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CBSE Class 12 Physics 2015 Delhi Set 1 Paper

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Question : 6 of 26
Marks: +1, -0
SECTION - B
A proton and an α\alpha-particle have the same deBroglie wavelength. Determine the ratio of (i) their accelerating potentials (ii) their speeds.
Solution:  
Ans. Determination of ratio (i) accelerating potential
(ii) speed
(i) λ  =  h2mqV\lambda\;=\;\frac{h}{\sqrt{2 m q V}}
V  =  h22mqλ2V\;=\;\frac{h^2}{2 m q \lambda^2}
mα  =4mpqα=2qpm_\alpha\;=4 m_p q_\alpha=2 q_p
⇒      VpVα  =  mαqαmpqp\Rightarrow\;\;\;\frac{V_p}{V_\alpha}\;=\;\frac{m_\alpha q_\alpha}{m_p q_p}
  =  4mp×2qpmpqp\;=\;\frac{4 m_p \times 2 q_p}{m_p q_p}
  =8:1\;=8: 1
(ii) λ  =  hmv\lambda\;=\;\frac{h}{m v}
⇒v  =  hmλ\Rightarrow v\;=\;\frac{h}{m \lambda}
⇒  vpvα  =  mαmp=4\Rightarrow \;\frac{v_p}{v_\alpha}\;=\;\frac{m_\alpha}{m_p}=4
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