An \alpha 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:

  • Option 1)

    1\: \: A

  • Option 2)

    0.1\: \: A

  • Option 3)

    10\: \: A

  • Option 4)

    0.01\: \: A

 

Answers (1)

As we learnt in

De - Broglie wavelength -

\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

 

 

 

Radius in a magnetic field is R=\frac{mv}{qB}\:\: \: or\:\: \: mv=qBR

the debroglic wavelength \lambda=\frac{h}{mv}=\frac{h}{qBR}=\frac{6.62 \times 10^{-34}}{(2\times 1.6 \times 10^{-19}\times 0.25 \times 8.3\times 10^{-3})}

\lambda=\frac{6.62 \times 10 ^{-34}}{6.64 \times 10^-22} \simeq 10^{-12} m=0.01\AA


Option 1)

1\: \: A

Incorrect

Option 2)

0.1\: \: A

Incorrect

Option 3)

10\: \: A

Incorrect

Option 4)

0.01\: \: A

Correct

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