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GRADE 11 PRACTICE

Physics Problems: Quantum physics, Grade 11

Photons and the photoelectric effect.

23 problems across 2 sub-topics
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Problem statements

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23 problems
Position in this set: 1 from 23

Problem #61

Quantum physics and photoelectric effect

Determine the energy, equivalent mass and momentum of a violet photon with wavelength 4\cdot10^{-7}\,\text{м}.

Answer options

  1. \left(1 \cdot 10^{-18}\,\text{Дж};\ 1{,}1 \cdot 10^{-35}\,\text{кг};\ 3{,}3 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с}\right)
  2. \left(5 \cdot 10^{-18}\,\text{Дж};\ 5{,}5 \cdot 10^{-35}\,\text{кг};\ 1{,}65 \cdot 10^{-26}\,\text{кг}\cdot\text{м/с}\right)
  3. \left(5 \cdot 10^{-19}\,\text{Дж};\ 5{,}5 \cdot 10^{-36}\,\text{кг};\ 1{,}65 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с}\right)
  4. \left(2{,}5 \cdot 10^{-19}\,\text{Дж};\ 2{,}75 \cdot 10^{-36}\,\text{кг};\ 8{,}25 \cdot 10^{-28}\,\text{кг}\cdot\text{м/с}\right)
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Given
\lambda
4\cdot 10^{-7}\,\text{м}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
c
3\cdot 10^{8}\,\text{м/с}
Find

Emp

  1. E = h\nu = \dfrac{hc}{\lambda}E = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{4\cdot 10^{-7}}5\cdot 10^{-19}\,\text{Дж} \approx 3{,}1\,\text{эВ}
  2. m = \dfrac{E}{c^{2}}m = \dfrac{5\cdot 10^{-19}}{9\cdot 10^{16}}m = 5{,}5\cdot 10^{-36}\,\text{кг}

    A photon has no rest mass; this is the equivalent mass corresponding to its energy.

  3. p = mc = \dfrac{h}{\lambda}p = 5{,}5\cdot 10^{-36}\cdot 3\cdot 10^{8}p = 1{,}65\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}

AnswerE = 5\cdot 10^{-19}\,\text{Дж}m = 5{,}5\cdot 10^{-36}\,\text{кг}p = 1{,}65\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}

Position in this set: 2 from 23

Problem #62

Quantum physics and photoelectric effect

Will a photoelectric effect be observed when a metal is illuminated with light of frequency 6\cdot10^{14}\,\text{Гц}, if the work function is equal 3\,\text{эВ}? Accept h=6{,}62\cdot10^{-34}\,\text{Дж}\cdot\text{с}, 1\,\text{эВ}=1{,}6\cdot10^{-19}\,\text{Дж}.

Answer options

  1. \text{Yes, always at visible light}
  2. \text{Yes, if increase intensity}
  3. \text{No, because what frequency too great}
  4. \text{No, energy photon not enough}
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Given
\nu
6\cdot10^{14}\,\text{Гц}
A
3\,\text{эВ}4{,}8\cdot10^{-19}\,\text{Дж}
Find

\text{possible whether photoeffect}

  1. E_\gamma=h\nuE_\gamma = 6{,}62\cdot10^{-34}\cdot6\cdot10^{14}E_\gamma = 3{,}972\cdot10^{-19}\,\text{Дж}
  2. E_\gamma<A

    The photon energy is about 2.48 eV, which is less than the work function of 3 eV. No photoelectric effect is observed.

Answer\text{photoelectric effect no: Photon energy less work exit}

Position in this set: 3 from 23

Problem #63

Quantum physics and photoelectric effect

Potassium is illuminated with light frequency 6\cdot10^{14}\,\text{Гц}. The work function is 2{,}2\,\text{эВ}. Determine the maximum kinetic energy and speed of photoelectrons, the retardation voltage, and the red limit of the photoelectric effect. Accept h=6{,}6\cdot10^{-34}\,\text{Дж}\cdot\text{с}, m_e=9{,}1\cdot10^{-31}\,\text{кг} and e=1{,}6\cdot10^{-19}\,\text{Кл}.

Answer options

  1. \left(4{,}4 \cdot 10^{-20}\,\text{Дж};\ 3{,}1 \cdot 10^{5}\,\text{м/с};\ 0{,}275\,\text{В};\ 5{,}6 \cdot 10^{-7}\,\text{м}\right)
  2. \left(2{,}2 \cdot 10^{-20}\,\text{Дж};\ 1{,}55 \cdot 10^{5}\,\text{м/с};\ 0{,}1375\,\text{В};\ 2{,}8 \cdot 10^{-7}\,\text{м}\right)
  3. \left(8{,}8 \cdot 10^{-20}\,\text{Дж};\ 6{,}2 \cdot 10^{5}\,\text{м/с};\ 0{,}55\,\text{В};\ 1{,}12 \cdot 10^{-6}\,\text{м}\right)
  4. \left(4{,}4 \cdot 10^{-19}\,\text{Дж};\ 3{,}1 \cdot 10^{6}\,\text{м/с};\ 2{,}75\,\text{В};\ 5{,}6 \cdot 10^{-6}\,\text{м}\right)
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Given
\nu
6\cdot10^{14}\,\text{Гц}
A
2{,}2\,\text{эВ}3{,}52\cdot 10^{-19}\,\text{Дж}
m
9{,}1\cdot 10^{-31}\,\text{кг}
e
1{,}6\cdot 10^{-19}\,\text{Кл}
h
6{,}6\cdot10^{-34}\,\text{Дж}\cdot\text{с}
Find

