INPhO 2023, Теория — Задача 2. Knock it off!

Indian National Physics Olympiad (INPhO)-2023 — Question Paper · 29 януари 2023 г. · 6 т.

Указания

Date: 29 January 2023

Time: 09:00-12:00 (3 hours) Maximum Marks: 60

Instructions Roll No.:

  1. This booklet consists of 6 pages and total of 5 questions. Write roll number at the top wherever asked.
  2. Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
  3. Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
  4. Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Table of Constants

Speed of light in vacuumcc3.00×108 m⋅s−13.00 \times 10^{8}\ \mathrm{m{\cdot}s^{-1}}
Planck’s constanthh6.63×10−34 J⋅s6.63 \times 10^{-34}\ \mathrm{J{\cdot}s}
ℏ\hbarh/2πh/2\pi
Universal constant of GravitationGG6.67×10−11 N⋅m2⋅kg−26.67 \times 10^{-11}\ \mathrm{N{\cdot}m^2{\cdot}kg^{-2}}
Magnitude of electron chargeee1.60×10−19 C1.60 \times 10^{-19}\ \mathrm{C}
Rest mass of electronmem_e9.11×10−31 kg9.11 \times 10^{-31}\ \mathrm{kg}
Value of 1/4πϵ01/4\pi\epsilon_09.00×109 N⋅m2⋅C−29.00 \times 10^{9}\ \mathrm{N{\cdot}m^2{\cdot}C^{-2}}
Avogadro’s numberNAN_A6.022×1023 mol−16.022 \times 10^{23}\ \mathrm{mol^{-1}}
Acceleration due to gravitygg9.81 m⋅s−29.81\ \mathrm{m{\cdot}s^{-2}}
Universal Gas ConstantRR8.31 J⋅K−1⋅mol−18.31\ \mathrm{J{\cdot}K^{-1}{\cdot}mol^{-1}}
RR0.0821 l⋅atm⋅mol−1⋅K−10.0821\ \mathrm{l{\cdot}atm{\cdot}mol^{-1}{\cdot}K^{-1}}
Boltzmann constantKBK_B1.3806×10−23 J⋅K−11.3806 \times 10^{-23}\ \mathrm{J{\cdot}K^{-1}}
Permeability constantμ0\mu_04π×10−7 H⋅m−14\pi \times 10^{-7}\ \mathrm{H{\cdot}m^{-1}}
1 Angstrom unit1Å1×10−10 m1 \times 10^{-10}\ \mathrm{m}
1 micro unit1μ1\mu1×10−61 \times 10^{-6} units
1 electron volt1 eV1.6×10−19 J1.6 \times 10^{-19}\ \mathrm{J}
Q No12345Total
Maximum Marks6616161660
Указания

Date: 29 January 2023

Time: 09:00-12:00 (3 hours) Maximum Marks: 60

Instructions Roll No.:

  1. This booklet consists of 19 pages and total of 5 questions. Write roll number at the top wherever asked.
  2. Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
  3. Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
  4. Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Table of Constants

Speed of light in vacuumcc3.00×108 m⋅s−13.00 \times 10^{8}\ \mathrm{m{\cdot}s^{-1}}
Planck’s constanthh6.63×10−34 J⋅s6.63 \times 10^{-34}\ \mathrm{J{\cdot}s}
ℏ\hbarh/2πh/2\pi
Universal constant of GravitationGG6.67×10−11 N⋅m2⋅kg−26.67 \times 10^{-11}\ \mathrm{N{\cdot}m^2{\cdot}kg^{-2}}
Magnitude of electron chargeee1.60×10−19 C1.60 \times 10^{-19}\ \mathrm{C}
Rest mass of electronmem_e9.11×10−31 kg9.11 \times 10^{-31}\ \mathrm{kg}
Value of 1/4πϵ01/4\pi\epsilon_09.00×109 N⋅m2⋅C−29.00 \times 10^{9}\ \mathrm{N{\cdot}m^2{\cdot}C^{-2}}
Avogadro’s numberNAN_A6.022×1023 mol−16.022 \times 10^{23}\ \mathrm{mol^{-1}}
Acceleration due to gravitygg9.81 m⋅s−29.81\ \mathrm{m{\cdot}s^{-2}}
Universal Gas ConstantRR8.31 J⋅K−1⋅mol−18.31\ \mathrm{J{\cdot}K^{-1}{\cdot}mol^{-1}}
RR0.0821 l⋅atm⋅mol−1⋅K−10.0821\ \mathrm{l{\cdot}atm{\cdot}mol^{-1}{\cdot}K^{-1}}
Boltzmann constantKBK_B1.3806×10−23 J⋅K−11.3806 \times 10^{-23}\ \mathrm{J{\cdot}K^{-1}}
Permeability constantμ0\mu_04π×10−7 H⋅m−14\pi \times 10^{-7}\ \mathrm{H{\cdot}m^{-1}}
1 Angstrom unit1Å1×10−10 m1 \times 10^{-10}\ \mathrm{m}
1 micro unit1μ1\mu1×10−61 \times 10^{-6} units
1 electron volt1 eV1.6×10−19 J1.6 \times 10^{-19}\ \mathrm{J}
Q No12345Total
Maximum Marks6616161660

Please note that alternate/equivalent methods and different way of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

Consider a 100 W small isotropic source of blue light of wavelength 4500Å. A metallic surface of 1.00 cm2^2 and work function 2.20 eV is kept at a distance of 1.00 m from the source and oriented to receive normal radiation.

(a) [2 marks] Assume that all the energy is uniformly absorbed by atoms on the top layer of the surface. Also, all the energy absorbed by an atom on the surface is taken up by one electron. The radius of the atom is 1.00Å. Estimate the time τe\tau_{\mathrm{e}} needed by the electron to receive 1.00 eV of energy.

(b) [1 marks] According to the above classical model, how many electrons are emitted by the metallic surface in time τe\tau_{\mathrm{e}}?

(c) [2 marks] In quantum theory, photons are emitted and absorbed as quanta. Assuming photoelectric efficiency of 1%, calculate the rate of emission of electrons (NeN_{\mathrm{e}}) from the surface.

(d) [1 marks] Assuming further that all the emitted photoelectrons move normal to the surface what would be the maximum current density (Jmax⁡J_{\max}) one may expect?

Отговори

Решение

Оригинал в Архива: INPhO2023-Question.pdf · официални решения: INPhO2023-Solution-20230215.pdf