IIT-JEE 2010 · Paper 1 · Physics Q76

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PhysicsInteger+3Question 76 of 84

An $\alpha$ -particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\dfrac{\lambda_p}{\lambda_\alpha}$ , to the nearest integer, is

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Answer and worked solution

Official answer: 3

Wavelength after acceleration. $\lambda=\dfrac h{\sqrt{2mqV}}$ .

Ratio. $m_\alpha=4m_p$ and $q_\alpha=2q_p$ :
$\frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha q_\alpha}{m_pq_p}}=\sqrt8\approx2.83,$
which is 3 to the nearest integer.

Answer: 3

Topics:de Broglie waves

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IIT-JEE 2010 · Paper 1, Physics question 76