E_{\text{км}}v_{\text{м}}U_{\text{з}}\lambda_0

  1. E_{\text{ф}}=h\nuE_{\text{ф}} = 6{,}6\cdot10^{-34}\cdot6\cdot10^{14}E_{\text{ф}} = 3{,}96\cdot10^{-19}\,\text{Дж}
  2. E_{\text{ф}} = A + E_{\text{км}}\ \Rightarrow\ E_{\text{км}} = E_{\text{ф}} - AE_{\text{км}} = 3{,}96\cdot 10^{-19} - 3{,}52\cdot 10^{-19}E_{\text{км}} = 4{,}4\cdot 10^{-20}\,\text{Дж}

    Einstein's equation for the photoelectric effect.

  3. E_{\text{км}} = \dfrac{mv_{\text{м}}^{2}}{2}\ \Rightarrow\ v_{\text{м}} = \sqrt{\dfrac{2E_{\text{км}}}{m}}v_{\text{м}} = \sqrt{\dfrac{2\cdot 4{,}4\cdot 10^{-20}}{9{,}1\cdot 10^{-31}}}v_{\text{м}} = 3{,}1\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}
  4. E_{\text{км}} = eU_{\text{з}}\ \Rightarrow\ U_{\text{з}} = \dfrac{E_{\text{км}}}{e}U_{\text{з}} = \dfrac{4{,}4\cdot 10^{-20}}{1{,}6\cdot 10^{-19}}U_{\text{з}} = 0{,}275\,\text{В}
  5. \lambda_0 = \dfrac{hc}{A}\lambda_0 = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{3{,}52\cdot 10^{-19}}\lambda_0 = 5{,}6\cdot 10^{-7}\,\text{м}

    The red boundary is the longest wavelength at which the photoelectric effect is still possible.

AnswerE_{\text{км}} = 4{,}4\cdot 10^{-20}\,\text{Дж}v_{\text{м}} = 3{,}1\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}U_{\text{з}} = 0{,}275\,\text{В}\lambda_0 = 5{,}6\cdot 10^{-7}\,\text{м}

Position in this set: 4 from 23

Problem #64

Quantum physics and photoelectric effect

Solar radiation power 4\cdot10^{23}\,\text{кВт}. How much does the mass of the Sun decrease per day?

Answer options

  1. 3{,}84 \cdot 10^{15}\,\text{кг}
  2. 3{,}84 \cdot 10^{14}\,\text{кг}
  3. 1{,}92 \cdot 10^{14}\,\text{кг}
  4. 7{,}68 \cdot 10^{14}\,\text{кг}
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Given
P
4\cdot 10^{23}\,\text{кВт}4\cdot 10^{26}\,\text{Вт}
t
1\,\text{сут}86\,400\,\text{с}
c
3\cdot 10^{8}\,\text{м/с}
Find

\Delta m

  1. \Delta E = Pt\Delta E = 4\cdot 10^{26}\cdot 86\,400\Delta E = 3{,}456\cdot 10^{31}\,\text{Дж}
  2. \Delta m = \dfrac{\Delta E}{c^{2}} = \dfrac{Pt}{c^{2}}\Delta m = \dfrac{4\cdot 10^{26}\cdot 86\,400}{9\cdot 10^{16}}\Delta m = 3{,}84\cdot 10^{14}\,\text{кг} = 3{,}84\cdot 10^{11}\,\text{т}

Answer\Delta m = 3{,}84\cdot 10^{14}\,\text{кг}

Position in this set: 5 from 23

Problem #65

Quantum physics and photoelectric effect

Determine the red limit of the photoelectric effect for zinc at work function 4{,}2\,\text{эВ}.

Answer options

  1. 5{,}9 \cdot 10^{-7}\,\text{м}
  2. 2{,}95 \cdot 10^{-6}\,\text{м}
  3. 2{,}95 \cdot 10^{-7}\,\text{м}
  4. 1{,}475 \cdot 10^{-7}\,\text{м}
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Given
A
4{,}2\,\text{эВ}6{,}72\cdot 10^{-19}\,\text{Дж}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
c
3\cdot 10^{8}\,\text{м/с}
Find

\lambda_0

  1. h\nu_0 = A\ \Rightarrow\ \nu_0 = \dfrac{A}{h}\nu_0 = \dfrac{6{,}72\cdot 10^{-19}}{6{,}6\cdot 10^{-34}}\nu_0 = 1{,}02\cdot 10^{15}\,\text{Гц}
  2. \lambda_0 = \dfrac{c}{\nu_0} = \dfrac{hc}{A}\lambda_0 = \dfrac{3\cdot 10^{8}}{1{,}02\cdot 10^{15}}\lambda_0 = 2{,}95\cdot 10^{-7}\,\text{м} = 295\,\text{нм}

Answer\lambda_0 = 2{,}95\cdot 10^{-7}\,\text{м}

Position in this set: 6 from 23

Problem #66

Quantum physics and photoelectric effect

What impulse is transmitted to the mirror by a photon with wavelength 2{,}95\cdot10^{-7}\,\text{м} with normal reflection?

