# Uh Fisika 12 Genap

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| By Luhur105sman
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Luhur105sman
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Quizzes Created: 1 | Total Attempts: 1,767
Questions: 12 | Attempts: 1,767

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• 1.

### Suatu benda hitam pada suhu 27o C memancarkan energi 200 J/s. Benda hitam tersebut dipanasi hingga menjadi 327oC. Energi yang dipancarkan menjadi....

• A.

400 J/S

• B.

800 J/S

• C.

1.600 J/S

• D.

3.200 J/S

• E.

6.400 J/S

D. 3.200 J/S
Explanation
When an object is heated, it emits thermal radiation in the form of energy. This energy is directly proportional to the temperature of the object. In this case, the initial temperature of the black object is 27oC and it emits energy at a rate of 200 J/s. When the object is heated to 327oC, the temperature has increased by 300oC. Since the energy emitted is directly proportional to the temperature, the energy emitted will also increase by a factor of 300. Therefore, the energy emitted becomes 300 times the initial energy, which is 300 x 200 = 60000 J/s or 6.400 J/S.

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• 2.

### Suatu balok besi dengan suhu 1270C dan luas permukaan 4 cm2 memiliki koefisien emisitas = 0,5. Daya yang diradiasikan besi tersebut adalah (σ=5,67 x 10-8W/m2K4)....

• A.

0,19 W

• B.

0,29 W

• C.

0,39 W

• D.

0,49 W

• E.

0,59 W

B. 0,29 W
Explanation
The power radiated by an object can be calculated using the Stefan-Boltzmann law, which states that the power radiated is proportional to the emissivity, surface area, and the fourth power of the temperature. In this case, the temperature of the iron block is given as 1270C, and the surface area is 4 cm2. The emissivity is given as 0.5. Plugging these values into the formula, we can calculate the power radiated by the iron block as P = (emissivity) x (surface area) x (Stefan-Boltzmann constant) x (temperature)^4. Evaluating this expression gives us a value of approximately 0.29 W.

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• 3.

### Suhu filamen suatu lampu pijar mencapai 8840C. Panjang gelombang dengan intesitas terkuat diberikan oleh panjang gelombang sebesar ....

• A.

2,5 nm

• B.

25 nm

• C.

250 nm

• D.

0,25 µm

• E.

2,5 µm

E. 2,5 µm
Explanation
The given question is asking for the wavelength with the highest intensity of a light bulb filament with a temperature of 8840C. According to Wien's displacement law, the wavelength of maximum intensity (peak wavelength) is inversely proportional to the temperature. As the temperature increases, the peak wavelength decreases. Since the temperature is very high (8840C), the peak wavelength will be in the infrared range. Among the given options, 2.5 µm is the only wavelength in the infrared range, making it the correct answer.

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• 4.

• A.

1,66 eV

• B.

0,6 eV

• C.

0,5 eV

• D.

0,4 eV

• E.

0,3 eV

A. 1,66 eV
• 5.

### Perhatikan pernyataan – penyataan berikut.1. Energi fotoelektron tidak bergantung pada intensitas cahaya2. Sel fotolistrik banyak menghasilkan cahaya putih3. Energi fotoelektron tidak bergantung pada intensitas cahaya4. Sel fotolistrik selalu menghasilkan arus listrik jika disinari cahaya apapun.Pernyataan yang berkaitan dengan fotolistrik ditunjukan oleh nomor…

• A.

1,2,3, dan 4

• B.

1,2 dan 3

• C.

1 dan 3

• D.

2 dan 4

• E.

3 dan 4

C. 1 dan 3
Explanation
The correct answer is 1 and 3. The first statement states that photoelectron energy is not dependent on the intensity of light, while the third statement repeats the same information. Therefore, both statements are related to the topic of photoelectric effect.

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• 6.

• A.

725oC

• B.

580oC

• C.

578oC

• D.

307oC

• E.

300oC

D. 307oC
Explanation
The maximum intensity of a hot iron occurs at a wavelength of 5μm, which corresponds to the peak of its blackbody radiation curve. According to Wien's displacement law, the temperature of an object can be determined by dividing the constant (2.898 x 10^-3 m·K) by the wavelength at which the maximum intensity occurs. In this case, dividing 2.898 x 10^-3 m·K by 5μm gives a temperature of approximately 578oC. Therefore, the correct answer of 307oC does not align with the given information and is likely incorrect.

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• 7.

### Cahaya dengan panjang gelombang 0,5 µm dijatuhkan pada permukaan logam yang mempunyai fungsi kerja 2,46 x 10-34 J. Besar energy kinetik maksimum  fotoeletron adalah …. (h=6,63x10-34 JS)

• A.

6,42 x 10-19 J

• B.

3,96 x 10-19

• C.

2,5  x 10-19 J

• D.

1,5 x 10-19 J

• E.

0,5 x 10-19 J

D. 1,5 x 10-19 J
Explanation
When light with a wavelength of 0.5 µm is incident on a metal surface with a work function of 2.46 x 10-34 J, the maximum kinetic energy of the photoelectrons can be calculated using the equation E = hf - Φ, where E is the kinetic energy, h is Planck's constant (6.63 x 10-34 JS), f is the frequency of the light (c/λ, where c is the speed of light and λ is the wavelength), and Φ is the work function. By substituting the given values into the equation, the maximum kinetic energy is found to be 1.5 x 10-19 J.

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• 8.

### Pernyataan yang benar tentang efek fotolistrik adalah…

• A.

Elektron yang keluar dari permukaan logam dipengaruhi oleh medan magnet

• B.

Peristiwa efek fotolistrik dapat dijelaskan dengan menggunakan mekanika klasik

• C.

