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
2√2λ0
λ02
λ02√2
λ0√2
A
λ02
B
2√2λ0
C
λ02√2
D
λ0√2
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Solution
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Proton is accelerated through a potential difference 100 V so,12mv2=e×100
mv=√2mpe×100
λo=h√2me×100
Debrogliewavelengthofα−particle
λα=h√2mαvα×100
λα=h√2×4m×2e×100
soλα=λ02√2
So, the answer is option (C).
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