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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

6 years ago


Answers : (1)


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. provides online iit jee courses and IIT JEE Test Series with IITians. Click here to get free online test series and check your status timely or you can join us as our registered user for getting best iit jee study material or iit jee test series.

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Akhilesh Shukla


6 years ago

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