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The energy that wold be changed of an electron to reduce its De broglie wavelength from $10^{- 10}$ m to $0.5 \times 10^{- 10} m$ will be :

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300 eV

6000 eV

450 eV

150 eV

answer is C.

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Detailed Solution

$λ_{1} = 1 A \Rightarrow v_{1} = \frac{150}{(1)^{2}} = 150 v$

$k E_{1} = 150 eV$

$λ_{2} = \frac{1}{2} A \Rightarrow v_{2} = \frac{150}{{(\frac{1}{2})}^{2}} = 600 V$

$k E_{2} = 600 eV$

$Δ K = K E_{2} - K E_{1}$

$= 600 - 150 = 450 eV$

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