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  • The ratio of de-Broglie wavelength of a  particle to that of a proton being subjected to the same magnetic fi

The ratio of de-Broglie wavelength of a \alpha particle to that of a proton being subjected to the same magnetic field so that the radi of their path are equal to each other assuming the field induction vector \vec{B} is perpendicular to the velocity vectors of the \alpha- particle and the proton, is

Option: 1

1


Option: 2

1 / 4


Option: 3

1 / 2


Option: 4

2


Answers (1)

best_answer

When a charged particle of charge q, mass m enters perpendicularly to the magnetic induction \vec{B} of a magnetic field, it will experience a magnetic force F=q(\vec{v} \times \vec{B})=q v B \sin 90=q v B

that provide a centripetal acceleration \frac{v^2}{\mathrm{r}}

\Rightarrow \quad r v B=\frac{ {mv}^2}{\mathrm{r}} \Rightarrow \, {mv}=\mathrm{qBr}

\Rightarrow \quad \text { The de-Broglie wavelength } \lambda=\frac{\mathrm{h}}{ {mv}}=\frac{\mathrm{h}}{\mathrm{qBr}}

\Rightarrow \quad \frac{\lambda_{\alpha \text {-particle }}}{\lambda_{\text {proton }}}=\frac{q_p r_p}{q_\alpha r_\alpha}

\text { Since } \frac{r_\alpha}{r_p}=1 \text { and } \frac{q_\alpha}{q_p}=2

\Rightarrow \quad \frac{\lambda_\alpha}{\lambda_p}=1 / 2

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Divya Prakash Singh

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