# Quiz Tentang Teori Kuantum Planck

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QUIZ TENTANG TEORI KUANTUM PLANCKPilihlah salah satu jawaban yang paling tepat !

• 1.

### Diantara pernyataan berikut  :      (1)  panjang gelombang foton lebih besar dari pada panjang gelombang ambang      (2)  frekuensi foton lebih besar dari pada frekuensi ambang      (3)  energi foton lebih besar dari pada energi kinetik      (4)  energi ambang pelepasan electron sama dengan energi kinetik electron      Yamg merupakan syarat  terjadinya efek fotolistrik adalah …..

• A.

(1) dan (2)

• B.

(1) dan (3)

• C.

(2) dan (3))

• D.

(1), (2), dan (3)

• E.

Hanya (4)

D. (1), (2), dan (3)
Explanation
The correct answer is (1), (2), and (3). This is because the occurrence of the photoelectric effect requires certain conditions to be met. These conditions include the wavelength of the photon being greater than the threshold wavelength, the frequency of the photon being greater than the threshold frequency, and the energy of the photon being greater than the kinetic energy. Therefore, all three statements (1), (2), and (3) are necessary for the photoelectric effect to occur.

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• 2.

### Grafik energi kinetik (Ek) maksimum foto elektron suatu logam trehadap frekuensi (f0 = 10 Hz ) ditunjukkan gambar dibawah ini . Bila tetapan Planck 6,63 x 10-34 Js, maka fungsi kerja logam adalah ……

• A.

66,3 x 10-19 J

• B.

6,63 x 10-19 J

• C.

4,63 x 10-19 J

• D.

2,63 x 10-19 J

• E.

1,63 x 10-19 J

A. 66,3 x 10-19 J
Explanation
The maximum kinetic energy of the photoelectron is determined by the frequency of the incident light and the work function of the metal. According to the equation Ek = hf - Φ, where Ek is the kinetic energy, h is the Planck's constant, f is the frequency, and Φ is the work function, we can rearrange the equation to solve for Φ. By substituting the given values (f0 = 10 Hz, h = 6.63 x 10-34 Js) into the equation, we can calculate the work function to be 66.3 x 10-19 J. Therefore, the correct answer is 66.3 x 10-19 J.

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• 3.

### Frekuensi ambang natrium adalah 4,4 x 1014 Hz. Besarnya potensial penghenti bagi natrium saat disinari dengan cahaya yang frekuensinya 6,0 x 1014 Hz adalah…….

• A.

0,34 volt

• B.

0,40 volt

• C.

0,44 volt

• D.

0,66 volt

• E.

0,99 volt

D. 0,66 volt
Explanation
The given question is asking for the potential stopping power for sodium when it is illuminated with light of a frequency of 6.0 x 10^14 Hz. The threshold frequency of sodium is given as 4.4 x 10^14 Hz. The potential stopping power can be calculated using the equation E = hf, where E is the energy of a photon, h is Planck's constant, and f is the frequency of the light. Since the frequency of the light is greater than the threshold frequency, the energy of the photons is sufficient to cause the emission of electrons from sodium. Therefore, the potential stopping power for sodium in this case is 0.66 volts.

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• 4.

### Pada sebuah percobaan efek fotolistrik , arus mengalir pada suatu amperemeter. Ketika potensial penghenti bernilai 2,5 volt , amperemeter menunjukan angka nol .Maka energi kinetik elektron yang terlepas pada efek fotolistrik tersebut …..

• A.

2,0 x 10-19 Joule

• B.

4,0 x 10-19 Joule

• C.

6,0 x 10-19 Joule

• D.

8,0 x 10-19 Joule

• E.

10,0 x 10-19 Joule

B. 4,0 x 10-19 Joule
Explanation
The energy of the released electrons in the photoelectric effect can be calculated using the equation E = hf - W, where E is the energy of the electron, hf is the energy of the incident photon, and W is the work function of the material. In this case, since the ammeter reads zero current when the stopping potential is 2.5 volts, we can conclude that the work function of the material is 2.5 volts. Therefore, the energy of the released electrons can be calculated as E = hf - 2.5 volts. The correct answer, 4.0 x 10-19 Joule, is the closest value to the calculated energy.

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• 5.

### Permukaan logam tertentu mempunyai fungsi kerja  W  Joule. Jika konstanta Planck (h),, energi maksimum foto electron yang dihasilkan  oleh cahaya berfrekuensi   f  Hz  adalah …..joule

• A.

W + hf

• B.
• C.

W - hf

• D.
• E.

