Saved Bookmarks
| 1. |
Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? |
|
Answer» Solution :The given ratio is `(KE^(2))/(Gm_(e) m_(p))` Where, G = Gravitational constant Its unit is `N m^(2) kg^(−2)`. `m_(e)` and `m_(p)` = Masses of electron and proton. Their unit is kg. e = ELECTRIC charge. Its unit is C. k = Aconstant `=(1)/(4piepsilon_(0))` `epsilon_(0)` =Permittivity of free space Its unit is `N m^(2) C^(−2)`. Therefore, unit of thegiven ratio `(ke^(2))/(Gm_(e)m_(p))=([NM^(2) C^(-2)][C^(-2)])/([Nm^(2) kg^(-2)][kg][kg])` `"" =M^(0)L^(0) T^(0)` Hence, the given ratio is dimensionless. `e = 1.6 × 10^(−19) C` `G = 6.67 × 10^(−11) N m_(2) kg^(-2)` `me = 9.1 × 10^(−31) kg` `mp = 1.66 × 10^(−27) kg` Hence, the numerical value of the given ratio is . `(ke^(2))/(Gm_(e)m_(p)) = (9 xx 10^(9)xx(1.6 xx 10^(-19))^(2))/(6.67 xx 10^(-11) xx 9.1 xx 10^(-3)xx 1.67 xx 10^(-22))~~2.3 xx 10^(39)` This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant. |
|