An electron is accelerated from rest, between two points A and B at which the potentials are 20V and 40V respectively. The de Broglie wavelength associated with the electron at B will be:

A) 0.75 Å
B) 7.5 Å
C) 2.75 Å
D) 2.75 m

To find the de Broglie wavelength associated with the electron at point B, we need to first understand the relationship between the electron's kinetic energy and the voltage difference it is accelerated through, and then use that to calculate its de Broglie wavelength.
Step 1: Understanding the Kinetic Energy
When an electron is accelerated by a potential difference VV, the kinetic energy gained by the electron is given by the work done on it by the electric field. The kinetic energy (K.E.K.E.) is:
K.E.=eV,K.E. = eV,
where:
• ee is the charge of the electron (e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}),
• VV is the potential difference through which the electron is accelerated.
The electron is accelerated between points A (20V) and B (40V), so the potential difference between these points is:
Vdiff=VB−VA=40V−20V=20V.V_{\text{diff}} = V_B - V_A = 40V - 20V = 20V.
Therefore, the kinetic energy gained by the electron is:
K.E.=e×20V=(1.6×10−19 C)×(20 V)=3.2×10−18 J.K.E. = e \times 20V = (1.6 \times 10^{-19} \, \text{C}) \times (20 \, \text{V}) = 3.2 \times 10^{-18} \, \text{J}.
Step 2: Finding the Electron's Speed
The kinetic energy of the electron is also related to its velocity (vv) by the equation:
K.E.=12mv2,K.E. = \frac{1}{2}mv^2,
where:
• mm is the mass of the electron (m=9.11×10−31 kgm = 9.11 \times 10^{-31} \, \text{kg}),
• vv is the velocity of the electron.
Rearranging the equation to solve for vv:
v=2×K.E.m.v = \sqrt{\frac{2 \times K.E.}{m}}.
Substituting the values:
v=2×3.2×10−18 J9.11×10−31 kg=6.4×10−189.11×10−31=7.02×1012≈2.65×106 m/s.v = \sqrt{\frac{2 \times 3.2 \times 10^{-18} \, \text{J}}{9.11 \times 10^{-31} \, \text{kg}}} = \sqrt{\frac{6.4 \times 10^{-18}}{9.11 \times 10^{-31}}} = \sqrt{7.02 \times 10^{12}} \approx 2.65 \times 10^6 \, \text{m/s}.
Step 3: de Broglie Wavelength
The de Broglie wavelength λ\lambda of a particle is given by:
λ=hmv,\lambda = \frac{h}{mv},
where:
• hh is Planck's constant (h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s}),
• mm is the mass of the electron,
• vv is the velocity of the electron.
Substituting the known values:
λ=6.63×10−34(9.11×10−31)×(2.65×106)=6.63×10−342.41×10−24≈2.75×10−10 m.\lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (2.65 \times 10^6)} = \frac{6.63 \times 10^{-34}}{2.41 \times 10^{-24}} \approx 2.75 \times 10^{-10} \, \text{m}.
Step 4: Converting to Angstroms
To express the wavelength in angstroms (A∘\text{A}^\circ), we convert from meters to angstroms:
1 m=1010 A∘.1 \, \text{m} = 10^{10} \, \text{A}^\circ.
Thus:
λ=2.75×10−10 m×1010 A∘/m=2.75 A∘.\lambda = 2.75 \times 10^{-10} \, \text{m} \times 10^{10} \, \text{A}^\circ / \text{m} = 2.75 \, \text{A}^\circ.
Final Answer:
The de Broglie wavelength associated with the electron at point B is 2.75 A∘2.75 \, \text{A}^\circ.
Thus, the correct answer is:
C)2.75 A∘.\boxed{C) 2.75 \, \text{A}^\circ}.