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 (Take h = 6.63 × 10-34 J s, e =1.6 ×10-19C)

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 (Take h = 6.63 × 10-34 J s, e =1.6 ×10-19C) (A) 1 mathringA (B) 0.1 mathringA (C) 10 mathringA (D) 0.01 mathringA

Tardigrade Admin · Accepted Answer

Radius of the circular path of a charged particle in a magnetic field is given by

R = \frac{m v}{Bq}

or

m v = RBq

Here,

R = 0.83 c m = 0.83 \times 1 0^{- 2} m

B = 0.25 Wb m^{2}

q = 2 e = 2 \times 1.6 \times 1 0^{- 19} C

∴ m v = (0.83 \times 1 0^{- 2}) (0.25) (2 \times 1.6 \times 1 0^{- 19})

de Broglie wavelength,

λ = \frac{h}{m v} = \frac{6.6 \times 1 0 ^{- 34}}{0.83 \times 1 0 ^{- 2} \times 0.25 \times 2 \times 1.6 \times 1 0 ^{- 19}}

= 0.01 \overset{˚}{A}

Q. An $α -$ particle moves in a circular path of radius $0.83 c m$ in the presence of a magnetic field of $0.25 Wb / m^{2} .$ The de Broglie wavelength associated with the particle will be
(Take $h = 6.63 \times 1 0^{- 34} J s$ , $e = 1.6 \times 1 0^{- 19} C$ )

$1 \overset{˚}{A}$

$0.1 \overset{˚}{A}$

$10 \overset{˚}{A}$

$0.01 \overset{˚}{A}$

Q. An α− particle moves in a circular path of radius 0.83cm in the presence of a magnetic field of 0.25Wb/m2. The de Broglie wavelength associated with the particle will be (Take h=6.63×10−34Js, e=1.6×10−19C)

1A˚

0.1A˚

10A˚

0.01A˚

Q. An $α -$ particle moves in a circular path of radius $0.83 c m$ in the presence of a magnetic field of $0.25 Wb / m^{2} .$ The de Broglie wavelength associated with the particle will be
(Take $h = 6.63 \times 1 0^{- 34} J s$ , $e = 1.6 \times 1 0^{- 19} C$ )

$1 \overset{˚}{A}$

$0.1 \overset{˚}{A}$

$10 \overset{˚}{A}$

$0.01 \overset{˚}{A}$