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A photoelectric experiment is carried out by students. They measure the threshold frequency of light required for photoemission to be 6.5 × 10\(^{14}\) Hz and the work function to be 3.2 × 10\(^{-19}\) J.
Using the students' measurements, what value would they calculate for Planck's constant? Outline your reasoning and show all your working. Give your answer in joule-seconds. (3 marks)
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\(h=4.9 \times 10^{-34}\ \text{J s}\)
\(f=6.5 \times 10^{14}\ \text{Hz},\ \ E=3.2 \times 10^{-19}\ \text{J}\)
\(E=hf\ \ \Rightarrow\ \ h=\dfrac{E}{f}\)
\(\therefore h=\dfrac{3.2 \times 10^{-19}}{6.5 \times 10^{14}} = 4.92 \times 10^{-34}\ \text{J s}\)
The James Webb Space Telescope (JWST) is sensitive to wavelengths from 6.0 \(\times\) 10\(^{-7}\) m to 2.8 \(\times\) 10\(^{-5}\) m. What is the minimum photon energy that it can detect? (3 marks) --- 6 WORK AREA LINES (style=lined) --- \(E_{\text{min}}= 7.1 \times 10^{-21}\) \( \text{J}\) → The minimum photon energy corresponds to the minimum frequency. → Minimum frequency occurs at the maximum wavelength as frequency and wavelength are inversely proportional.
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\(E_{\text{min}}\)
\(=hf\)
\(=\dfrac{hc}{ \lambda}\)
\(=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.8 \times 10^{-5}} \)
\(=7.1 \times 10^{-21}\) \(\text J\)
Applying the law of conservation of energy, explain why `K_max = h f-phi`. (3 marks)
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→ The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed into different forms.
→ The initial energy of a photon is equal to `hf`.
→ If this photon strikes a metal surface, a photoelectron may be released. Some energy is required to remove the electron from the metal surface which is equal to the work function (`phi`) of the metal.
→ The electron will also possess kinetic energy, `K_(max)`.
→ Applying the law of conservation of energy, the energy before equals the energy after, or `hf=K_(max)+phi`.
→ Rearranging this gives `K_(max)=hf-phi`.
→ The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed into different forms.
→ The initial energy of a photon is equal to `hf`.
→ If this photon strikes a metal surface, a photoelectron may be released. Some energy is required to remove the electron from the metal surface which is equal to the work function (`phi`) of the metal.
→ The electron will also possess kinetic energy, `K_(max)`.
→ Applying the law of conservation of energy, the energy before equals the energy after, or `hf=K_(max)+phi`.
→ Rearranging this gives `K_(max)=hf-phi`.
What is the wavelength, in metres, of a photon with an energy of 3.5 eV?
`B`
| `E` | `=3.5 xx1.602 xx10^(-19)` | |
| `=5.607 xx10^(-19) text{J}` |
`E=hf=(hc)/(lambda)\ \ => \ \ lambda=(hc)/(E)`
| `:.lambda` | `=(6.626 xx10^(-34)xx3xx10^(8))/(5.607 xx10^(-19))` | |
| `=3.5 xx10^(-7) text{m}` |
`=>B`
A laser emits light of wavelength 550 nm.
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a. `f=5.45 xx10^(14) text{Hz}`
| b. | `E` | `=3.63 xx10^(-19) text{J}` |
| `<5 xx10^(-19) text{J}` |
`text{So, no electrons are ejected.}`
| a. | `v` | `=flambda` |
| `f` | `=(v)/(lambda)` | |
| `=(3xx10^(8))/(550 xx10^(-9))` | ||
| `=5.45 xx10^(14) text{Hz}` |
| b. | `E` | `=hf` |
| `=6.62 xx10^(-34)xx5.46 xx10^(14)` | ||
| `=3.63 xx10^(-19) text{J}` |
→ The work function of the metal sample is `5 xx10^(-19)\ text{J}`.
→ Since the incident photon energy of `3.63 xx10^(-19) text{J}` is less than `5 xx10^(-19)\ text{J}`, they are unable to eject electrons from the metal.
A photon has an energy of `9.0 xx10^(-24)\ text{J}`.
What is the frequency of this radiation?
`C`
| `E` | `=hf` | |
| `f` | `=(E)/(h)` | |
| `=(9.0 xx10^(-24))/(6.626 xx10^(-34))` | ||
| `=1.36 xx10^(10) text{Hz}` |
`=>C`