Answer:
The correct option is (4) - 0.153 nm
de-Broglie wavelength is inversely proportional to the potential difference
Explanation:
de-Broglie wavelength:
λ = h/√(2Km) K = qV [kinetic energy]
λ = h/√(2qVm)
λ = 1.227 / (√V) (here, λ is in nm)
given V = 64V
∴ λ = 1.227 / √64
= 1.227 / 8
= 0.153 nm
Hence, The de-Broglie wavelength associated with an electron accelerated through a potential difference of 64 V is, λ = 0.153nm
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