An electron at ground state in the atom (-10.38eV)is struck by a passing electron with 7.00eV of energy. What is the velocity of the passing electron afrer the collision?
To find the velocity of the passing electron after the collision, we need to use the conservation of energy and momentum principles.
1. We know that the total energy before the collision is the sum of the energies of both electrons: -10.38eV + 7.00eV = -3.38eV
2. Assuming an elastic collision, the total energy after the collision remains the same: -3.38eV
3. We can convert the total energy to kinetic energy using the equation: E = 0.5 * m * v^2, where E is the kinetic energy, m is the mass, and v is the velocity.
4. We can assume the electrons have the same mass, so we can set up the following equation: (0.5 * m * v1^2) + (0.5 * m * v2^2) = -3.38eV
5. Since the ground state electron is initially at rest, its velocity (v1) is zero. The equation simplifies to: 0 + (0.5 * m * v2^2) = -3.38eV
6. Rearranging the equation, we find: v2^2 = (-3.38eV) / (0.5 * m)
7. The mass of an electron is approximately 9.10938356 × 10^-31 kilograms.
8. Plugging in the values, we get: v2^2 = (-3.38eV) / (0.5 * 9.10938356 × 10^-31 kg)
9. Calculate v2^2: v2^2 = -3.38eV / 2 * 9.10938356 × 10^-31 kg
10. The square root of v2^2 will give us the velocity of the passing electron after the collision.