If λ0 is the de Broglie wavelength for a proton accelerated through a potential difference of 100V, the de Broglie wavelength for α -particle accelerated through the same potential difference is
A
2√2λ0
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B
λ02
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C
λ02√2
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D
λ0√2
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Solution
The correct option is Cλ02√2 Proton is accelerated through a potential difference 100 V so,12mv2=e×100