Question

An electron is accelerated under a potential difference of \(64 \mathrm{~V}\), the de Broglie wave length associated with electron is $=\ldots \ldots \ldots \ldots . \AA$ (Use charge of election $=1.6 \times 10^{-19} \mathrm{C},=9.1 \times 10^{-31} \mathrm{Kg}$, mass of electrum \(\mathrm{h}=6.6623 \times 10^{-43} \mathrm{~J}\).sec \()\) (A) \(4.54\) (B) \(3.53\) (C) \(2.53\) (D) \(1.534\)

Short answer

The de Broglie wavelength associated with the electron is approximately \(3.53 \, \AA \), which corresponds to option (B).

Step-by-step solution

Calculate the kinetic energy of the electron

Using the given potential difference (V) and the charge of the electron (e), we can calculate the kinetic energy (KE) of the electron: \(KE = eV\) where e = 1.6 x 10^-19 C V = 64 V Plug in the values: KE = (1.6 x 10^-19 C)(64 V) = 1.024 x 10^-17 J

Calculate the momentum of the electron

Now, we can find the momentum (p) of the electron using its kinetic energy (KE) and mass (m): \(p = \sqrt{2mKE}\) where m = 9.1 x 10^-31 kg KE = 1.024 x 10^-17 J Plug in the values: p = \(\sqrt{2(9.1 \times 10^{-31} \mathrm{ kg})(1.024 \times 10^{-17} \mathrm{ J})}\) = 1.897 x 10^-24 kg m/s

Calculate the de Broglie wavelength

Finally, we can find the de Broglie wavelength (λ) using Planck's constant (h) and the momentum (p): \(λ = \dfrac{h}{p}\) where h = 6.6623 x 10^-34 Js p = 1.897 x 10^-24 kg m/s Plug in the values: λ = \(\dfrac{6.6623 \times 10^{-34} \mathrm{ Js}}{1.897 \times 10^{-24} \mathrm{ kg ~m/s}}\) ≈ 3.51 x 10^-10 m = 3.51 Å Comparing our result to the given options of (A) 4.54, (B) 3.53, (C) 2.53, and (D) 1.534, option (B) 3.53 is the closest to our calculated de Broglie wavelength. Therefore, the correct answer is: (B) 3.53 Å