Question

The De Broglie wavelength of a 10 KeV neutron whose mass if equal to 1.675X10^-27 Kg. *
1 point
A. 0.00486 angstroms
B. 0.00386 angstroms
C. 0.00286 angstroms
D. 0.00186 angstroms

Answer 1

Answer:

Option C (0.00286 angstroms)

Explanation:

1 electron volt is equal to 1.6×10^-19 C

Given,

E = 10KeV = 10 × 10^3 × (1.6 × 10^-19) = 1.6 × 10^-15 V

de Broglie's wavelength of neutron is calculated by:

λ = h / √(2mE)

λ = 6.63×10^-34 / √(2 × 1.675×10^-27 × 1.6×10^-15)

λ = 6.63×10^-34 / 2.315×10^-21

λ = 2.864×10^-13 m

Therefore, convert meters to angstroms (1 m = 1 × 10^10 ang)

So, (2.864×10^-13 m) × (1 × 10^10 )

==> 0.00286 angstroms (OPTION C)

Thank you:)

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