Answer options

  1. 2{,}235 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с}
  2. 8{,}94 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с}
  3. 4{,}47 \cdot 10^{-26}\,\text{кг}\cdot\text{м/с}
  4. 4{,}47 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с}
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Given
\lambda
2{,}95\cdot 10^{-7}\,\text{м}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
Find

\Delta p

  1. \Delta p = 2p = \dfrac{2h}{\lambda}\Delta p = \dfrac{2\cdot 6{,}6\cdot 10^{-34}}{2{,}95\cdot 10^{-7}}\Delta p = 4{,}47\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}

    During mirror reflection, the photon momentum changes sign, so double the momentum is transferred to the mirror.

Answer\Delta p = 4{,}47\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}

Position in this set: 7 from 23

Problem #67

Quantum physics and photoelectric effect

Solar radiation power 4\cdot10^{26}\,\text{Вт}. Determine the flux density of solar radiation on Earth. Distance 1{,}5\cdot10^{11}\,\text{м}.

Answer options

  1. 1400\,\text{Вт/м}^{2}
  2. 700\,\text{Вт/м}^{2}
  3. 2800\,\text{Вт/м}^{2}
  4. 14000\,\text{Вт/м}^{2}
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Given
P
4\cdot 10^{26}\,\text{Вт}
r
1{,}5\cdot 10^{8}\,\text{км}1{,}5\cdot 10^{11}\,\text{м}
Find

I

  1. I = \dfrac{E}{St} = \dfrac{P}{S}

    Flux density is the energy coming to 1 m² in 1 s.

  2. S = 4\pi r^{2}

    All energy passes through a sphere of radius r.

  3. I = \dfrac{P}{4\pi r^{2}}I = \dfrac{4\cdot 10^{26}}{4\cdot 3{,}14\cdot\left(1{,}5\cdot 10^{11}\right)^{2}}I = 1400\,\tfrac{\text{Вт}}{\text{м}^{2}}

    This quantity is called the solar constant.

AnswerI = 1400\,\tfrac{\text{Вт}}{\text{м}^{2}}

Position in this set: 8 from 23

Problem #68

Quantum physics and photoelectric effect

What wavelength of light should be directed onto a tungsten plate so that electrons are ejected at the speed 8\cdot10^{5}\,\text{м/с}? Determine red limit and delay voltage when operating output 4{,}95\,\text{эВ}.

Answer options

  1. \left(1{,}83 \cdot 10^{-6}\,\text{м};\ 2{,}5 \cdot 10^{-6}\,\text{м};\ 18{,}2\,\text{В}\right)
  2. \left(1{,}83 \cdot 10^{-7}\,\text{м};\ 2{,}5 \cdot 10^{-7}\,\text{м};\ 1{,}82\,\text{В}\right)
  3. \left(9{,}15 \cdot 10^{-8}\,\text{м};\ 1{,}25 \cdot 10^{-7}\,\text{м};\ 0{,}91\,\text{В}\right)
  4. \left(3{,}66 \cdot 10^{-7}\,\text{м};\ 5 \cdot 10^{-7}\,\text{м};\ 3{,}64\,\text{В}\right)
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Given
v
8\cdot 10^{5}\,\text{м/с}
A
4{,}95\,\text{эВ}7{,}92\cdot 10^{-19}\,\text{Дж}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
m
9{,}1\cdot 10^{-31}\,\text{кг}
e
1{,}6\cdot 10^{-19}\,\text{Кл}
Find

\lambda\lambda_0U_{\text{з}}

  1. E_{\text{ф}} = A + \dfrac{mv^{2}}{2}E_{\text{ф}} = 7{,}92\cdot 10^{-19} + \dfrac{9{,}1\cdot 10^{-31}\cdot\left(8\cdot 10^{5}\right)^{2}}{2}E_{\text{ф}} = 1{,}083\cdot 10^{-18}\,\text{Дж}
  2. E_{\text{ф}} = \dfrac{hc}{\lambda}\ \Rightarrow\ \lambda = \dfrac{hc}{E_{\text{ф}}}\lambda = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{1{,}083\cdot 10^{-18}}\lambda = 1{,}83\cdot 10^{-7}\,\text{м}
  3. \lambda_0 = \dfrac{hc}{A}\lambda_0 = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{7{,}92\cdot 10^{-19}}\lambda_0 = 2{,}5\cdot 10^{-7}\,\text{м}
  4. \dfrac{mv^{2}}{2} = eU_{\text{з}}\ \Rightarrow\ U_{\text{з}} = \dfrac{mv^{2}}{2e}U_{\text{з}} = \dfrac{9{,}1\cdot 10^{-31}\cdot\left(8\cdot 10^{5}\right)^{2}}{2\cdot 1{,}6\cdot 10^{-19}}U_{\text{з}} = 1{,}82\,\text{В}

    The holding voltage depends only on the speed of the electrons.

