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Explain quantitatively the order of magnitude diff ere nee between the diamagnetic susceptibility of N_(2) (-5 xx 10^(-9) ) (at STP) and Cu (-10^(-5) ). |
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Answer» Solution :MAGNETIC susceptibility is the property of the SUBSTANCE which shows how easily a substance can be magnetised. Density of `N_(2)`, `(rho)N_(2) =- (28 g) /( 22.4 L) = (28 g)/( 22,400 cm^(3) ) ""…(1)` Density of copper, `(rho)_(Cu) = (8g)/( cm^(3) )""...(2)` Taking the ratio of equation (1) and (2), `therefore ((rho)N_2 )/( (rho ) Cu) = (((28)/( 22,400)))/( 8)` `=0.0001562` `=1.562 xx 10^(-4)` `((rho) N)_(2) ) /( (rho)_(Cu) )= 1.6 xx 10^(-4)""...(3)` Magnetic susceptibility of given MATERIAL, `CHI = ("Magnetisation (M)")/( " Magnetic intensity (H)")` `= ("Magnetic dipole moment per volume")/( " H")` `((("Magnetic dipole moment" m) /( "Volume V")))/(H)"` `chi = (m) /( HV)""...(4)` but density `rho = ("MASS" m.)/( "volume V")` `therefore V= (m.)/( rho) ""...(5)` `therefore chi = ((m)/( H_m) .)rho ` Taking terms of bracket as constant, Thus `chi xx rho ""...(6)` `therefore ((chi) N_2) /( (chi ) Cu)= ((rho) N_2) /((rho) (Cu) ) ""...(7)` Putting the value of equation (3) in equation (7), `((chi) N_2 )/( (chi) Cu ) = 1.6 xx 10^(-4) ""...(8)` |
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