QJEEM16Ph29be: JEE Main 2016 Physics - Class 11 JEEM16Ph29be Solution: A neutron moving with a speed ‘v’ makes a head on collision…

Let, velocity offer collision = v₁

$$\therefore$$   From conservation of momentum,

mv = (m + m) v₁

$$\Rightarrow$$    v₁ = $$\frac{v}{2}$$

$$\therefore$$   Loss in kinetic energy

= $$\frac{1}{2}$$ mv² $$-$$ $$\frac{1}{2}$$ (2m) $$\times$$ $${\left(\frac{v}{2}\right)}^2$$

= $$\frac{1}{4}$$mv²

lost kinetic energy is used by the electron to jump from first orbit to second orbit.

$$\therefore$$   $$\frac{1}{4}$$mv² = (13.6 $$-$$ 3.4) eV = 10.2 eV

$$\Rightarrow$$  $$\frac{1}{2}$$mv² = 20.4 eV