Answer\lambda = 1{,}83\cdot 10^{-7}\,\text{м}\lambda_0 = 2{,}5\cdot 10^{-7}\,\text{м}U_{\text{з}} = 1{,}82\,\text{В}

Position in this set: 9 from 23

Problem #69

Quantum physics and photoelectric effect

Red border for metal 5{,}3\cdot10^{14}\,\text{Гц}. Determine the maximum speed of photoelectrons at wavelength 5\cdot10^{-7}\,\text{м}. Will there be a photoelectric effect when 6\cdot10^{-7}\,\text{м}?

Answer options

  1. \left(3{,}2\cdot10^5\,\text{м/с};\ \text{фотоэффекта нет в обоих случаях}\right)
  2. \left(1{,}6\cdot10^5\,\text{м/с};\ \text{при }6\cdot10^{-7}\text{ м фотоэффект есть}\right)
  3. \left(3{,}2\cdot10^5\,\text{м/с};\ \text{при }6\cdot10^{-7}\text{ м фотоэффекта нет}\right)
  4. \left(6{,}4\cdot10^5\,\text{м/с};\ \text{фотоэффект есть в обоих случаях}\right)
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Given
\nu_0
5{,}3\cdot 10^{14}\,\text{Гц}
\lambda_1
5\cdot 10^{-7}\,\text{м}
\lambda_2
6\cdot 10^{-7}\,\text{м}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
m
9{,}1\cdot 10^{-31}\,\text{кг}
Find

v_{\text{м}1}v_{\text{м}2}

  1. A = h\nu_0A = 6{,}6\cdot 10^{-34}\cdot 5{,}3\cdot 10^{14}A = 3{,}5\cdot 10^{-19}\,\text{Дж}
  2. E_{\text{ф}1} = \dfrac{hc}{\lambda_1}E_{\text{ф}1} = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{5\cdot 10^{-7}}E_{\text{ф}1} = 3{,}96\cdot 10^{-19}\,\text{Дж}
  3. E_{\text{к}1} = E_{\text{ф}1} - AE_{\text{к}1} = 3{,}96\cdot 10^{-19} - 3{,}5\cdot 10^{-19}E_{\text{к}1} = 4{,}6\cdot 10^{-20}\,\text{Дж}
  4. v_1 = \sqrt{\dfrac{2E_{\text{к}1}}{m}}v_1 = \sqrt{\dfrac{2\cdot 4{,}6\cdot 10^{-20}}{9{,}1\cdot 10^{-31}}}v_{\text{м}1} = 3{,}2\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}
  5. E_{\text{ф}2} = \dfrac{hc}{\lambda_2}E_{\text{ф}2} = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{6\cdot 10^{-7}}3{,}3\cdot 10^{-19}\,\text{Дж} < A

    There is not enough photon energy: at this wavelength there is no photoelectric effect.

Answerv_{\text{м}1} = 3{,}2\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}v_{\text{м}2}\ \text{- фотоэффекта нет}

Position in this set: 10 from 23

Problem #70

Quantum physics and photoelectric effect

At what speed must an electron move so that its momentum is the same as that of a photon with a wavelength 4\cdot10^{-7}\,\text{м}?

Answer options

  1. 906{,}5\,\text{м/с}
  2. 3626\,\text{м/с}
  3. 18130\,\text{м/с}
  4. 1813\,\text{м/с}
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Given
\lambda
4\cdot 10^{-7}\,\text{м}
m
9{,}1\cdot 10^{-31}\,\text{кг}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
Find

v

  1. p_{\text{ф}} = p_{\text{эл}}:\quad \dfrac{h}{\lambda} = mv

    The photon momentum is h/λ, the electron momentum is mv.

  2. v = \dfrac{h}{\lambda m}v = \dfrac{6{,}6\cdot 10^{-34}}{4\cdot 10^{-7}\cdot 9{,}1\cdot 10^{-31}}v = 1813\,\tfrac{\text{м}}{\text{с}}

Answerv = 1813\,\tfrac{\text{м}}{\text{с}}

Position in this set: 11 from 23

Problem #71

Quantum physics and photoelectric effect

Wavelength of light in glass 5\cdot10^{-7}\,\text{м}, refractive index 1{,}5. What is the wavelength in air and frequency?

