The Total solution for NCERT class 6-12
Answer - 31 : -
(a) For n = 4Total number of electrons = 2n2 = 2 × 16 = 32Half out of these will have ms = —1/2∴ Total electrons withms (-1/2) = 16(b) For n = 3l= 0 ; ml = 0, ms +1/2, -1/2 (two e–)
Answer - 32 : -
Thus, thecircumference (2πr) of the Bohr orbit for hydrogen atom is an integral multipleof the de Broglie wavelength.
Answer - 33 : -
Answer - 34 : -
For H atom (Z = 1), En =2.18× 10-18 × (l)2 J atom-1 (given)For He+ ion (Z = 2), En =2.18 × 10-18 ×(2)2 = 8.72 × 10-18 J atom-1 (oneelectron species)
Answer - 35 : -
Answer - 36 : - The length of the arrangement = 2.4 cmTotal number of carbon atoms present = 2 ×108
Radius of each carbonatom = 12(1.2 × 10-8) = 6.0 × 10-9cm = 0.06 nm
Answer - 37 : -
Answer - 38 : -
Answer - 39 : -
We have studied thatin Rutherford’s experiment by using heavy metals like gold and platinum, alarge number of a-particles sufferred deflection while a very few had toretrace their path.
If a thin foil oflighter atoms like aluminium etc. be used in the Rutherford experiment, thismeans that the obstruction offered to the path of the fast moving a-particleswill be comparatively quite less.
As a result, thenumber of a-particles deflected will be quite less and the particles which aredeflected back will be negligible.
Answer - 40 : -
In the symbol BAX ofan element :A denotes the atomic number of the elementB denotes the mass number of the element.The atomic number of the element can be identified from its symbol because notwo elements can have the atomic number. However, the mass numbers have to bementioned in order to identify the elements. Thus,Symbols 7935Br and 79Br are accepted because atomicnumber of Br will remain 35 even if not mentioned. Symbol 3579Br isnot accepted because atomic number of Br cannot be 79 (more than the massnumber = 35). Similarly, symbol 35Br cannot be accepted because mass number hasto be mentioned. This is needed to differentiate the isotopes of an element.