SATHEE: Modern Physics 5 Question 33

Modern Physics 5 Question 33

36. The potential energy of a particle varies as

$\begin{aligned} U (x) & = E_{0} & for 0 \leq x \leq 1 \\ = 0 for x > 1 \end{aligned}$

For $0 \leq x \leq 1$ , de-Broglie wavelength is $λ_{1}$ and for $x > 1$ the de-Broglie wavelength is $λ_{2}$ . Total energy of the particle is $2 E_{0}$ . Find $\frac{λ_{1}}{λ_{2}}$ .

$(2005, 2 M)$

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Solution:

(a) $A - 4 = 228$

$\begin{matrix} A = 232 \\ 92 - 2 = Z or Z = 90 \end{matrix}$

(b) From the relation, $r = \frac{\sqrt{2 K m}}{B q}$

$\begin{aligned} K_{α} & = \frac{r^{2} B^{2} q^{2}}{2 m} = \frac{(0.11)^{2} (3)^{2} {(2 \times 1.6 \times 10^{- 19})}^{2}}{2 \times 4.003 \times 1.67 \times 10^{- 27} \times 1.6 \times 10^{- 13}} \\ = 5.21 M e V \end{aligned}$

From the conservation of momentum,

$\begin{aligned} p_{Y} & = p_{α} or \sqrt{2 K_{γ} m_{Y}} = \sqrt{2 K_{α} m_{α}} \\ ∴ K_{Y} & = \frac{m_{α}}{m_{Y}} K_{α} = \frac{4.003}{228.03} \times 5.21 \\ = 0.09 M e V \end{aligned}$

$∴$ Total energy released $= K_{α} + K_{Y} = 5.3 M e V$

Total binding energy of daugther products

$= [92 \times ($ mass of proton $) + (232 - 92) ($ mass of neutron $)$

$- (m_{γ}) - (m_{α})] \times 931.48 M e V$

$= [(92 \times 1.008) + (140) (1.009) - 228.03$

-4.003] $931.48 M e V$

$= 1828.5 M e V$ $∴$ Binding energy of parent nucleus $=$ binding energy of daughter products - energy released

$= (1828.5 - 5.3) M e V = 1823.2 M e V$

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Modern Physics 5 Question 33

36. The potential energy of a particle varies as

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Modern Physics 5 Question 33

36. The potential energy of a particle varies as

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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