An $α$ particle is moving in a circular path of radius in the presence of magnetic field B. The de-Broglie wavelength associated with the particle will be $(q \to c h a r g e o n α p a r t i c l e)$

\frac{h r}{q B}

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\frac{h B}{q r}

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\frac{h}{q B r}

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\frac{h r^{2}}{q B}

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Solution

The correct option is C $\frac{h}{q B r}$
We know,

Radius $= r = \frac{m v}{q B}$

and, $λ = \frac{h}{m v}$

$\Rightarrow λ = \frac{h}{q B r}$

$∴$ De-Broglie wavelength $= λ = \frac{h}{q B r}$

Hence, option "C" is correct.

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An α particle is moving in a circular path of radius in the presence of magnetic field B. The de-Broglie wavelength associated with the particle will be (q→chargeonαparticle)

The correct option is C hqBrWe know,Radius =r=mvqB and, λ=hmv⇒ λ=hqBr∴ De-Broglie wavelength =λ=hqBrHence, option "C" is correct.

An $α$ particle is moving in a circular path of radius in the presence of magnetic field B. The de-Broglie wavelength associated with the particle will be $(q \to c h a r g e o n α p a r t i c l e)$

The correct option is C $\frac{h}{q B r}$
We know,

Radius $= r = \frac{m v}{q B}$

and, $λ = \frac{h}{m v}$

$\Rightarrow λ = \frac{h}{q B r}$

$∴$ De-Broglie wavelength $= λ = \frac{h}{q B r}$

Hence, option "C" is correct.