Quantum mech
- 1. Quantum Mechanics 1. How many Photons of red light λ = 6.7× have the same energy as one Photon of γ-rays λ =1. 6.× Sol. = Let n be the required no. of photons of red light which have the same energy as that of one photon of γ-rays, then = 2. Show that the energy of a photon of wavelength λ is given by . Find the range of photon energies in the visible portion of electromagnetic spectrum. Sol. = =12422/λ The Wavelength range for visible light is 4000-7000 Therefore range of energies for visible photons as 1.8 to 3.1 eV 3. The maximum energy given to an electron during Compton scattering is 35 KeV. Find the wavelength of the incident photon. Sol. The maximum energy is given to an electron when photon is backscattered, that is when . = =0.048 . The energy gained by the electron = energy of the incident photon - energy of the scattered photon =
- 2. Or 4. Calculate the percent change in photon energy for a Compton collision with scattering angle equal to for the radiation in (a) the microwave range, with λ=3.0 cm, (b) the visible range, with λ = 5000 , (c) the X-ray range, with λ =1 and (d) the γ- ray range, with λ = 0.012 . Sol. Fractional loss of energy = = Thus the percentage change in photon energy Therefore the percentage change in photon energy for the radiation in (a) the microwave range, with λ=3.0 cm =8 . (b) The visible range, with λ = 5000 , =5 (c) The X-ray range, with λ =1 =2.3 (d) The γ-ray range, with λ = 0.012 . =66.7 5. Calculate the de Broglie wavelength of (a) a rock of mass 50 g thrown with the speed of 40 , and (b) an electron accelerated through 50 V. Sol. (a) for a rock, λ= =3.31 (b) For an electron,
- 3. λ= 6. Find the uncertainty in position for (a) a ball of mass 45 g with a speed of 40 m/s measured to a precision of 1.5% and (b) an electron moving with a speed of m/s, measured to a precision of 1.5%. Sol. (a) (b) P = 9.1 =1.82 Hence 10 Find the first three energy levels of (a) a marble of mass 10 g in a box of width 10 cm and (b) an electron in a box of width 1Å.` Sol. The marble or electron cannot leave the box, therefore the problem is like that of a particle in an infinite potential well. The energy level is given by (a) For a marble =5.5 Thus (b) For an electron
- 4. Thus