You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Matter Waves

An α partic...

Question

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{h B}{q r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{h}{q B r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

An $α$ particle moves in circular path of radius $0.83 c m$ in the presence of a magnetic field of $0.25 W b / m^{2}$ . Find the De Broglie wavelength associated with the particle

Q. An particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb

m^{2}

. The de Broglie wavelength associated with the particle will be :

Q. The radius of an

α

-particle moving in a circle in a constant magnetic field is half of the radius of an electron moving in circular path in the same field. The de Broglie wavelength of

α

-particle is

x

times that of the electrons. Find

x

(an integer).

Q. An alpha particle moving with an initial velocity

(u j)

enters a region of uniform magnetic field

B = B k

. the de-Broglie wavelength of the alpha particle.

Q. An

α - p a r t i c l e

and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The

α - p a r t i c l e

and the proton move such that radius of curvature of their paths is same. Find the ratio of their de Broglie wavelengths :

Join BYJU'S Learning Program

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.