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1)an electron moves in an electric field with a kinetic energy of 2.5ev.the associated debroglie wave length is [h=6.626x10^-27 erg sec,me=9.108x106-28 grams,1ev=1.602x106-12 ergs] 2)the conductivity of a standard solution of MSO4 is 8.04x10^-10 Ohm^-1 cm^-1 and its equivalent conductance is 4.02 Ohm^-1 cm^2 equiv^-1.the Ksp for MSO4 will be

1)an electron moves in an electric field with a kinetic energy of 2.5ev.the associated debroglie wave length is


[h=6.626x10^-27 erg sec,me=9.108x106-28 grams,1ev=1.602x106-12 ergs]


 


 


2)the conductivity of a standard solution of MSO4 is 8.04x10^-10 Ohm^-1 cm^-1 and its equivalent conductance is 4.02 Ohm^-1 cm^2 equiv^-1.the Ksp for MSO4 will be

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1 Answers

Deepak Patra
askIITians Faculty 471 Points
13 years ago

Dear SRi,

Since according to the formula de broglie wavelengh is given as:

lamda = h(planks contant)/p(momentum)   i.e lamba = h/mv      --------------  eq . 1                          

where as  : p= (m)mass of electron*v(velocity of electron).

and         : K.E(kinetic enegy) = (1/2)mv2.

 so          :  v = underoot(2*K.E/m) . now sustitue value of  'v' is eq . 1.

so the debroglie wavelength: lambda = h/underoot(2*K.E*m).

Now put all the given values in the final equation

                                                    the final answer is        7.7*10-3 m.

 

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