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Check that the ratio (k e^(2))/( Gm_em_p)is dimensionless. Look upa Table of Physical Constant and determine the value of this ratio. What does the ratio signify? |
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Answer» Solution :From the RELATION `F= ( kq_1q_2)/(r^(2)) ` we find units of k as `Nm^(2) C^(2) `. Again units of F are `Nm^(2) kg^(-2) ` ` THEREFORE`Units of ` (ke^(2))/(Gm_rm_p)= ( Nm^(2) C^(-2) (C^(2)))/( (Nm^(2) kg^(-2) )(kg)(kg)) = `units less and hence dimensionless. We know that ` k= 9 xx 10 ^(9) Nm ^(2) C^(2) .,e= 1.6 xx 10 ^(-19)C , G = 6.67 xx 10 ^(-11)Nm ^(2)kg ^(-2), m _e= 9.1 xx 10 ^(-31) kgand m_p = 1.67 xx 10 ^(-27)kg ` ` therefore` value of `(k e^(2))/( Gm_e,m_p) =( 9xx 10^(9) xx (1.6 xx 10 ^(-19))^(2) )/(6.67 xx 10 ^(-11)xx 9.1 xx10 ^(-31)xx 1. 67 xx 10 ^(-27))= 2.4 xx 10 ^(39 ) ` The ratio signifies the ratio of electricforce and the GRAVITATIONAL FORCE ACTING between an electron and a proton separated by a finite distance. |
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