Modern Physics 5 Question 33
36. The potential energy of a particle varies as
$$ \begin{array}{rlrl} U(x) & =E _0 & \text { for } 0 \leq x \leq 1 \\ & =0 \quad \text { for } x>1 \end{array} $$
For $0 \leq x \leq 1$, de-Broglie wavelength is $\lambda _1$ and for $x>1$ the de-Broglie wavelength is $\lambda _2$. Total energy of the particle is $2 E _0$. Find $\frac{\lambda _1}{\lambda _2}$.
$(2005,2 M)$
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Solution:
- (a) $A-4=228$
$$ \begin{gathered} A=232 \\ 92-2=Z \text { or } Z=90 \end{gathered} $$
(b) From the relation, $r=\frac{\sqrt{2 K m}}{B q}$
$$ \begin{aligned} K _{\alpha} & =\frac{r^{2} B^{2} q^{2}}{2 m}=\frac{(0.11)^{2}(3)^{2}\left(2 \times 1.6 \times 10^{-19}\right)^{2}}{2 \times 4.003 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-13}} \\ & =5.21 MeV \end{aligned} $$
From the conservation of momentum,
$$ \begin{aligned} p _Y & =p _{\alpha} \text { or } \sqrt{2 K _{\gamma} m _Y}=\sqrt{2 K _{\alpha} m _{\alpha}} \\ \therefore \quad K _Y & =\frac{m _{\alpha}}{m _Y} K _{\alpha}=\frac{4.003}{228.03} \times 5.21 \\ & =0.09 MeV \end{aligned} $$
$\therefore \quad$ Total energy released $=K _{\alpha}+K _Y=5.3 MeV$
Total binding energy of daugther products
$=[92 \times($ mass of proton $)+(232-92)($ mass of neutron $)$
$$ \left.-\left(m _{\gamma}\right)-\left(m _{\alpha}\right)\right] \times 931.48 MeV $$
$=[(92 \times 1.008)+(140)(1.009)-228.03$
-4.003] $931.48 MeV$
$=1828.5 MeV$ $\therefore$ Binding energy of parent nucleus $=$ binding energy of daughter products - energy released
$=(1828.5-5.3) MeV=1823.2 MeV$
