Final answer:
The ionization energy of krypton, calculated using the energy of a photon from a helium lamp and the kinetic energy of ejected electrons, is approximately 13.98 eV.
Step-by-step explanation:
To find the ionization energy of krypton, we first need to use the photoelectric effect equation:
E = K + W
where E is the energy of the incoming photon, K is the maximum kinetic energy of the ejected electrons, and W is the work function or ionization energy of the material.
The energy of a photon (E) can be calculated using the equation:
E = (hc) / λ
where h is Planck's constant (6.626 x 10-34 J·s), c is the speed of light (3 x 108 m/s), and λ is the wavelength of the incoming UV radiation. Substituting h, c, and the given λ of 58.4 nm into the equation:
E = (6.626 x 10-34 J·s x 3 x 108 m/s) / 58.4 x 10-9 m = 3.39 x 10-18 J
The kinetic energy (K) of an ejected electron is given by:
K = 0.5mv2
where m is the mass of an electron (9.109 x 10-31 kg) and v is the speed of the ejected electron. Substituting the given v into the equation:
K = 0.5 x 9.109 x 10-31 kg x (1.59 x 106 m/s)2 = 1.15 x 10-18 J
The work function (W), or ionization energy of krypton, is E - K:
W = 3.39 x 10-18 J - 1.15 x 10-18 J = 2.24 x 10-18 J
To express W in electron volts (eV), we use the conversion 1 eV = 1.602 x 10-19 J:
W = 2.24 x 10-18 J / 1.602 x 10-19 J/eV = 13.98 eV
Therefore, the ionization energy of krypton is approximately 13.98 eV.