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sri valli Grade:
        

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

7 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
474 Points
										

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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thanks and regards.


Akhilesh Shukla


 

7 years ago
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