Answer options

  1. \left(7{,}5 \cdot 10^{-7}\,\text{м};\ 4 \cdot 10^{14}\,\text{Гц}\right)
  2. \left(3{,}75 \cdot 10^{-7}\,\text{м};\ 2 \cdot 10^{14}\,\text{Гц}\right)
  3. \left(1{,}5 \cdot 10^{-6}\,\text{м};\ 8 \cdot 10^{14}\,\text{Гц}\right)
  4. \left(7{,}5 \cdot 10^{-6}\,\text{м};\ 4 \cdot 10^{15}\,\text{Гц}\right)
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Given
\lambda_{\text{ст}}
5\cdot 10^{-7}\,\text{м}
n
1{,}5
c
3\cdot 10^{8}\,\text{м/с}
Find

\lambda\nu

  1. \lambda = \lambda_{\text{ст}}n\lambda = 5\cdot 10^{-7}\cdot 1{,}5\lambda = 7{,}5\cdot 10^{-7}\,\text{м}

    In the air the wave is longer - this is red light.

  2. \nu = \dfrac{c}{\lambda}\nu = \dfrac{3\cdot 10^{8}}{7{,}5\cdot 10^{-7}}\nu = 4\cdot 10^{14}\,\text{Гц}

Answer\lambda = 7{,}5\cdot 10^{-7}\,\text{м}\nu = 4\cdot 10^{14}\,\text{Гц}

Position in this set: 12 from 23

Problem #72

Quantum physics and photoelectric effect

X-ray tube is energized 45\,\text{кВ}. What is the minimum wavelength of X-rays?

Answer options

  1. 2{,}75 \cdot 10^{-10}\,\text{м}
  2. 2{,}75 \cdot 10^{-11}\,\text{м}
  3. 1{,}375 \cdot 10^{-11}\,\text{м}
  4. 5{,}5 \cdot 10^{-11}\,\text{м}
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Given
U
45\,\text{кВ}45\,000\,\text{В}
e
1{,}6\cdot 10^{-19}\,\text{Кл}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
c
3\cdot 10^{8}\,\text{м/с}
Find

\lambda_{\min}

  1. E_{\text{ф}} = E_{\text{к}} = A = eU

    The minimum wavelength occurs when the electron transfers all its kinetic energy to one photon.

  2. \dfrac{hc}{\lambda} = eU\ \Rightarrow\ \lambda = \dfrac{hc}{eU}\lambda = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{1{,}6\cdot 10^{-19}\cdot 45\,000}\lambda_{\min} = 2{,}75\cdot 10^{-11}\,\text{м}

Answer\lambda_{\min} = 2{,}75\cdot 10^{-11}\,\text{м}

Position in this set: 13 from 23

Problem #73

Quantum physics and photoelectric effect

The photoelectric effect occurs only when the momentum of the incident photon is no less than 9\cdot10^{-28}\,\text{кг}\cdot\text{м/с}. At what speed will electrons be ejected if the light frequency is twice the threshold?

Answer options

  1. 1{,}54 \cdot 10^{6}\,\text{м/с}
  2. 7{,}7 \cdot 10^{6}\,\text{м/с}
  3. 7{,}7 \cdot 10^{5}\,\text{м/с}
  4. 3{,}85 \cdot 10^{5}\,\text{м/с}
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Given
p_0
9\cdot 10^{-28}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}
\nu
2\nu_0
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
m
9{,}1\cdot 10^{-31}\,\text{кг}
Find

v

  1. p_0 = \dfrac{h}{\lambda_0}\ \Rightarrow\ A = \dfrac{hc}{\lambda_0} = cp_0A = 3\cdot 10^{8}\cdot 9\cdot 10^{-28}A = 2{,}7\cdot 10^{-19}\,\text{Дж}

    The threshold of the photoelectric effect is specified through the pulse, but this is the same work function.

  2. E_{\text{ф}} = h\cdot 2\nu_0 = 2AE_{\text{ф}} = 2\cdot 2{,}7\cdot 10^{-19}E_{\text{ф}} = 5{,}4\cdot 10^{-19}\,\text{Дж}
  3. E_{\text{к}} = E_{\text{ф}} - AE_{\text{к}} = 5{,}4\cdot 10^{-19} - 2{,}7\cdot 10^{-19}E_{\text{к}} = 2{,}7\cdot 10^{-19}\,\text{Дж}
  4. v = \sqrt{\dfrac{2E_{\text{к}}}{m}}v = \sqrt{\dfrac{2\cdot 2{,}7\cdot 10^{-19}}{9{,}1\cdot 10^{-31}}}v = 7{,}7\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}

Answerv = 7{,}7\cdot 10^{5}\,\tfrac{\text{м}}{\text{с}}

Position in this set: 14 from 23

Problem #74

Quantum physics and photoelectric effect

The atom went from a state with energy -1{,}5\,\text{эВ} in a state -5\,\text{эВ}. Determine the energy, frequency, wavelength, momentum, and equivalent mass of the emitted photon.

