Book Chosen

Physics Part II

Subject Chosen

Physics

Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
zigya tab

Boron has two stable isotopes, B510 and B511. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.81u. Find the abundances of B510 and B511.


Let, the relative abundance of B510 be x% 

  Relative abundance of B511 = (100-x)% 

Atomic weight = weighted average of the isotopes,

  10.811 = 10.01294(x)+11.00931×(100-x)100

i.e.,          x = 19.9%   and (100 - x) = 80.1%.

866 Views

The three stable isotopes of neon: Ne1020,  Ne1021  and  Ne1022 have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Given, the masses of isotopes of neon are, 19.99u, 20.99u and 21.99u.
Relative abundance of the isotopes are 90.51 %, 0.27 % and 9.22 % . 
Therefore, 
Average atomic mass of neon is, 

m = 90.51×19.99+0.27×20.99+9.22×21.99(90.51+0.27+9.22) 
 
   = 18.9.29+5.67+202.75100 

   =2017.7100 = 20.17 u
         



 
829 Views

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of Cu2963 atoms (of mass 62.92960 u).

Given, mass of the coin = 3.0 g 

Mass of atom = 62.92960 u

Each atom of the copper contains 29 protons and 34 neutrons.

Mass of 29 electrons = 29 x 0.000548 u
                                = 0.015892 u

Mass of nucleus  = (62.92960 - 0.015892) u

                          = 62.913708 u

Mass of 29 protons = 29 x 1.007825 u
                             = 29.226925 u 

Mass of 34 neutrons = 34 x 1.008665 u 

                                = 34.29461 u
Total mass of protons and neutrons  = (29.226925 + 34.29461) u
                                                       = 63.521535 u

Binding energy  = (63.521535 - 62.913708) x 931.5 MeV
                        = 0.607827 x 931.5 MeV 

 Required energy = 6.023 × 102363×3×0.607827× 931.5 MeV
                              = 1.6 × 1025 MeV = 2.6 ×1012J.

1508 Views

Two stable isotopes of lithium Li36 and Li37 have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.


Given, 
Mass of isotope 1 = 6.01512 u
Mass of isotope 2 = 7.01600 u 
Abundance of isoptope 1 = 7.5 %
Abundance of isotope 2 = 92.5 %

Atomic mass of Li =  weighted average of the isotopes 

                           = 6.01512×7.5+7.01600×92.5(7.5+92.5) 

                           = 45.1134+648.98100= 6.941 u

1239 Views

Obtain the binding energy (in MeV) of a nitrogen nucleus 
N714, given m N714 = 14.00307 u.


N714 nucleus is made up of 7 protons and 7 neutrons.

Mass of nucleons forming nucleus = 7mp + 7m

       = Mass of 7 protons + Mass of 7 neutrons 

       = 7 × 1.00783 + 7 × 1.00867 u 

       = 7.05431 + 7.06069 u 

       = 14.11550.u 

Mass of nucleus, mN = 14.00307 u 

Therefore,

Mass defect = 14.11550 - 14.00307 = 0.11243 a.m.u. 

Energy equivalent to mass defect = 0.11243 × 931 = 104.67 MeV 

∴   Binding energy = 104.67 MeV.
1039 Views