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Interpretation: The monochromatic light shining on cesium metal is just above the threshold frequency is to be explained.
Concept Introduction: Single-
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Answer to Problem 65A
When the monochromatic light shining on the cesium metal is just slightly above the threshold frequency, the electrons are released with a low velocity.
Explanation of Solution
The
Monochromatic light means the light of one frequency.
When the monochromatic light is just above the threshold frequency, the electrons will be emitted. The electron will be having low velocity.
Interpretation: The intensity of the light increases, but the frequency remains the same is to be explained.
Concept Introduction: Single-wavelength light sources are known as monochromatic lights, where mono stands for only one chroma for color. The threshold frequency is a minimum frequency below which, regardless of the strength of the incident radiation, photoelectric emission is not possible.
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Answer to Problem 65A
A greater number of electrons are emitted at a low velocity when the intensity of the light is increased.
Explanation of Solution
The photoelectric effect takes place when the frequency of light is higher than the threshold frequency.
Since the intensity is increased, there is the emission of more electrons but at a low velocity.
Interpretation: Monochromatic light of a shorter wavelength is used is to be explained.
Concept Introduction: Single-wavelength light sources are known as monochromatic lights, where mono stands for only one chroma for color. The threshold frequency is a minimum frequency below which, regardless of the strength of the incident radiation, photoelectric emission is not possible.
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Answer to Problem 65A
There is an emission of electrons at a faster velocity due to the usage of a shorter wavelength.
Explanation of Solution
The photoelectric effect takes place when the frequency of light is higher than the threshold frequency.
Monochromatic light means the light of one frequency.
The wavelength and frequency are inversely proportional to each other.
Since the shorter wavelength is used, that means the higher frequency, so the emission of electrons occurs with a high velocity.
Chapter 5 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
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- Use the literature Ka value of the acetic acid, and the data below to answer these questions. Note: You will not use the experimental titration graphs to answer the questions that follow. Group #1: Buffer pH = 4.35 Group #2: Buffer pH = 4.70 Group #3: Buffer pH = 5.00 Group #4: Buffer pH = 5.30 Use the Henderson-Hasselbalch equation, the buffer pH provided and the literature pKa value of acetic acid to perform the following: a) calculate the ratios of [acetate]/[acetic acid] for each of the 4 groups buffer solutions above. b) using the calculated ratios, which group solution will provide the best optimal buffer (Hint: what [acetate]/[acetic acid] ratio value is expected for an optimal buffer?) c) explain your choicearrow_forwardHow would you prepare 1 liter of a 50 mM Phosphate buffer at pH 7.5 beginning with K3PO4 and 1 M HCl or 1 M NaOH? Please help and show calculations. Thank youarrow_forwardDraw the four most importantcontributing structures of the cation intermediate thatforms in the electrophilic chlorination of phenol,(C6H5OH) to form p-chlorophenol. Put a circle aroundthe best one. Can you please each step and also how you would approach a similar problem. Thank you!arrow_forward
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