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Question: The energy that should be added to an electron, to reduce its de-Broglie wavelengths from \(10^{- 10...

The energy that should be added to an electron, to reduce its de-Broglie wavelengths from 10−1010^{- 10}m to 0.5×10−100.5 \times 10^{- 10}m, will be.

A

Four times the initial energy

B

Thrice the initial energy

C

Equal to the initial energy

D

Twice the initial energy

Answer

Thrice the initial energy

Explanation

Solution

λ=h2mE⇒λ∝1E⇒λ1λ2=E2E1\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow \lambda \propto \frac{1}{\sqrt{E}} \Rightarrow \frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}

⇒10−100.5×10−10=E2E1⇒E2=4E1\Rightarrow \frac{10^{- 10}}{0.5 \times 10^{- 10}} = \sqrt{\frac{E_{2}}{E_{1}}} \Rightarrow E_{2} = 4E_{1}

Hence added energy =E2−E1=3E1= E_{2} - E_{1} = 3E_{1}