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A proton and an alpha particle after accelerating through a potential difference V1 and V2 volts respectively from rests have the same wavelength. The ratio of V1/V2 is

Ananya , 6 Years ago
Grade 11
anser 3 Answers
Pooja
Relation of de Broglie wavelength and energy is
Lambda = h/p
E = p2/2m =  qV
therefore p = (2mqV)1/2
 So lambda * (mqV)1/2 = constant
 
Applying this for proton and alpha particle
mpqpV1 = maqaV2
V1/V2 = 4mp × 2qp / mp × qp
V1/V = 8
 
 
Last Activity: 6 Years ago
Khimraj
 
mpqpV1 = maqaV2
V1/V2 = 4mp × 2qp / mp × qp
V1/V = 8
…...................................................
Last Activity: 6 Years ago
Khimraj
 
mpqpV1 = maqaV2
V1/V2 = 4mp × 2qp / mp × qp
V1/V = 8
….............................................................
Last Activity: 6 Years ago
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