Numerical problems of Planck's quantum theory

1. By, Jayam chemistry adda

2. In 1900, Max Planck explained the particle nature of light with his theory known as Planck's quantum theory. Planck quantum formula Each discrete tiny packet of energy is quantum. And the bundle of light energy is called photon. (? = ??) For a single photon (? = ???) For “n” no of photons Units of Planck quantum ? =?? ? = ??Ῡ (Or) ? SI Unit---- Joule (Or) ? = (6.626 × 10−34) × (3 × 108) × Ῡ ? =(6.626 × 10−34) × (3 × 108) CGS unit--- Erg λ ? = (??.??? × ??−??) × Ῡ Atomic unit --- electron volt (eV) ??.??? × ??−?? ? Planck wavenumber equation ? = 2 Planck wavelength equation Jayam chemistry adda

3. Problem-1: Calculate the energy of a photon at 525 nm. Solution: The Wavelength of photon=525 nm The formula to measure the energy of photon is; ? =??.??? × ??−?? ? 10−26 −26 10−9 −9 E=19.878 19.878× ×10 525 525× ×10 E= joule 10−17 −17joules E=0.0378 E=0.0378× ×10 ? = ?.?? × ??−???????? 3 Jayam chemistry adda

4. Problem-2: The energy absorbed by each molecule (A2) of a compound is 8.4 X 10-19joules. And bond energy per molecule is 8.0 X 10-19joules. What is the kinetic energy per atom of the molecule? Solution: Energy absorbed by a molecule = 8.4 X 10-19joules Bond energy of each molecule of the substance = 8.0 X 10-19joules The formula to calculate the kinetic energy per atom of the molecule is; Energy absorbed − (bond energy per molecule) number of atoms in the molecule Kinetic energy of each atom of a molecule = 8.4 × 10−19− 8.0 × 10−19?????? 2 ?? = KE = 0.2 × 10−19joules = 2 × 10−20joules 4 Jayam chemistry adda

5. Problem-3: Find the number of photons of light at 4000 pm having 1 joule of energy ? Solution: Energy of light = 1 joule Wavelength of light = 4000 pm= 4000 x 10-12m The formula to measure the number of photons of light is; E = nhν =nhc λ (By substituting the values of h and c) n =λ? λ? ℎ?= 19.878 × 10−26?? 4000 × 10−12? × 1 ? 19.878 × 10−26 ? = ?? ? = 201.22 × 1014= 2.0122 × 1016 5 Jayam chemistry adda

6. Problem-4: A 242 nm electromagnetic light can ionize a sodium atom. What is the ionization energy of sodium in KJ/mole? Solution: The wavelength of electromagnetic light = 242 nm = 242 x 10-9m The formula to calculate the ionization energy of one sodium atom is; 19.878 × 10−26Jm λ ? = 19.878 × 10−26Jm 242 × 10−9m E = E = 0.0821 × 10−17Joule Ionization energy of one mole of a substance = ionization energy of one atom x Avogadro's number Ionization energy of one mole of sodium = (0.0821 × 10−17joule) × (6.023 × 1023) 6 Ionization energy of one mole of sodium = 0.494 × 106joule/mole = 494 KJ/mole Jayam chemistry adda

7. Problem-5: A nitrogen laser produces electromagnetic radiation at 337.1 nm and emits 5.6 X 1024 photons. Calculate the energy of the radiation. Solution: Wavelength of electromagnetic radiation = 337.1 nm = 337.1 x 10-9m Number of photons emitted by the laser = 5.6 X 1024 The formula to calculate the energy of radiation is; =n 19.878 × 10−26jm E =nhc λ λ 5.6 × 1024× 19.878 × 10−26jm 337.1 × 10−9m E = E = 0.3302 x 107 joules E = 3.302 × 106joules 7 Jayam chemistry adda

8. Problem-6: Calculate the energy of the photon having frequency of 4.5 X 1012Hz. Solution: The frequency of photon = 4.5 X 1012Hz The formula to measure the energy of photon is: E = hν E = 6.626 × 10−34js × 4.5 × 1012Hz ? = 29.8 × 10−22joule E = 2.98 × 10−21joule 8 Jayam chemistry adda

9. Problem-7: Find the distance traveled by light in a vacuum in 20 seconds. Solution: Time period of light in vacuum= 20 seconds Velocity of light in vacuum = 3 x 108m/sec The formula to calculate the distance travelled by light in vacuum= velocity x distance Distance travelled by light in vacuum in 20 seconds is; ? = 3 × 108? × 20 ??? ??? d= 6 x 109m 9 Jayam chemistry adda

10. Problem-8 : Calculate the energy of a photon whose frequency is 6 x 1020Hz Solution: The frequency of photon = 6 x 1020Hz The formula to calculate the energy of photon is; E = hν E = 6.626 × 10−34js × 6 × 1020Hz E = 39.75 × 10−14joules 1 ?? = 1.602 × 10−19????? 39.75 × 10−14 1.602 × 10−19eV E = E = 24.8 × 105eV = 2.48 × 106eV = 2.48 MeV 10 Jayam chemistry adda

11. Problem-9: What is the energy of a mole of photons having a wavelength of 715 nm? Solution: The wavelength of photon=715 nm=715 x 10-9nm The formula to measure the energy of one mole of photons is; ? =19.878 × 10−26?????? × ????????′? ?????? λ ? =19.878 × 10−26?????? 715 × 10−9 × (6.023 × 1023) ? = 0.0278 × 10−17?????? × (6.023 × 1023) E=0.1674 x 106joules E = 1.674 x 105 joules 11 Jayam chemistry adda

12. Problem-10: Electromagnetic radiation has a frequency of 8.3 x 1014Hz. Then find its energy in eV. Solution: The frequency of electromagnetic radiation = 8.3 x 1014Hz The formula to calculate the energy of electromagnetic radiation is; E = hν h = 4.136 x 10-15eV second E = 4.136 × 10−15eV second × 8.3 × 1014Hz ? = 34.32 × 10−1eV E = 3.432 eV 12 Jayam chemistry adda

13. Problem-11: The value of Planck’s constant is 6.626 x 10-34Js. The velocity of light in a vacuum is 3 x 108m/sec. What is the wavelength of light at 8 x 1015Hz? Solution: The frequency of light = 8 x 1015Hz The formula to calculate the wavelength of light is; λ =c Where, (c is velocity of light in vacuum) ν λ =3 × 108m/sec 8 × 1015Hz λ = 0.375 × 10−7m λ = 37.5 × 10−9m = 37.5 nm 13 Jayam chemistry adda

14. Problem-12: The energies of the two photons are E1and E2. The wavelengths associated with the two photons are 9000 nm and 18000nm. What is the relationship between E1 and E2? Solution: The wavelength of photon-1=λ1= 9000nm The wavelength of photon-2=λ2= 18000 nm By Planck quantum law, the relationship between photon’s energy and its wavelengths is E =hc E1 E2 =18000 9000= 2 λ E ∝1 E1= 2E2 λ ?1 ?2 =λ2 λ1 14 Jayam chemistry adda