Answer options

  1. \left(1{,}75\,\text{эВ};\ 4{,}24 \cdot 10^{14}\,\text{Гц};\ 1{,}77 \cdot 10^{-7}\,\text{м};\ 9{,}35 \cdot 10^{-28}\,\text{кг}\cdot\text{м/с};\ 3{,}1 \cdot 10^{-36}\,\text{кг}\right)
  2. \left(7\,\text{эВ};\ 1{,}696 \cdot 10^{15}\,\text{Гц};\ 7{,}08 \cdot 10^{-7}\,\text{м};\ 3{,}74 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с};\ 1{,}24 \cdot 10^{-35}\,\text{кг}\right)
  3. \left(35\,\text{эВ};\ 8{,}48 \cdot 10^{15}\,\text{Гц};\ 3{,}54 \cdot 10^{-6}\,\text{м};\ 1{,}87 \cdot 10^{-26}\,\text{кг}\cdot\text{м/с};\ 6{,}2 \cdot 10^{-35}\,\text{кг}\right)
  4. \left(3{,}5\,\text{эВ};\ 8{,}48 \cdot 10^{14}\,\text{Гц};\ 3{,}54 \cdot 10^{-7}\,\text{м};\ 1{,}87 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с};\ 6{,}2 \cdot 10^{-36}\,\text{кг}\right)
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Given
E_1
-1{,}5\,\text{эВ}
E_2
-5\,\text{эВ}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
c
3\cdot 10^{8}\,\text{м/с}
Find

E_{\text{ф}}\nu\lambdapm

  1. E_{\text{ф}} = E_1 - E_2-1{,}5 - \left(-5\right)E_{\text{ф}} = 3{,}5\,\text{эВ} = 5{,}6\cdot 10^{-19}\,\text{Дж}

    1 eV = 1.6·10⁻¹⁹ J.

  2. \nu = \dfrac{E_{\text{ф}}}{h}\nu = \dfrac{5{,}6\cdot 10^{-19}}{6{,}6\cdot 10^{-34}}\nu = 8{,}48\cdot 10^{14}\,\text{Гц}
  3. \lambda = \dfrac{c}{\nu}\lambda = \dfrac{3\cdot 10^{8}}{8{,}48\cdot 10^{14}}\lambda = 3{,}54\cdot 10^{-7}\,\text{м}
  4. p = \dfrac{h}{\lambda}p = \dfrac{6{,}6\cdot 10^{-34}}{3{,}54\cdot 10^{-7}}p = 1{,}87\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}
  5. m = \dfrac{E_{\text{ф}}}{c^{2}}m = \dfrac{5{,}6\cdot 10^{-19}}{9\cdot 10^{16}}m = 6{,}2\cdot 10^{-36}\,\text{кг}

    Check: p = mc gives the same value of momentum.

AnswerE_{\text{ф}} = 3{,}5\,\text{эВ}\nu = 8{,}48\cdot 10^{14}\,\text{Гц}\lambda = 3{,}54\cdot 10^{-7}\,\text{м}p = 1{,}87\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}m = 6{,}2\cdot 10^{-36}\,\text{кг}

Position in this set: 15 from 23

Problem #75

Quantum physics and photoelectric effect

Having energy -4{,}5\,\text{эВ}, the atom emits a photon with frequency 5\cdot10^{14}\,\text{Гц}. What is the final energy of an atom?

Answer options

  1. -6{,}56\,\text{эВ}
  2. -3{,}28\,\text{эВ}
  3. -13{,}12\,\text{эВ}
  4. -65{,}6\,\text{эВ}
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Given
E_1
-4{,}5\,\text{эВ}
\nu
5\cdot 10^{14}\,\text{Гц}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
Find

E_2

  1. E_{\text{ф}} = h\nuE_{\text{ф}} = 6{,}6\cdot 10^{-34}\cdot 5\cdot 10^{14}3{,}3\cdot 10^{-19}\,\text{Дж} = 2{,}06\,\text{эВ}
  2. E_2 = E_1 - E_{\text{ф}}E_2 = -4{,}5 - 2{,}06E_2 = -6{,}56\,\text{эВ}

    By emitting a photon, the atom descends to a lower level.

AnswerE_2 = -6{,}56\,\text{эВ}

Position in this set: 16 from 23

Problem #77

Quantum physics and photoelectric effect

Gas laser wavelength 632{,}8\,\text{нм}. Determine the frequency, energy, momentum, and equivalent mass of the photon.

Answer options

  1. \left(9{,}48 \cdot 10^{14}\,\text{Гц};\ 6{,}26 \cdot 10^{-19}\,\text{Дж};\ 2{,}08 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с};\ 7 \cdot 10^{-36}\,\text{кг}\right)
  2. \left(4{,}74 \cdot 10^{15}\,\text{Гц};\ 3{,}13 \cdot 10^{-18}\,\text{Дж};\ 1{,}04 \cdot 10^{-26}\,\text{кг}\cdot\text{м/с};\ 3{,}5 \cdot 10^{-35}\,\text{кг}\right)
  3. \left(4{,}74 \cdot 10^{14}\,\text{Гц};\ 3{,}13 \cdot 10^{-19}\,\text{Дж};\ 1{,}04 \cdot 10^{-27}\,\text{кг}\cdot\text{м/с};\ 3{,}5 \cdot 10^{-36}\,\text{кг}\right)
  4. \left(2{,}37 \cdot 10^{14}\,\text{Гц};\ 1{,}565 \cdot 10^{-19}\,\text{Дж};\ 5{,}2 \cdot 10^{-28}\,\text{кг}\cdot\text{м/с};\ 1{,}75 \cdot 10^{-36}\,\text{кг}\right)
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Given
\lambda
632{,}8\,\text{нм}632{,}8\cdot 10^{-9}\,\text{м}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
c
3\cdot 10^{8}\,\text{м/с}
Find

