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An α-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 wavelengthassociated with the particle will be :  [2012]
  • a)
    1 Å
  • b)
    0.1 Å
  • c)
    0.01 Å 
  • d)
    10 Å
Correct answer is option 'C'. Can you explain this answer?
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An α-particle moves in a circular path of radius 0.83 cm in the ...
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An α-particle moves in a circular path of radius 0.83 cm in the ...
Given data:
- Radius of circular path (r) = 0.83 cm = 0.83 x 10^-2 m
- Magnetic field intensity (B) = 0.25 Wb/m^2

Formula:
- The de-Broglie wavelength (λ) associated with a charged particle moving in a magnetic field is given by the formula:
λ = h / (mvr)

Calculations:
- The charge of an α-particle is 2e (where e is the charge of an electron)
- Mass of an α-particle (m) = 4 times the mass of a proton = 4 x 1.67 x 10^-27 kg
- Plank's constant (h) = 6.63 x 10^-34 Js
- Substituting the values in the formula, we get:
λ = (6.63 x 10^-34) / (4 x 1.67 x 10^-27 x 0.83 x 10^-2) = 0.01 Å
Therefore, the de-Broglie wavelength associated with the α-particle will be 0.01 Å (Option C).
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