Use the exact values you enter in previous answer(s) to make later calculation(s
ID: 1519476 • Letter: U
Question
Use the exact values you enter in previous answer(s) to make later calculation(s). The ionization energy of a hydrogen atom is 13.6 eV. When a hydrogen atom absorbs a photon with this energy, the electron is ejected from the atom. (a) What are the frequency and wavelength of a photon with this energy? frequency 3.28e15 Correct: Your answer is correct. . seenKey 3.28e+15 Hz wavelength 91.4 Correct: Your answer is correct. . seenKey 91.5 nm Where does this photon lie in the electromagnetic spectrum? (Use data from this figure.) correct ultraviolet visible light X-rays microwave gamma rays Correct: Your answer is correct. . . Could you see light with this frequency? Yes correct No Correct: Your answer is correct. . . (b) When a photon has an energy greater than 13.6 eV, the "extra" energy (the energy in excess of 13.6 eV) goes into kinetic energy of the ejected electron. If a hydrogen atom absorbs a photon with an energy of 13.9 eV, what are the kinetic energy and speed of the ejected electron? kinetic energy 2.364 Incorrect: Your answer is incorrect. . seenKey 0.3 eV speed 3.688 Incorrect: Your answer is incorrect. . seenKey 3.25e+05 m/s (c) When the light intensity is very high, it is possible for a hydrogen atom to absorb two photons simultaneously. If a hydrogen atom absorbs two photons of equal energy and the atom is just barely ionized, what are the frequency and wavelength of these photons? frequency 1.64e15 Correct: Your answer is correct. . seenKey 1.64e+15 Hz wavelength 183 Correct: Your answer is correct. . seenKey 183 nm Could you see this light? Yes correct No Correct: Your answer is correct. .
Explanation / Answer
a) Frequency = 13.6 * 1.6 * 10-19/(6.63 * 10-34)
= 3.28 * 1015 Hz
=> wavelength = 3 * 108/3.28 * 1015
= 9.14 * 10-8 m
=> this photon lie in the X ray region .
=> No, we cant see light with this frequency .
b) kinetic energy of the ejected electron = 13.9 - 13.6 = 0.3 eV
speed of the ejected electron = sqrt(0.3 * 1.6 * 10-19/9.11 * 10-31)
= 229541.61 m/sec
c) Frequency = (13.6/2) * 1.6 * 10-19/(6.63 * 10-34)
= 1.64 * 1015 Hz
=> wavelength = 3 * 108/1.64 * 1015
= 18.3 * 10-8 m
= 183 nm
=> No, we cant see light with this frequency .
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