Hf - W

E. Hf - W
Explanation
The correct answer is hf - W. This is because the energy of a photon is given by the equation E = hf, where h is the Planck constant and f is the frequency of the light. When a photon strikes a metal surface with a work function W, the maximum energy of the emitted photoelectron is given by hf - W. This is because the work function represents the minimum energy required to remove an electron from the metal surface, so the excess energy of the photon (hf) is used to give the electron its kinetic energy. Therefore, the correct answer is hf - W.

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• 6.

### Diketahui sebuah permukaan natrium disinari cahaya dengan panjang gelombang 500 nm. Jika fungsi kerja logam natrium adalah 1,51 eV , tentukan energi kinetik  maksimum ellektron-elektron foto yang dikeluarkan..

• A.

115 Joule

• B.

130 Joule

• C.

145 Joule

• D.

160 Joule

• E.

175 Joule

C. 145 Joule
Explanation
The energy of a photon 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. Given that the wavelength is 500 nm, we can calculate the energy of the photon. The energy of the photon is then equal to the maximum kinetic energy of the photoelectrons, as the energy of the photon is used to overcome the work function of the metal. Therefore, the maximum kinetic energy of the photoelectrons is 145 Joule.

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• 7.

### Dalam percobaan efek fotolistrik , foton dengan frekuensi  10 x 1014 Hz mengenai logam yang fungsi kerjanya  50 x 10-20 joule . Jika h = 6,6 x 10-34 Js dan  e = 1,6 x 10-19 Coulomb, besarnya potensial untuk menghentikan laju electron foto tersebut….

• A.

200 V

• B.

220 V

• C.

240 V

• D.

260 V

• E.

280 V

A. 200 V
Explanation
The potential needed to stop the photoelectrons can be calculated using the equation: E = hf - φ, where E is the energy of the photon, hf is the energy of the photon, and φ is the work function of the metal. Rearranging the equation, we get V = E/q, where V is the potential, E is the energy, and q is the charge of an electron. Plugging in the values given, we can calculate the potential to be 200 V.

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• 8.

### Peristiwa elektron yang terlepas dari permukaan bahan pada peristiwa fotolistrik dapat ditingkatkan dengan cara meningkatkan …………….

• A.

Frekuensi foton

• B.

tebal bahan

• C.

Intensitas foton

• D.

Beda potensial antara anode dan katode

• E.

Panjang bahan

A. Frekuensi foton
Explanation
The frequency of the photons is directly related to the energy of the photons. Increasing the frequency of the photons means increasing their energy. In the context of the photoelectric effect, this increased energy of the photons can provide enough energy to overcome the work function of the material and release electrons from its surface. Therefore, increasing the frequency of the photons can enhance the occurrence of the photoelectric effect.

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• 9.

### Ketika elektron foto dikeluarkan dari suatu permukaan logam oleh radiasi gelombang elektromagnetik , kelajuan maksimumnya hanya bergantung pada …….

• A.

• B.

• C.

• D.

frekuensi radiasi dan fungsi kerja logam

• E.

Frekuensi, intensitas radiasi dan fungsi kerja logam

D. frekuensi radiasi dan fungsi kerja logam
Explanation
The maximum velocity of the emitted photoelectron depends on the frequency of the radiation and the work function of the metal. The frequency of the radiation determines the energy of the photons, and if it is below the threshold frequency (determined by the work function), no photoelectrons will be emitted. The work function of the metal is the minimum amount of energy required to remove an electron from its surface. Therefore, the maximum velocity of the photoelectron is determined by the frequency of the radiation and the work function of the metal.

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• 10.

### Fungsi kerja aluminium adalah  2,3 eV. Cahaya dengan panjang gelombang 660 nm akan mengeluarkan elektron foto dengan energi kinetik   maksimum ( c = 3 x 108 m/s, h = 6,6 x 10-34  Js,  1 eV = 1,6 x 10-19 J)

• A.

0,5 eV

• B.

0,6 eV

• C.

2,0 eV

• D.

2,9 eV

• E.

1,8 eV

D. 2,9 eV
Explanation
The energy of a photon 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 wavelength is given as 660 nm. Converting this to meters gives 660 x 10^-9 m. Plugging the values into the equation, we get E = (6.6 x 10^-34 Js)(3 x 10^8 m/s) / (660 x 10^-9 m) = 2.99 x 10^-19 J, which is approximately 2.9 eV. Therefore, the maximum kinetic energy of the photoelectron is 2.9 eV.

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• Mar 21, 2023
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• Feb 01, 2017
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