Modern Physics 7 Question 83

13. A nucleus A, with a finite de-Broglie wavelength λA, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same directions as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths λB and λC of B and C respectively

(Main 2019, 8 April II)

(a) 2λA,λA

(b) λA2,λA

(c) λA,2λA

(d) λA,λA2

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Solution:

  1. Let m be the mass of nuclei B and C.

So, the given situation can be shown in the figure below

Now, according to the conservation of linear momentum, Initial momentum = Final momentum

pA=pB+pC or 2mvA=mvB+mvC2mvA=mvBmvB22vA=12vBvB=4vA and vC=vB2=2vA

So, momentum of B and C respectively, can now be given as

pB=mBvB=m4vA=2(2mvA)

or pB=2pA

and pC=mCvC=m2vA [using Eq. (ii)]

or pC=pA

From the relation of de-Broglie wavelength, i.e. λ=hp

where, p is momentum and h is Planck’s constant.

So, for A,λA=hpA or pA=hλA

From Eq. (v), λB can be written as

λB=h2×hλA=λA2

Similarly, for C,λC=hpC=hpA

[using Eq. (iv)]

Similarly, from Eq. (v), λC can be written as

λC=hhλA=λA



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