Question

An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m². The de-Broglie wavelength associated with the particle will be
1. 10 Å
2. 0.01 Å
3. 1 Å
4. 0.1 Å

Answer 1

Correct Answer - Option 2 : 0.01 Å

CONCEPT:

• The radius of the circle swept by a charged particle under an applied magnetic field is given by

\(⇒ r = \frac{mV}{qB}\)

Where V = Velocity, m=  mass. q = charge , B = magnetic field 

• De Broglie wavelength of a particle connects the wave nature and particle nature of an object or accounts for wave-particle duality, is given by

\(⇒ \lambda = \frac{h}{mV}\)

Where h = Plancks constant, V = Velocity, m = mass 

CALCULATION :

Given - r = 0.0083 m, B = 0.25 Wb/m² q = 2 × charge of proton = 2 × 1.69× 10^-19C

• The radius of a circle swept by charged particle under an applied magnetic field is given by

\(⇒ r = \frac{mV}{qB}\)

The above equation can be rewritten as

⇒ mV = qBr

Substituting the given values in the  above equation

⇒ mV = 2 × 1.69 × 10^-19 × 0.25 × 0.0083 = 0.664 × 10^-21 N

• The wavelength of the alpha particle is given by

\(⇒ \lambda = \dfrac{h}{mv}\)

Substituting the given values in the above equation

\(⇒\lambda = \dfrac{ 6.626\times 10^{-34}}{0.664\times 10^{-21}}\approx 0.01 A^{0}\)

• Hence, option 2 is the answer