An electron enters perpendicularly into uniform magnetic field having magnitude \(0.5 \times 10^{-5} \mathrm{~T}\). If it moves on a circular path of radius \(2 \mathrm{~mm}\), its de - Broglie wavelength is $\ldots \ldots . . . \AA$ (A) 3410 (B) 4140 (C) 2070 (D) 2785

When the electron moves in a circular path, the centripetal force required for the circular motion is provided by the magnetic force acting on it. The magnetic force on an electron can be given by the equation: \(F = e(vB)\), where e is the charge of an electron, v is its velocity, and B is the magnetic field strength. The centripetal force required for circular motion can be given by: \(F = \dfrac{mv^2}{r}\), where m is the mass of an electron and r is the radius of the circular path. Since the magnetic force provides the centripetal force, we can write: \(\dfrac{mv^2}{r} = e(vB)\)

The de Broglie wavelength formula relates the wavelength of a particle to its momentum: \(\lambda = \dfrac{h}{mv}\), where λ is the de Broglie wavelength, h is the Planck's constant, and mv is the momentum of the electron. To find the de Broglie wavelength, we need to determine the velocity of the electron. From the above relation in step 1, we can find the velocity as follows: \(\dfrac{mv^2}{r} = e(vB)\) => \(v = \dfrac{eBr}{m}\)

Now, we can substitute the expression for velocity into the de Broglie wavelength equation: \(\lambda = \dfrac{h}{m (\frac{eBr}{m})}\) Given values: B = \(0.5 \times 10^{-5} T\), r = \(2 \times 10^{-3} m\), e = \(1.6 \times 10^{-19} C\), m = \(9.11 \times 10^{-31} kg\), h = \(6.626 \times 10^{-34} Js\). \(\lambda = \dfrac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(\frac{(1.6 \times 10^{-19})(0.5 \times 10^{-5})(2 \times 10^{-3})}{9.11 \times 10^{-31}})}\) After calculating, we get: \(\lambda=2.07 \times 10^{-10} m\) Now, we need to convert the wavelength to Angstrom (Å): \(1 \mathrm{~Å} = 10^{-10} \mathrm{~m}\), \(\lambda = 2.07 \times 10^{-10} \mathrm{~m} = 2070 \mathrm{~Å}\) This corresponds to answer (C).