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An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is:

1. 1 / c √(2mE)
2. (E / 2m) ½
3. 1 / √(2mE)
4. (2mE / E) ½

Correct answer: 1 / c √(2mE)

Solution

The de-Broglie wavelength for a particle is given by λ = h / p, where p is momentum. For an electron, p = √(2mE). For a photon, p = E / c. Taking the ratio of their wavelengths, λ_electron / λ_photon = (h / √(2mE)) / (h / (E / c)) = 1 / c √(2mE).

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