Correct Answer - Option 2 : (1/√2) times
CONCEPT:
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de Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
- The wavelength of material waves is also known as the de Broglie wavelength.
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de Broglie wavelength (λ) of electrons can be calculated from Planks constant h divided by the momentum of the particle.
λ = h/p
where λ is de Broglie wavelength, h is Plank's const, and p is the momentum.
- In terms of Energy, de Broglie wavelength of electrons:
\(λ=\frac{h}{\sqrt{2mE}}\)
where λ is de Broglie wavelength, h is Plank's const, m is the mass of an electron, and E is the energy of the electron.
CALCULATION:
Given that E2 = 2E1
\(λ_1=\frac{h}{\sqrt{2mE_1}}\) .....(i)
\(λ_2=\frac{h}{\sqrt{2mE_2}}\) .....(ii)
\(\frac{λ_1}{λ_2}=\sqrt\frac{E_2}{E_1}=\sqrt\frac{2E_1}{E_1}\)
\(\frac{λ_2}{λ_1}=\sqrt\frac{1}{2}\)
\({λ_2}={λ_1}\frac{1}{\sqrt2}\)
So the correct answer is option 2.
