This comprehensive revision guide focuses on key concepts of modern physics, including the photoelectric effect, blackbody radiation, and wave-particle duality. It covers the calculation of threshold frequency and wavelength, work function, and kinetic energy of photoelectrons. Additionally, it explores the relationship between temperature and radiation, the properties of photons, and the de Broglie wavelength of macroscopic objects.
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اختر إجابة واحدة لكل سؤال. عند الاختيار ستظهر النتيجة فورًا: الأخضر صحيح، والأحمر خطأ، وسيظهر تفسير الإجابة مباشرة إن كان متوفرًا. وبعد آخر سؤال ستظهر الدرجة النهائية تلقائيًا.
Incident radiation falls on a tin surface. The threshold frequency of tin is: f0 = 1.2 × 1015 Hz. What is the threshold wavelength of tin?
Explanation
The threshold wavelength is calculated using the formula \(\lambda_0 = \frac{c}{f_0}\). Substituting the values: \(\lambda_0 = \frac{3 \times 10^8 \text{ m/s}}{1.2 \times 10^{15} \text{ Hz}} = 2.5 \times 10^{-7} \text{ m}\).
Question 2
DB ID: 420
Points: 1
Incident radiation falls on a tin surface. The threshold frequency of tin is: f0 = 1.2 × 1015 Hz. What is the work function of tin?
Explanation
The work function W is given by W = h f0. Using h = 6.63 × 10-34 J\cdots, \(W = (6.63 \times 10^{-34} \times 1.2 \times 10^{15}) / (1.6 \times 10^{-19}) \approx 4.97 \text{ eV}\), which rounds to 5.0 eV.
Question 3
DB ID: 421
Points: 1
Incident radiation falls on a tin surface (f0 = 1.2 × 1015 Hz). The incident radiation has wavelength \(\lambda = 167 \text{ nm}\). What is the maximum kinetic energy of the emitted electrons?
Explanation
Energy of incident photon \(E = \frac{1240}{\lambda(\text{nm})} = \frac{1240}{167} \approx 7.42 \text{ eV}\). Using the photoelectric equation KEmax = E - W = 7.42 - 5.0 = 2.42 eV.
Question 4
DB ID: 422
Points: 1
Two iron bars are heated. One glows dark red, while the other glows bright orange. Which bar is hotter?
Explanation
According to Wien's displacement law, as temperature increases, the peak wavelength decreases. Orange light has a shorter wavelength and higher frequency than red light, indicating a higher temperature.
Question 5
DB ID: 423
Points: 1
Two iron bars are heated. One glows dark red, while the other glows bright orange. Which bar is radiating more energy?
Explanation
According to the Stefan-Boltzmann law, the total energy radiated per unit area is proportional to the fourth power of the absolute temperature (\(P \propto T^4\)). Since the orange bar is hotter, it radiates more energy.
Question 6
DB ID: 424
Points: 1
Will high-frequency light eject a greater number of electrons than low-frequency light (both above threshold)?
Explanation
The number of emitted electrons per second depends on the intensity of the incident light (number of photons), not the frequency, provided the frequency is above the threshold.
Question 7
DB ID: 425
Points: 1
Potassium emits electrons under blue light, while tungsten requires ultraviolet light. Which metal has the higher threshold frequency?
Explanation
Ultraviolet light has a higher frequency than blue light. Since tungsten requires a higher frequency photon to start emitting electrons, it has a higher threshold frequency.
Question 8
DB ID: 426
Points: 1
Based on the fact that potassium emits electrons under blue light while tungsten requires ultraviolet light, which metal has the larger work function?
Explanation
The work function is directly proportional to the threshold frequency (W = h f0). Since tungsten has a higher threshold frequency, it also has a larger work function.
Question 9
DB ID: 427
Points: 1
A baseball of diameter 0.10 m moves at 21 m/s. How does its de Broglie wavelength compare to its size?
Explanation
Due to the macroscopic mass of the baseball, the de Broglie wavelength (\(\lambda = h/mv\)) is extremely small (on the order of 10-34 m), which is negligible compared to its diameter of 0.10 m.
Question 10
DB ID: 428
Points: 1
Just barely visible light has intensity: I = 1.5 × 10-11 W/m2. The pupil diameter is 7.0 mm. What is the power entering the eye?
Explanation
Power is intensity \times area: \(P = I \times \pi r^2 = 1.5 \times 10^{-11} \times \pi \times (0.0035)^2 \approx 5.77 imes 10^{-16} \text{ W}\).
Question 11
DB ID: 429
Points: 1
Light of wavelength: \(\lambda = 550 \text{ nm}\). What is the energy of one photon?
Explanation
Energy of one photon is \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}} \approx 3.62 \times 10^{-19} \text{ J}\).
Question 12
DB ID: 430
Points: 1
Using the results from the previous questions (Power entering eye = 5.8 × 10-16 W, Photon energy = 3.62 × 10-19 J), what is the number of photons per second entering the eye?
Explanation
The number of photons per second is the total power divided by the energy per photon: \(n = \frac{P}{E} = \frac{5.8 \times 10^{-16}}{3.62 \times 10^{-19}} \approx 1602 \text{ photons/s}\).
Question 13
DB ID: 431
Points: 1
Light of wavelength 443 nm ejects electrons with kinetic energy 1.56 eV. What is the work function of the metal?
Explanation
Incident photon energy \(E = \frac{1240}{443} \approx 2.80 \text{ eV}\). Work function W = E - KE = 2.80 - 1.56 = 1.24 eV.
Question 14
DB ID: 432
Points: 1
A researcher illuminates a sample of metal and finds the longest wavelength to eject electrons is 273 nm. Use the table below to identify the most likely metal.
Explanation
The longest wavelength corresponds to the threshold wavelength. In the provided table, Silver has a threshold wavelength of 270 nm, which is the closest value to 273 nm.
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