AP SSC 10th Class Physical Science Chapter 8 Solution – Structure of Atom
Andhra Pradesh SSC 10th Class Physical Science Chapter 8 Structure of Atom Solution for AP SSC 10th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 10 Physical Science Textbook Solution or Study Material for AP SSC 10th exam. Here you search will end! Here in this page we have provided for all question answer for Chapter 8 Structure of Atom.
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AP 10th Class Physics Chapter 8 Structure of Atom Solution
(1) What information does the electronic configuration of an atom provide? (AS1)
Ans: –In electronic configuration the element distributed among the shells, subshell and into orbital, by these properties we came to know about the position of the atom in the space.
(2) (a) How many maximum number of electrons that can be accommodated in a principal energy shell?
Ans: –In a principal energy shell there are 8 electron.
(b) How many maximum number of electrons that can be accommodated in a sub shell?
Ans: –In an orbital the maximum electron is 2. As we all know s subshell has two orbital whereas p, d, f has 3, 5, 7 subshell respectively.
(c) How many maximum number of electrons can be accommodated in an orbital?
Ans: –In an orbital maximum no of electron will be 2.
(d) How many sub shells present in a principal energy shell?
Ans: –There are four subshell in a principal energy shell which are s, p, d, and f.
(e) How many spin orientations are possible for an electron in an orbital?
Ans: –Only two spin orientation are possible for an electron in an orbital.
(3) In an atom the number electrons in M-shell is equal to the number of electrons in the K and L shell. Answer the following questions. (AS1)
(a) Which is the outer most shell?
(b) How many electrons are there in its outermost shell?
(c) What is the atomic number of element?
(d) Write the electronic configuration of the element.
Ans:-
(a) M shell is the outer most shell.
(b) In k and L shell the no of electron is 2 and 8. So the total no of electron in outermost shell is 10.
(c) The atomic no of the element is 20.
(d) The electronic configuration of the element is 1s^22s^22p^63s^23p^64s^2.
(4) Rainbow is an example for continuous spectrum – explain. (AS1)
Ans:- The dispersion of light cause create the natural phenomenon of different spectrum of light which is rainbow.There is no boundaries in between the different colours present in the rainbow and also it consist of different wavelength with no gap . So for this rainbow is an example for continuous spectrum.
(5) How many elliptical orbits are added by Somerfield in third Bohr’s orbit? What was the purpose of adding these elliptical orbits? (AS1)
Ans: –Somerfield add two more orbital in Bhor orbital. The purpose of the splitting orbit is to study the line spectrum.
(6) What is absorption spectrum?
Ans: –For the absorption spectrum the dark band shows when it transmitted electro magnetic radiation through the substance.
(7) What is an orbital? How is it different from Bohr’s orbit? (AS1)
Ans: –The nucleus where the maximum no of electron present that is orbital. In Bohr’s orbital the path is fixed. As it is according to their energy.
(8) Explain the significance of three Quantum numbers in predicting the positions of an electron in an atom. (AS1)
Ans: –The three quantum no that can predict the position of an electron is n, l, m. Where n is called the principal quantum no, l is called the angular quantum no and m which is called the magnetic quantum no. These described about the shell, subshell and orbital in the subshell.
(9) What is nlx method? How it is useful? (AS1)
Ans: –The nl* method is use for the notation of electronic configuration. As it describe the four quantum value of an element it has wide uses. I represent the principal quantum no, l denote the subshell, x denote the no of electron.
(10) Following orbital diagram shows the electron configuration of nitrogen atom. Which rule does not support this? (AS1) N (Z = 7).
Ans: –The orbital diagram of the N doesn’t support the hunds principal where it clear that electron fill in the orbital equal energy. But in this case the third orbital is empty which violet the principal.
(11) Which rule is violated in the electronic configuration 1s0 2s2 2p4?
Ans: –In this configuration the Aufbau principal is violated. As the 2s orbital has more energy than the 1s.
(12) Write the four quantum numbers for the differentiating electron of sodium (Na) atom? (AS1)
Ans: –The electronic configuration of Na is 1s^22s^22p^63s^1
From the configuration we can measure the four quantum no that is n= 3; l= 0 for s orbital; m=0 as its ranges is +l to -l; s is +1/2 and -1/2.
(13) Is emission spectrum?
Ans: –When the substance heated the electromagnetic radiations happen. The spectrum which seen by this radiation is called the emission spectrum.
(14) (i) An electron in an atom has the following set of four quantum numbers to which orbital it
n | l | ml | ms |
2 | 0 | 0 | + 1/2 |
(ii) Write the four quantum numbers for 1s1 electron.(AS1)
Ans: –(i) For l=0 and n=2 it’s clear that the orbital is 2s orbital.
(ii) For 1s^1 orbital n= 1; l= 0; m= 0; s= +1/2.
(15) Which electronic shell is at a higher energy level K or L? (AS2)
Ans: –From Bhors quantum theory we came to know that which shell is closer to the nucleus has low energy level so L is higher energy level.
(16) Collect the information regarding wave lengths and corresponding frequencies of three primarycolours red, blue and green. (AS4)
Ans: –The wavelength and the frequency of red, blue and green is (40p to 420 Hz; 720 to 620 to 750 nm),(631 to 668 Hz; 450 to 475 nm) and (526 to 606 Hz to 495 to 570 nm).
(17) The wave length of a radio wave is 1.0 m. Find its frequency.(AS7)
Ans:- As we know that c = v× y ( c= speed of light , v= frequency, y= wavelength)
We know that c = 3× 10^8; so v = 3×10^8 /1 = 3× 10^8Hz
FILL IN THE BLANKS
(1) If n = 1 then angular momentum quantum number (l) =…..0……….
(2) If a sub-shell is denoted as 2p then its magnetic quantum number values are… 0, 1, 2….
(3) Maximum number of electrons that an M-shell contain is/are …18…………..
(4) For ‘n’, the minimum value is ……1……… and the maximum value is …infinity…
(5) For ‘l’, the minimum value is ………0…..and the maximum value is ……n-1…
(7)’ the minimum value is ………-1… and the maximum value is ……+1
MULTIPLE CHOICE QUESTIONS
(1) An emission spectrum consists of bright spectral lines on a dark back ground. Which one of the following does not correspond to the bright spectral lines? – (d) Velocity of lights
(2) The maximum number of electrons that can be accommodated in the L – shell of an atom is: (c) 8.
(3) If l = 1 for an atom then the number of orbitals in its sub – shell is 3.
(4) The quantum number which explains about size and energy of the orbit or shell is: n.