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Dual Nature of Radiation and Matter question

2012 · Q146
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Dual Nature of Radiation and Matter question

2012 · Q146

NEETPhysicsDual Nature of Radiation and MatterMCQ+4 / −1
An α\alphaα-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The de Broglie wavelength associated with the particle will be
  1. A
    1 A∘\mathop A\limits^ \circA∘​
  2. B
    0.1 A∘\mathop A\limits^ \circA∘​
  3. C
    10 A∘\mathop A\limits^ \circA∘​
  4. D
    0.01 A∘\mathop A\limits^ \circA∘​
View written solutionFree

Correct answer: D

Wavelength λ\lambda λ = hp{h \over p}ph​

⇒\Rightarrow⇒ hp{h \over p}ph​ = hmv{h \over {mv}}mvh​

For circular motion = Fc = qvB

⇒\Rightarrow⇒ mv2r{{m{v^2}} \over r}rmv2​ = qvB

⇒\Rightarrow⇒ mv = rBq

⇒\Rightarrow⇒ mv = (0.83 × 10–2)(0.25)(2 × 1.6 × 10–19)

de Broglie wavelength,

λ\lambda λ = hmv{h \over {mv}}mvh​

= 6.6×10−340.83×10−2×0.25×2×1.6×10−19{{6.6 \times {{10}^{ - 34}}} \over {0.83 \times {{10}^{ - 2}} \times 0.25 \times 2 \times 1.6 \times {{10}^{ - 19}}}}0.83×10−2×0.25×2×1.6×10−196.6×10−34​

= 0.01 A∘\mathop A\limits^ \circ A∘​

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