4. The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 is:
Solution
p-orbitals (l=1) can have six electrons
p-orbitals (l=1) can have six electrons
Q5.Which possesses an inert gas configuration?
Solution
Cl^- has 3s^2 3p^6 configuration, i.e., of Ar.
Cl^- has 3s^2 3p^6 configuration, i.e., of Ar.
Q6. The relation between energy of a radiation and its frequency was given by:
Solution
According to Planck, E/photon=hv.
According to Planck, E/photon=hv.
Q7.Which of the following radial distribution graphs correspond to n=3,l=2 for an atom?
Solution
The electron density is directly proportional to Ψ^2. The larger the electron density, the larger the value of Ψ^2 and more is the probability of finding the electrons
The electron density is directly proportional to Ψ^2. The larger the electron density, the larger the value of Ψ^2 and more is the probability of finding the electrons
Q8.In the absence of magnetic field p-orbitals are known as..fold degenerate
Solution
p-orbital are three, i.e.,p_x,p_y and p_z each having same energy level, i.e., degenerate orbitals.
p-orbital are three, i.e.,p_x,p_y and p_z each having same energy level, i.e., degenerate orbitals.
Q9.The magnitude of the spin angular momentum of an electron is given by
Solution
Q10. The longest λ for the Lyman series is : (Given R_H=109678 cm^(-1))
Solution
For longest λ of Lyman series n_1=1 and n_2=2, 1/λ=R_H [1/(n_1^2 )-1/(n_2^2 )] Because ΔE=hc/λ is minimum when λ is longest Thus,ΔE=E_2-E_1
For longest λ of Lyman series n_1=1 and n_2=2, 1/λ=R_H [1/(n_1^2 )-1/(n_2^2 )] Because ΔE=hc/λ is minimum when λ is longest Thus,ΔE=E_2-E_1