\nuE_{\text{ф}}pm

  1. \nu = \dfrac{c}{\lambda}\nu = \dfrac{3\cdot 10^{8}}{632{,}8\cdot 10^{-9}}\nu = 4{,}74\cdot 10^{14}\,\text{Гц}
  2. E_{\text{ф}} = h\nuE_{\text{ф}} = 6{,}6\cdot 10^{-34}\cdot 4{,}74\cdot 10^{14}E_{\text{ф}} = 3{,}13\cdot 10^{-19}\,\text{Дж}
  3. p = \dfrac{h}{\lambda}p = \dfrac{6{,}6\cdot 10^{-34}}{632{,}8\cdot 10^{-9}}p = 1{,}04\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}
  4. m = \dfrac{E_{\text{ф}}}{c^{2}}m = \dfrac{3{,}13\cdot 10^{-19}}{\left(3\cdot 10^{8}\right)^{2}}m = 3{,}5\cdot 10^{-36}\,\text{кг}

Answer\nu = 4{,}74\cdot 10^{14}\,\text{Гц}E_{\text{ф}} = 3{,}13\cdot 10^{-19}\,\text{Дж}p = 1{,}04\cdot 10^{-27}\,\tfrac{\text{кг}\cdot\text{м}}{\text{с}}m = 3{,}5\cdot 10^{-36}\,\text{кг}

Position in this set: 17 from 23

Problem #90

Quantum physics and photoelectric effect

Ionization energy of a hydrogen atom 13{,}6\,\text{эВ}. Is a photon with a wavelength capable of 4\cdot10^{-8}\,\text{м} ionize an atom?

Answer options

  1. \text{No, Photon energy less energy ionization}
  2. \text{No, photons not ionize atoms}
  3. \text{Yes, but only at doubling intensity}
  4. \text{Yes, Photon energy more energy ionization}
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Given
E_i
13{,}6\,\text{эВ}2{,}176\cdot 10^{-18}\,\text{Дж}
\lambda
4\cdot 10^{-8}\,\text{м}
h
6{,}6\cdot 10^{-34}\,\text{Дж}\cdot\text{с}
c
3\cdot 10^{8}\,\text{м/с}
Find

E_{\text{ф}}

  1. E_{\text{ф}} = \dfrac{hc}{\lambda}E_{\text{ф}} = \dfrac{6{,}6\cdot 10^{-34}\cdot 3\cdot 10^{8}}{4\cdot 10^{-8}}E_{\text{ф}} = 4{,}95\cdot 10^{-18}\,\text{Дж}
  2. E_{\text{ф}} > E_i\text{ionization possible}

    The energy of a photon is enough to remove an electron from a hydrogen atom.

AnswerE_{\text{ф}} = 4{,}95\cdot 10^{-18}\,\text{Дж} > E_i\ \text{- способен}

Position in this set: 18 from 23

Problem #140

Photons and the photoelectric effect

Light frequency 500000000000000 Hz Find the photon energy at h = 6{,}626\cdot 10^{-34} J·s.

Answer options

  1. 3{,}313\cdot 10^{-19}\,\text{Дж}
  2. 6{,}626\cdot 10^{-19}\,\text{Дж}
  3. 1{,}6565\cdot 10^{-19}\,\text{Дж}
  4. 9{,}939\cdot 10^{-19}\,\text{Дж}
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Given
f
500000000000000\,\text{Гц}
h
6{,}626\cdot 10^{-34}\,\text{Дж·с}
Find

E

  1. E=h\nuE=6{,}626\cdot 10^{-34} \cdot 500000000000000E=3{,}313\cdot 10^{-19}\,\text{Дж}

    The energy of one photon is proportional to frequency.

Answer3{,}313\cdot 10^{-19}\,\text{Дж}

Position in this set: 19 from 23

Problem #141

Photons and the photoelectric effect

Light frequency 750000000000000 Hz Find the photon energy at h = 6{,}626\cdot 10^{-34} J·s.

Answer options

  1. 9{,}939\cdot 10^{-19}\,\text{Дж}
  2. 4{,}9695\cdot 10^{-19}\,\text{Дж}
  3. 2{,}48475\cdot 10^{-19}\,\text{Дж}
  4. 1{,}49085\cdot 10^{-18}\,\text{Дж}
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Given
f
750000000000000\,\text{Гц}
h
6{,}626\cdot 10^{-34}\,\text{Дж·с}
Find

E

  1. E=h\nuE=6{,}626\cdot 10^{-34} \cdot 750000000000000E=4{,}9695\cdot 10^{-19}\,\text{Дж}

    The energy of one photon is proportional to frequency.

