Q24 of 43 Page 409

In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)

Given:


Total energy of system, E = 10.2 BeV = 10.2 × 109 × 1.6 × 10-19J


E = 16.32 × 10-10J


Energy in each ray, E’ = E/2


E’ = 8.16 × 10-10J


We know that Energy in a photon is given by,


…(1)


Where,


E = Energy of photon


h = Planck’s constant = 6.6 × 10-34 Js


c = 3 × 108 ms-1


λ = wavelength of photon


From equation (1) we can get,




λ = 2.34 × 10-16m


Hence, the wavelength of each ray is 2.3 × 10-16 m.


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(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?

(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

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Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in the barely detectable light.

(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.


(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10–10 W m–2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz.

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Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼105 W m–2) red light of wavelength 6328 Å produced by a He-Ne laser?