When irradiated with a radiation of wavelength 58.4 nm the electron of H atom was ejected
with kinetic energy 7.6 eV. Calculate the ionization energy.
convert the photon wavelenght to energy, in ev.
subtract the 7.5eV, and you have it.
E=PlancksConstant*speedlight/lambda
h=4.15)×10−15 eV s
To calculate the ionization energy, we first need to understand what information is given: the wavelength of the radiation (λ) and the kinetic energy of the ejected electron (KE).
1. Convert the wavelength of the radiation to energy using the equation: E = hc/λ.
- Planck's constant (h) = 6.626 x 10^(-34) J·s
- Speed of light (c) = 3.00 x 10^8 m/s
E = (6.626 x 10^(-34) J·s * 3.00 x 10^8 m/s) / (58.4 x 10^(-9) m)
2. Convert the energy from Joules to electron volts (eV) using the conversion factor: 1eV = 1.6 x 10^(-19) J.
- Energy (in Joules) / (1.6 x 10^(-19) J/eV)
Now, we have the energy in electron volts (eV).
3. The ionization energy (IE) is the difference between the energy of the ejected electron (KE) and the energy level of the initial state of the electron in the hydrogen atom.
IE = KE - Energy of the initial state of the electron
Substituting the value of KE into the equation:
IE = 7.6 eV - Energy of the initial state of the electron
Note: The energy of the initial state in the hydrogen atom is the ground state energy, which is -13.6 eV.
IE = 7.6 eV - (-13.6 eV)
By simplifying the above expression, we get:
IE = 7.6 eV + 13.6 eV
Thus, the ionization energy of the hydrogen atom is 21.2 eV.