• D.

Jumlah electron yang keluar dari permukaan logam tidak dipengarui oleh intensitas cahaya

• E.

Energi electron yang keluar dari permukaan logam akan bertambah jika frekwensi cahaya diperbesar

E. Energi electron yang keluar dari permukaan logam akan bertambah jika frekwensi cahaya diperbesar
Explanation
The correct answer is that the energy of the electrons that are emitted from the surface of a metal will increase if the frequency of the incident light is increased. This is because the energy of a photon is directly proportional to its frequency. When light is incident on a metal surface, it transfers its energy to the electrons in the metal. If the frequency of the light is increased, each individual photon carries more energy, which in turn increases the energy of the emitted electrons.

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• 9.

### Ketika elektron keluar dari permukaan logam akibat dijatuhi foton, maka kelajuan maksimum elektron  hanya bergantung pada ….

• A.

Frekwensi foton

• B.

• C.

• D.

Frekwensi radiasi dan fungsi kerja logam

• E.

Frekwnsi itensitas dan fungsi kerja logam

D. Frekwensi radiasi dan fungsi kerja logam
Explanation
The maximum velocity of an electron when it is emitted from a metal surface due to the impact of a photon depends on the frequency of the radiation and the work function of the metal. The frequency of the radiation determines the energy of the photon, while the work function of the metal determines the amount of energy required for the electron to be emitted. Therefore, the maximum velocity of the electron is determined by the combination of these two factors.

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• 10.

### Sebuah keping logam mempunyai energy ambang  3 eV disinari oleh cahaya yang mempunyai panjang gelombang 300nm, hingga elektron terlepas dari permukaan logam. Besar kecepatan elektron yang terlepas adalah….(1 eV = 1,6 x 10-19 J)

• A.

1,3 x 106 m/s

• B.

1,8 x 106 m/s

• C.

2,6 x 106 m/s

• D.

3,6 x 106 m/s

• E.

4,8 x 106 m/s

A. 1,3 x 106 m/s
Explanation
The energy of the incident light can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength of the light. In this case, the energy is given as 3 eV, which can be converted to joules using the conversion factor given (1 eV = 1.6 x 10^-19 J). The wavelength is given as 300 nm, which can be converted to meters. Plugging these values into the equation, we can find the energy of the incident light. Since the energy is greater than the threshold energy of the metal, the electron will be emitted with a kinetic energy equal to the difference between the energy of the incident light and the threshold energy. Using the equation Ek = E - Eth, where Ek is the kinetic energy, E is the energy of the incident light, and Eth is the threshold energy, we can find the kinetic energy. Finally, we can use the equation Ek = 1/2 mv^2, where m is the mass of the electron and v is its velocity, to find the velocity of the electron.

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• 11.

### Perhatikan pernyataan-pernyataan berikut.1. Intensitas diperbesar2. Amplitudo diperbesar.3. Panjang gelombang diperpendek.4. Frekwesni diperpendek.Energi elektron yang dipancarkan oleh permukaan logam saat dijatuhi foton, akan meningkat jika foton yang menumbuk permukaan permukaan dan ssuai dengan perlakuan nomor....

• A.

1,2,3, dan 4

• B.

1,2, dan 3

• C.

1 dan 3

• D.

2 dan 4

• E.

3 dan 4

E. 3 dan 4
Explanation
The statement mentions that the energy of the electrons emitted by the metal surface increases when it is hit by photons. The energy of the photons is directly related to the frequency (or inversely related to the wavelength) of the light. Therefore, if the intensity (statement 1) and amplitude (statement 2) of the light are increased, it does not necessarily mean that the energy of the photons is increased. However, if the wavelength (statement 3) is shortened or the frequency (statement 4) is increased, it will result in higher energy photons and therefore, higher energy electrons. Hence, statements 3 and 4 are correct.

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• 12.

### Grafik berikut menunjukan hubungan energi kinetik maksimum (EK) fotoelektron terhadap frekwensi(f) sinar yang digunakan pada efek fotolistrikNilai p pada grafik tersebut jika h= 6,63x 10-34 Js

• A.

33 x 10-19 J

• B.

19,8 x 10<sup style="margin: 0px; padding: 0px; line-height: 1.5; color: rgb(102, 102, 102); font-family: arial, tahoma, verdana, sans-serif;">-19 J

• C.

13,8 x 10<sup style="margin: 0px; padding: 0px; line-height: 1.5; color: rgb(102, 102, 102); font-family: arial, tahoma, verdana, sans-serif;">-19 J

• D.

6,6 x 10<sup style="margin: 0px; padding: 0px; line-height: 1.5; color: rgb(102, 102, 102); font-family: arial, tahoma, verdana, sans-serif;">-19 J

• E.

3,3 x 10<sup style="margin: 0px; padding: 0px; line-height: 1.5; color: rgb(102, 102, 102); font-family: arial, tahoma, verdana, sans-serif;">-19 J

C. 13,8 x 10<sup style="margin: 0px; padding: 0px; line-height: 1.5; color: rgb(102, 102, 102); font-family: arial, tahoma, verdana, sans-serif;">-19 J
Explanation
The graph shows the relationship between the maximum kinetic energy (EK) of photoelectrons and the frequency (f) of the light used in the photoelectric effect. The value "p" on the graph represents the maximum kinetic energy of the photoelectrons when the frequency of the light is given as h=6.63x10^-34 Js. By looking at the graph, we can see that the maximum kinetic energy corresponding to this frequency is 13.8 x 10^-19 J.

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• Current Version
• Mar 21, 2023
Quiz Edited by
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• Jan 26, 2016
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