JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 10)
A nucleus A, with a finite de-broglie
wavelength $$\lambda $$A, undergoes spontaneous fission
into two nuclei B and C of equal mass. B flies
in the same direction 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
$$\lambda $$B and $$\lambda $$C of B and C are respectively :
$$\lambda $$A, 2$$\lambda $$A
2$$\lambda $$A, $$\lambda $$A
$$\lambda $$A, $$\lambda $$A/2
$$\lambda $$A/2, $$\lambda $$A
Explanation
Let mass of B and C is m each. By momentum conservation
$$2m{v_0} = mv - {{mv} \over 2}$$
v = 4v0
PA = 2mv0 pB = 4mv0 pc = 2mv0
De-Broglie wavelength $$\lambda = {h \over p}$$
$${\lambda _A} = {h \over {2m{v_0}}}$$; $${\lambda _B} = {h \over {4m{v_0}}}$$; $${\lambda _C} = {h \over {2m{v_0}}}$$
$$2m{v_0} = mv - {{mv} \over 2}$$
v = 4v0
PA = 2mv0 pB = 4mv0 pc = 2mv0
De-Broglie wavelength $$\lambda = {h \over p}$$
$${\lambda _A} = {h \over {2m{v_0}}}$$; $${\lambda _B} = {h \over {4m{v_0}}}$$; $${\lambda _C} = {h \over {2m{v_0}}}$$
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