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If the kinetic energy of an electron gets doubled, its de Broglie wavelength will become _________.1. doubled2. (1/√2) times3. squared4. √2 times |
Answer» Correct Answer - Option 2 : (1/√2) times CONCEPT:
λ = h/p whereλ is de Broglie wavelength,h is Plank's const, and p is the momentum.
\(λ=\frac{h}{\sqrt{2mE}}\) whereλ 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 isoption 2. |
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