an electron in the second energy level absorbs a photon of energy 6 ev what's the speed of the ejected electron?
I have tried to manipulate and use the kinetic energy equation, c= lamba x f, and many other ways, but I don't know why can't get the answer, which is v= 8.38 x 10^5 m/s
To determine the speed of the ejected electron, you can start by using the energy conservation principle. The energy of the absorbed photon is equal to the difference in energy between the initial and final states of the electron.
The energy of an electron in a particular energy level can be calculated using the formula: E = -13.6 eV / n^2, where n is the principal quantum number.
Given that the electron is in the second energy level (n = 2), the initial energy of the electron is:
E_initial = -13.6 eV / (2^2) = -3.4 eV
Now, since the electron absorbs a photon of energy 6 eV, the final energy of the electron can be calculated by adding the absorbed energy to the initial energy:
E_final = E_initial + 6 eV = -3.4 eV + 6 eV = 2.6 eV
To find the speed of the ejected electron, you can use the relativistic equation:
E = (γ - 1)mc^2
where E is the total energy of the electron, m is the rest mass of the electron (9.11 x 10^-31 kg), c is the speed of light (3 x 10^8 m/s), and γ is the Lorentz factor.
To calculate γ, rearrange the equation to solve for γ:
γ = E/(mc^2) + 1
Now, plug in the energy value (converted to joules) in place of E:
γ = (2.6 eV * 1.6 x 10^-19 J/eV) / (9.11 x 10^-31 kg * (3 x 10^8 m/s)^2) + 1
γ ≈ 1.000000000963
Since the ejected electron has non-relativistic speed (γ ≈ 1), you can calculate its speed using the kinetic energy formula:
KE = 1/2 mv^2
Rearranging the formula to solve for v:
v = √((2 * KE) / m)
Substituting the energy value in place of KE:
v = √((2 * 2.6 eV * 1.6 x 10^-19 J/eV) / (9.11 x 10^-31 kg))
v ≈ 8.38 x 10^5 m/s
Therefore, the approximate speed of the ejected electron is 8.38 x 10^5 m/s.
To calculate the speed of the ejected electron, we need to use the concept of conservation of energy. When an electron absorbs a photon and transitions to a higher energy level, the absorbed energy is equal to the difference in energy between the initial and final levels.
In this case, we know that the energy absorbed by the electron is 6 eV. Since 1 eV is equivalent to 1.6 x 10^-19 joules, we can convert the energy absorbed to joules:
6 eV x (1.6 x 10^-19 J/eV) = 9.6 x 10^-19 J
Now, let's consider the kinetic energy of the ejected electron, which depends on its speed (v). The kinetic energy (KE) can be calculated using the formula:
KE = (1/2)mv^2
Where:
- KE is the kinetic energy
- m is the mass of the electron (9.11 x 10^-31 kg)
Since the initial energy of the electron is at the second energy level, we can use the equation for calculating the energy levels in electrons:
E = (-13.6 eV)/n^2
Where:
- E is the energy of the level
- n is the principal quantum number (in this case, n = 2 for the second energy level)
Plugging in the values, we find:
Energy of second level = (-13.6 eV)/(2^2) = -3.4 eV
To calculate the energy gained by the electron, we subtract the initial energy from the absorbed energy:
Energy gained by the electron = 6 eV - (-3.4 eV) = 9.4 eV
Now, we need to convert this energy to joules:
9.4 eV x (1.6 x 10^-19 J/eV) = 1.504 x 10^-18 J
Next, we set the kinetic energy of the ejected electron equal to the energy gained by the electron:
(1/2)mv^2 = 1.504 x 10^-18 J
Rearranging this equation, we can solve for the speed (v) of the ejected electron:
v^2 = (2 x 1.504 x 10^-18 J) / m
Plugging in the values, we have:
v^2 = (2 x 1.504 x 10^-18 J) / (9.11 x 10^-31 kg)
Solving for v, we find:
v = √[(2 x 1.504 x 10^-18 J) / (9.11 x 10^-31 kg)]
Calculating this, we get:
v ≈ 8.38 x 10^5 m/s
So, the speed of the ejected electron is approximately 8.38 x 10^5 m/s.