Homework 8 solutions

Homework 8 solutions - 28.1 AC 28.2 B Intensity of incident...

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28.1 - AC 28.2 - B Intensity of incident light refers to the number of photons falling upon a surface, while the frequency of light corresponds with the energy each photon carries. Increasing the number of photons increases the number of emitted electrons, while increasing the frequency of each photon increases the energy transferred to each electron resulting in greater electron speed. 28.3 - The distance between adjacent electron orbits gets progressively larger and larger. According to the Bohr model, the radius of an electron orbit in a hydrogen atom is proportional to the square of its state n. Therefore, as n increases, the distance between adjacent orbits increases. 28.4 - The difference in energy between adjacent electron orbits gets progressively smaller and smaller. According to the Bohr model, the magnitude of the energy of a hydrogen atom in state n is inversely proportional to the square of n. Therefore, as n increases, the difference between adjacent energy states decreases. 28.5 - If the photon's wavelength were doubled, its energy would be 1 2 ˙ E A photon's energy is inversely proportional to its wavelength. Doubling the wavelength halves its energy. 28.6 - If an electron makes a transition from the n = 3 to the n = 1 shell, it will give up 72.0 eV of energy Higher shells correspond to higher energy states. An atom transitioning from a higher to a lower shell must release an amount of energy equal to the difference between energy levels. 28.7 - 5.0 eV A table of common elements and their work functions is provided on page 935 in your text book. 28.9 - 81.0 E. E n = hcR n 2 Ground state: E 1 = hcR 1 2 = hcR therefore: E n = E 1 n 2 E 1 = E n ˙ n 2 E 1 = E 9 ˙ 9 2 E 1 = 81 E n
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28.10 - wavelength decreases = h mv Wavelength and velocity are inversely proportional, so a greater voltage V results in greater electron velocity v. 28.16 - greater than 2K The kinetic energy of a released electron is equal to the energy of the photon that released it minus the energy it took to release it: K 1 = E − ф doesn't change when E is doubled, so: K 2 = 2 ˙ E − the difference between K 1 and K 2 : K = 2 E −− E − K = E ф is always positive, so: E E − E K 1 if K had doubled, the difference between K 1 and K 2 would be K 1 , but we found the difference to be E, which is greater than K 1 . 28.13 - the wavelength of the photon increases. When the photon collides with a stationary electron, it loses some of its energy. we saw before that the energy and wavelength of a photon are inversely proportional, so a decrease in energy corresponds with an increase in wavelength.
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