Question

(b) Light oF a certain wavelength is just able to cause the emission of photoelectrons with close to zero velocity from silver whose work-function is 4,73 electron-volts. The same light is now shone on to a sodium surface whose work-function is 2,46 e-V. Calculate the energy of the electrons which are emitted, in electron-volts. [6] (c) The electron in a Bohr hydrogen atom has been excited to the n = 3 state. (I) Sketch on a labelled energy level diagram the possible transitions by which it can return to the ground state; [4] (ii) Calculate the three possible wavelengths of the photons which might be observed. Take the Rydberg constant in the form RH = (13,6e/hc). [4]

Answer 1

Part B:

The light is shone on silver, photo electrons emerge eith zero velocity. So the energy of photon which is absorbed by an electron of silver, is just enough to release it from the surface of silver. In others words the energy of a incident photon is equal to the work function of silver = 4.73 eV.

Same light when shone on sodium, whose work function is 2.46 eV. As light is always absorbed in integer multiples, the photon cannot eject two photo electrons from surface. So when an electron absorbs a photon, the excess energy of photon is observed as kinetic energy of the electron= 4.73eV - 2.46eV= 2.27eV

Part C:

The energy of a nth stationary level in a hydrogen atom, is given by

$E_n=[\frac{2\pi^2 me^4}{h^2}]\ \frac{1}{n^2}=\frac{-13.6}{n^2}\ eV$

where m and e are mass and charge of an electron and h is planck's constant

The magnitude of energy of a stationary level decreases, with increasing n. Pictorially

A dexcitation is a phenomena in which an electron moves from high energy level (excited state) to low energy state. So we the number of transitions would be the equal to ways to select any two levels out of 3 total levels.

Thus we have $3_{C_2}$ = 3 transitions totally, from n=3 to n=2 shown as red ; n=3 to n=1 shown as blue ; n=2 to n=1 shown as green

$E_{photon\ in\ eV}X q_{electron}=E_{photon\ in\ joule}=\frac{hc}{\lambda}\Rightarrow \lambda=\frac{hc}{qE_{ev}}$

This a handy conversion formula $\lambda_{A^o}=\frac{12406}{E_{ev}}\ and\ 1A^o=10nm$

The red code line emission has an energy =1.89 eV which corresponds to 656.4nm

The green coded line emission has an energy= 10.20 eV which corresponds to 121.6nm

The blue coded line emission corresponds to= 12.09 eV which corresponds to 102.6nm

Answer 2

b) When the electrons are emitted with almost zero velcoity the work function is equal to the incident energy. so incident energy is 4.73 eV

now using einstein photoelectric equation

        E = W + K

so kinetic energy k = E - W = 4.73 - 2,46 = 2.27 eV

C) II)

The three possible transistions are n=3 to n=2, n=3 to n=1 & n=2 to n=1.

the wavelength equation is 1/ $\lambda$ = R [ (1/n₃²) - (1/n₂²) ] = 656.3nm

                                       1/ $\lambda$ = R [ (1/n₃²) - (1/n₁²) ] = 102.6nm

                                       1/ $\lambda$ = R [ (1/n₂²) - (1/n₁²) ] = 121.6nm