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The de - Broglie wavelength `lambda` associated with an electron having kinetic energy `E` is given by the expression

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The de - Broglie wavelength `lambda` associated with an electron having kinetic energy `E` is given by the expression
A. `(h)/(sqrt(2mE))`
B. `(2h)/(mE)`
C. `2mhE`
D. `(2sqrt(2mE))/(h)`
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2 Answers

0 votes
by (25.2k points)
Correct Answer - A
+1 vote
by (20 points)
edited
1/2mv²=E
mv=√2mE
Therefore, 
E=h/mv
= h/√2mE
Therefore, Option A
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