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Question

An $$\alpha-particle$$ and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The $$\alpha-particle$$ and the proton move such that radius of curvature of their paths is same. Find the ratio of their de Broglie wavelengths :


A
2:3
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B
3:4
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C
5:7
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D
1:2
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Solution

The correct option is D $$1:2$$
Magnetic force, $$\vec{F_B}=q(\vec{v}\times \vec B)=qvB\sin\theta$$

As magnetic field is perpendicular to velocity so $$\theta=90^o$$ so $$F_B=qvB$$

or, $$\dfrac{mv^2}{r}=qvB$$ or $$mv=qBr$$ 

So, momentum $$p=mv=qBr$$

de Broglie wavelength for proton , $$\lambda_p=\dfrac{h}{p_{p}}=\dfrac{h}{q_pBr_p}$$ and 

de Broglie wavelength for alpha particle , $$\lambda_{\alpha}=\dfrac{h}{p_{\alpha}}=\dfrac{h}{q_{\alpha}Br_{\alpha}}$$

Given, $$\dfrac{r_{\alpha}}{r_p}=1$$ and $$q_{\alpha}=2q_p$$

Thus, $$\dfrac{\lambda_{\alpha}}{\lambda_p}=\dfrac{r_pq_p}{r_{\alpha}q_{\alpha}}=1/2$$

Physics

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