JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 8)
A neutron moving with a speed ‘v’ makes a head on collision with a stationary
hydrogen atom in ground state. The minimum kinetic energy of the neutron for
which inelastic collision will take place is :
10.2 eV
16.8 eV
12.1 eV
20.4 eV
Explanation
Let, velocity offer collision = v1
$$ \therefore $$ From conservation of momentum,
mv = (m + m) v1
$$ \Rightarrow $$ v1 = $${v \over 2}$$
$$ \therefore $$ Loss in kinetic energy
= $${1 \over 2}$$ mv2 $$-$$ $${1 \over 2}$$ (2m) $$ \times $$ $${\left( {{v \over 2}} \right)^2}$$
= $${1 \over 4}\,$$mv2
lost kinetic energy is used by the electron to jump from first orbit to second orbit.
$$ \therefore $$ $${1 \over 4}$$mv2 = (13.6 $$-$$ 3.4) eV = 10.2 eV
$$ \Rightarrow $$ $${1 \over 2}$$mv2 = 20.4 eV
$$ \therefore $$ From conservation of momentum,
mv = (m + m) v1
$$ \Rightarrow $$ v1 = $${v \over 2}$$
$$ \therefore $$ Loss in kinetic energy
= $${1 \over 2}$$ mv2 $$-$$ $${1 \over 2}$$ (2m) $$ \times $$ $${\left( {{v \over 2}} \right)^2}$$
= $${1 \over 4}\,$$mv2
lost kinetic energy is used by the electron to jump from first orbit to second orbit.
$$ \therefore $$ $${1 \over 4}$$mv2 = (13.6 $$-$$ 3.4) eV = 10.2 eV
$$ \Rightarrow $$ $${1 \over 2}$$mv2 = 20.4 eV
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