Q.

An electron accelerated through a potential difference  has a de-Broglie wavelength of λ. When the potential is changed to V2. Its de-Broglie wavelength increases by 50%. The value of V1V2 is equal to

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a

3/2

b

9/4

c

3

d

4

answer is D.

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Detailed Solution

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de-Broglie wavelength =hmv=h2m(ev)
⇒∝1V λ2λ1=V1V2

     V1V2=λ2λ12=94

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