Answer4{,}9695\cdot 10^{-19}\,\text{Дж}

Position in this set: 20 from 23

Problem #142

Photons and the photoelectric effect

Photon energy 5 eV, metal work function 2 eV. Find the delay voltage in volts. For an electron, energy of 1 eV corresponds to work at a voltage of 1 V.

Answer options

  1. 6\,\text{В}
  2. 3\,\text{В}
  3. 1{,}5\,\text{В}
  4. 9\,\text{В}
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Given
E
5\,\text{эВ}
A
2\,\text{эВ}
Find

U_{\text{з}}

  1. K=E-AK=5 - 2K=3\,\text{эВ}

    Part of the energy goes to the exit of the electron.

  2. U_{\text{з}}=\frac{K}{|e|}U_{\text{з}}=3U_{\text{з}}=3\,\text{В}

    The number of electron volts for one electron is equal to the number of volts of retardation voltage.

Answer3\,\text{В}

Position in this set: 21 from 23

Problem #143

Photons and the photoelectric effect

Photon energy 6 eV, metal work function 2{,}5 eV. Find the delay voltage in volts. For an electron, energy of 1 eV corresponds to work at a voltage of 1 V.

Answer options

  1. 7\,\text{В}
  2. 1{,}75\,\text{В}
  3. 3{,}5\,\text{В}
  4. 10{,}5\,\text{В}
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Given
E
6\,\text{эВ}
A
2{,}5\,\text{эВ}
Find

U_{\text{з}}

  1. K=E-AK=6 - 2{,}5K=3{,}5\,\text{эВ}

    Part of the energy goes to the exit of the electron.

  2. U_{\text{з}}=\frac{K}{|e|}U_{\text{з}}=3{,}5U_{\text{з}}=3{,}5\,\text{В}

    The number of electron volts for one electron is equal to the number of volts of retardation voltage.

Answer3{,}5\,\text{В}

Position in this set: 22 from 23

Problem #144

Photons and the photoelectric effect

The light pulse lasts 0{,}001 s at power 0{,}003313 Tue Light frequency 500000000000000 Hz, h = 6{,}626\cdot 10^{-34} J·s. Find the number of photons.

Answer options

  1. 20000000000000\,\text{фотонов}
  2. 5000000000000\,\text{фотонов}
  3. 10000000000000\,\text{фотонов}
  4. 30000000000000\,\text{фотонов}
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Given
t
0{,}001\,\text{с}
P
0{,}003313\,\text{Вт}
f
500000000000000\,\text{Гц}
h
6{,}626\cdot 10^{-34}\,\text{Дж·с}
Find

N

  1. E=h\nuE=6{,}626\cdot 10^{-34} \cdot 500000000000000E=3{,}313\cdot 10^{-19}\,\text{Дж}

    Finding the energy of one photon.

  2. W=PtW=0{,}003313 \cdot 0{,}001W=0{,}000003313\,\text{Дж}

    We find the energy of the entire impulse.

  3. N=\frac{W}{E}N=\frac{0{,}000003313}{3{,}313\cdot 10^{-19}}N=10000000000000\,\text{фотонов}

    We divide the total energy by the photon energy.

Answer10000000000000\,\text{фотонов}

Position in this set: 23 from 23

Problem #145

Photons and the photoelectric effect

The light pulse lasts 0{,}002 s at power 0{,}0049695 Tue Light frequency 750000000000000 Hz, h = 6{,}626\cdot 10^{-34} J·s. Find the number of photons.

Answer options

  1. 40000000000000\,\text{фотонов}
  2. 10000000000000\,\text{фотонов}
  3. 60000000000000\,\text{фотонов}
  4. 20000000000000\,\text{фотонов}
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Given
t
0{,}002\,\text{с}
P
0{,}0049695\,\text{Вт}
f
750000000000000\,\text{Гц}
h
6{,}626\cdot 10^{-34}\,\text{Дж·с}
Find

N

  1. E=h\nuE=6{,}626\cdot 10^{-34} \cdot 750000000000000E=4{,}9695\cdot 10^{-19}\,\text{Дж}

    Finding the energy of one photon.

  2. W=PtW=0{,}0049695 \cdot 0{,}002W=0{,}000009939\,\text{Дж}

    We find the energy of the entire impulse.

  3. N=\frac{W}{E}N=\frac{0{,}000009939}{4{,}9695\cdot 10^{-19}}N=20000000000000\,\text{фотонов}

    We divide the total energy by the photon energy.

Answer20000000000000\,\text{фотонов}

The theory behind these problems

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Photons and the photoelectric effectPhoton energy and momentum, the photoelectric equation and the threshold wavelength.

Formulas by topic

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WorkA · ДжFrequencyν · ГцEnergyE, W · Дж