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Grade 11Physical Chemistry

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

Profile image of Ananya
7 Years agoGrade 11
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3 Answers

Profile image of Pooja
7 Years ago
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
 
 
Profile image of Khimraj
7 Years ago
 
mpqpV1 = maqaV2
V1/V2 = 4mp × 2qp / mp × qp
V1/V = 8
…...................................................
Profile image of Khimraj
7 Years ago
 
mpqpV1 = maqaV2
V1/V2 = 4mp × 2qp / mp × qp
V1/V = 8
….............................................................