In the Bohr model of the atom, the ground-state electron in hydrogen has an orbital speed of 2190 km/s.The mass of an electron is 9.11×10−31kg. What is its kinetic energy?
what is 1/2 mass*velocity^2 ?
change km/s to m/s
13,6ev
To find the kinetic energy of the electron, we can use the formula:
Kinetic Energy (KE) = (1/2) * mass * velocity^2
Given:
Mass of electron (m) = 9.11×10^(-31) kg
Velocity of electron (v) = 2190 km/s
First, we need to convert the velocity from km/s to m/s:
1 km = 1000 m
1 s = 1 s
Thus, the velocity of the electron in m/s is:
v = 2190 km/s * (1000 m/km) * (1 s/1 s) = 2.19 × 10^6 m/s
Now, we can substitute the values into the formula to find the kinetic energy:
KE = (1/2) * m * v^2
= (1/2) * (9.11×10^(-31) kg) * (2.19 × 10^6 m/s)^2
Calculating this expression will give us the answer for the kinetic energy of the electron.
To determine the kinetic energy of the ground-state electron in hydrogen, we need to use the formula for kinetic energy:
K.E. = (1/2)mv^2
Where K.E. is the kinetic energy, m is the mass of the electron, and v is the orbital speed. Given that the mass of an electron is 9.11×10−31 kg and the orbital speed is 2190 km/s, we can plug in these values into the formula:
K.E. = (1/2) * (9.11×10−31 kg) * (2190 km/s)^2
Now, we need to convert the orbital speed from km/s to m/s because the mass is in kg:
1 km = 1000 m
1 s = 1 s
Therefore, we can convert km/s to m/s by multiplying by 1000:
K.E. = (1/2) * (9.11×10−31 kg) * (2190 km/s * 1000 m/km)^2
Simplifying the equation:
K.E. = (1/2) * (9.11×10−31 kg) * (2190000 m/s)^2
Now, we can calculate the kinetic energy:
K.E. = (1/2) * (9.11×10−31 kg) * (4784100000 m^2/s^2)
Multiplying and simplifying further:
K.E. ≈ 2.081×10−17 J
Therefore, the kinetic energy of the ground-state electron in hydrogen is approximately 2.081×10−17 J.