Section

= / – / R v eB v c . We have e V = (m v2/ ) and R = (m v/e B) which gives (e/m) = (2V/R B ); using the given data (e/m) = . × C kg – . .

(a) . keV (b) of the order of kV . Use #= (hc/E) with E = . × .

× –28J. Number of photons emitted per second = 104J s– / . × –28J × s– We see that the energy of a radiophoton is exceedingly small, and the number of photons emitted per second in a radio beam is enormously large. There is, therefore, negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave as continuous.

(b) For += × Hz, E × –19J. Photon flux corresponding to minimum intensity = – W m– / × –19J = . × m– s– Number of photons entering the pupil per second = . × × .

× – s– = s– . Though this number is not as large as in (a) above, it is large enough for us never to ‘sense’ or ‘count’ individual photons by our eye. . - = h + – e V0 = .

The photo-cell will not respond howsoever high be the intensity of laser light. . Use e V0 = h + – - for both sources. From the data on the first source, - = .

eV. Use this value to obtain for the second source V0 = . V. .

Obtain V0 versus + plot. The slope of the plot is (h/e) and its intercept on the +-axis is + . The first four points lie nearly on a straight line which intercepts the +-axis at + = . × Hz (threshold frequency).

The fifth point corresponds to + < + ; there is no photoelectric emission and therefore no stopping voltage is required to stop the current. Slope of the plot is found to be . × – V s. Using e = .

. It is found that the given incident frequency +is greater than + (Na), and + (K); but less than + (Mo), and + (Ni). Therefore, Mo and Ni will not give photoelectric emission. If the laser is brought closer, intensity of radiation increases, but this does not affect the result regarding Mo and Ni.

However, photoelectric current from Na and K will increase in proportion to intensity. . Assume one conduction electron per atom. Effective atomic area ~ – m2 Number of electrons in layers = = Incident power = – W m– × × – m2 = × – W In the wave picture, incident power is uniformly absorbed by all the electrons continuously.

Consequently, energy absorbed per second per electron = × – / = × – W Time required for photoelectric emission = × . × –19J/ × – W = . × s which is about . year.

Implication: Experimentally, photoelectric emission is observed nearly instantaneously ( – s): Thus, the wave picture is in gross disagreement with experiment. In the photon-picture, energy of the radiation is not continuously shared by all the electrons in the top layers. Rather, energy comes in discontinuous ‘quanta’. and absorption of energy does not take place gradually.

A photon is either not absorbed, or absorbed by an electron nearly instantly. . For #= Å, electron’s energy = eV; photon’s energy = . keV.

Thus, for the same wavelength, a photon has much greater energy than an electron. . (a) h h p m K Thus, for same K, # decreases with m as ( / m ). Now (mn/me) = .

; therefore for the same energy, ( eV) as in Ex. . , wavelength of neutron = . >?

× – m = . × – m. The interatomic spacing is about a hundred times greater. A neutron beam of eV energy is therefore not suitable for diffraction experiments.

(b) #= . × – m [Use #=( / ) h m k T ] which is comparable to interatomic spacing in a crystal.