ATOMIC STRUCTURE QUIZ-5
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 number of electrons and protons in an atoms of third alkaline earth metal is
Solution
The third alkaline metal is 〖_20^40〗Ca. It contains 20 protons and 20 electrons.
The third alkaline metal is 〖_20^40〗Ca. It contains 20 protons and 20 electrons.
Q2.In ground state of chromium atom (Z=24) the total number of orbitals populated by one or more electrons is:
Solution
The configuration of _24 Cr is 1s^2,2s^2 2p^6,3s^2 3p^6 3d^5,4s^1 ∴ Total s-orbitals =4 Total p-orbitals =6 Total d-orbitals =5 and thus Total orbitals =4+6+5=15
The configuration of _24 Cr is 1s^2,2s^2 2p^6,3s^2 3p^6 3d^5,4s^1 ∴ Total s-orbitals =4 Total p-orbitals =6 Total d-orbitals =5 and thus Total orbitals =4+6+5=15
Q3. The energy per mole of photon of electromagnetic radiation of wavelength 4000 â„« is:
Solution
E=(N∙hc)/λ
E=(N∙hc)/λ
Q4. The number of electrons in an atom with atomic number 105 having (n+l)=8 are:
Solution
The electronic configuration of element with at. no. 105 is: 1s^2,2s^2 2p^6,3s^2 3p^6 3d^10,4s^2 4p^6 4d^10 4f^14, 5s^2 5p^6 5d^10 5f^14,6s^2 6p^6 6d^3,7s^2 for 5f (n+l)=5+3=8 for 6d (n+l)=6+2=8
The electronic configuration of element with at. no. 105 is: 1s^2,2s^2 2p^6,3s^2 3p^6 3d^10,4s^2 4p^6 4d^10 4f^14, 5s^2 5p^6 5d^10 5f^14,6s^2 6p^6 6d^3,7s^2 for 5f (n+l)=5+3=8 for 6d (n+l)=6+2=8
Q5.According to Bohr's theory, the angular momentum for an electron of 5th orbit is:
Solution
Angular momentum=(n∙h)/2Ï€=(5∙h)/2Ï€=(2.5 h)/Ï€ .
Angular momentum=(n∙h)/2Ï€=(5∙h)/2Ï€=(2.5 h)/Ï€ .
Q6. The hydrogen spectrum from an incandescent source of hydrogen is:
Solution
Elements show characteristics line spectrum which is finger print of atom.
Elements show characteristics line spectrum which is finger print of atom.
Q7.The maximum number of 3d-electrons having spin quantum number s=+1/2 are:
Solution
3d-subshell has five orbitals. Each orbital can have one electron with spin +1/2.
3d-subshell has five orbitals. Each orbital can have one electron with spin +1/2.
Q8.Compared to the lightest atom the heaviest atom weighs:
Solution
Heaviest atom has mass no. 238,(i.e.,_92 U^238) and lighter one is 〖_1〗H^1.
Heaviest atom has mass no. 238,(i.e.,_92 U^238) and lighter one is 〖_1〗H^1.
Q9.Choose the arrangement which shows the increasing value of e/m for e,p,n and α-particles
Solution
Mass of electron =9.1×10^(-31) kg, Mass of proton =1.67×10^(-27) kg Mass of neutron =1.675×10^(-27) kg Mass of α-particle =6.67×10^(-27) kg So, increasing order of e/m for e,p,n and α-particle is e>p>α>n(∵ neutron has no charge)
Mass of electron =9.1×10^(-31) kg, Mass of proton =1.67×10^(-27) kg Mass of neutron =1.675×10^(-27) kg Mass of α-particle =6.67×10^(-27) kg So, increasing order of e/m for e,p,n and α-particle is e>p>α>n(∵ neutron has no charge)
Q10. The de Broglie wavelength associated with a material particle is:
Solution
de Broglie equation is λ=h/mu
de Broglie equation is λ=h/mu
