A 10 eV electron is circulating in a plane at right angles to a uniform magnetic field of 1.0 × 10⁻⁴ WB/m² (= 1.0 gauss). The orbital radius of the electron is:
(A) 11 cm
(B) 18 cm
(C) 12 cm
(D) 16 cm

To find the orbital radius of an electron moving in a magnetic field, we need to apply the principles of motion of charged particles in a magnetic field.
Given:
• The energy of the electron is 10 eV.
• The magnetic field induction B=10−4 WB/m2=1.0 gaussB = 10^{-4} \, \text{WB/m}^2 = 1.0 \, \text{gauss}.
• The electron moves in a plane at right angles to the magnetic field, implying circular motion.
We need to calculate the orbital radius of the electron.
Step 1: Convert the Energy of the Electron to Joules
The energy given is in electron volts (eV), and we need to convert it to joules (J) because the SI unit of energy is the joule.
1 eV = 1.6×10−19 J1.6 \times 10^{-19} \, \text{J}.
Therefore, the energy of the electron is:
E=10 eV=10×1.6×10−19 J=1.6×10−18 J.E = 10 \, \text{eV} = 10 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-18} \, \text{J}.
Step 2: Find the Speed of the Electron
The kinetic energy of the electron is given by:
E=12mv2E = \frac{1}{2} m v^2
Where:
• mm is the mass of the electron, which is m=9.11×10−31 kgm = 9.11 \times 10^{-31} \, \text{kg},
• vv is the velocity of the electron.
Rearranging to solve for vv:
v=2Em=2×1.6×10−189.11×10−31v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-18}}{9.11 \times 10^{-31}}} v=3.2×10−189.11×10−31v = \sqrt{\frac{3.2 \times 10^{-18}}{9.11 \times 10^{-31}}} v≈1.88×107 m/s.v \approx 1.88 \times 10^7 \, \text{m/s}.
Step 3: Use the Formula for the Radius of Circular Motion
For an electron moving in a circular orbit under the influence of a magnetic field, the radius rr is given by the formula:
r=mvqBr = \frac{mv}{qB}
Where:
• mm is the mass of the electron (9.11×10−31 kg9.11 \times 10^{-31} \, \text{kg}),
• vv is the velocity of the electron (1.88×107 m/s1.88 \times 10^7 \, \text{m/s}),
• qq is the charge of the electron (q=1.6×10−19 Cq = 1.6 \times 10^{-19} \, \text{C}),
• BB is the magnetic field induction (B=10−4 TB = 10^{-4} \, \text{T}).
Substituting the values into the formula:
r=9.11×10−31×1.88×1071.6×10−19×10−4r = \frac{9.11 \times 10^{-31} \times 1.88 \times 10^7}{1.6 \times 10^{-19} \times 10^{-4}} r=1.71×10−231.6×10−23r = \frac{1.71 \times 10^{-23}}{1.6 \times 10^{-23}} r≈1.07 m.r \approx 1.07 \, \text{m}.
Thus, the radius of the electron's orbit is approximately 1.07 m.
Since the question asks for the orbital radius in centimeters, we convert meters to centimeters:
r≈107 cm.r \approx 107 \, \text{cm}.
Conclusion:
The orbital radius of the electron is approximately 107 cm. Therefore, the correct answer is none of the options provided in the list. There might be an issue with the options or further clarification needed in the problem statement.