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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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An electron of mass $'m'$ moving with velocity $'v'$ collids head on with an atom of mass M. As a result of the collision a certain fixed amount of energy $\Delta E$ is stored internally in the atom. The minimum initial velocity possessed by the electron is

$a)\;\sqrt {\large\frac{2(M+m) \Delta E}{Mm}}$

$b)\;\sqrt {\large\frac{2(M-m) \Delta E}{Mm}}$

$c)\;\sqrt {\large\frac{2M \Delta E}{Mm}}$

$d)\;zero$

Can you answer this question?
 
 

1 Answer

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Using conservation of momentum before and after collision
$ mv+M(\partial)=(M+m)v'$
$v'=\large\frac{mv}{M+m}$
now kinetic energy of the combined mass is
$\large\frac{1}{2} \bigg(\large\frac{mM}{M+m}\bigg)$$v^2=\Delta E$
$=>v^2=2 \Delta E \bigg[\large\frac{M+m}{mM}\bigg]$
$v=\sqrt {\large\frac{2 \Delta E(M+m)}{mM}}$
Hence a is the correct answer. 
 

 

answered Aug 8, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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