Plot a graph showing variation of de Broglie wavelength λ associated with a charged particle of mass m , versus 1 V , where V is the potential difference through which the particle is accelerated. How does this graph gives us the information regarding the magnitude of the charge of the particle?

Let us consider a charged particle carrying charge q , mass m , then the associated de Broglie wavelength is given by, λ = h 2 mqV where h is Planck ' s constant and V is the potential difference through which the particle is accelerated. ⇒ λ ∝ 1 V — equation 1 and slope = h 2 mq — equation 2 So the plot of λ versus 1 V is shown below : from equation 2 , charge, q = h 2 2 m slope 2 We can determine slope from the graph, and this graph gives us the information regarding the magnitude of the charge of the particle as q ∝ 1 slope 2 . So, greater the charge, slope will be lesser.

The variation of the stopping potential V ∘ with the frequency ν of the light incident on two different photosensitive surfaces M1 and M2 is shown in the figure. Identify the surface which has greater value of the work function.

Here, ν is frequency of the metal, for the given plot, the v intercept of M 1 and M 2 resembles their corresponding threshold frequency. ν 1 < ν 2 The relation between threshold frequency and binding energy B E is: ν = B E h , here h is Planck's constant. So, M 2 surface is having greater threshold frequency in comparison to M 1 (i.e. ν 2 > ν 1 ). This also tells that the amount of energy binding the electron to the metal M 2 is greater than M 1 (i.e. B E 2 > B E 1 ). Since, binding energy of electron in metal is also referred to be work function of the metal. Hence, M 2 has greater value of the work function.

A proton and an electron have equal speeds. Find the ratio of de Broglie wavelengths associated with them.

The de Broglie wavelength is λ = h m v Here, h is Planck's constant and m v is momentum of a given particle. Both electron and proton are moving with same speed then their ratio of wavelength: λ p λ e = h m p v h m e v = m e m p = 1 1836 Hence, the ratio of de Broglie wavelengths associated with proton and electron is 1 : 1836 .

If the light of wavelength 412 . 5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why? Metal Work Function(eV) Na 1.92 eV K 2.15 eV Ca 3.20 eV Mo 4.17 eV

We are given, Wavelength of the incident light, λ = 412 . 5 nm = 412 . 5 × 10 - 9 m Thus, the energy of the incident light, E = h c λ E = 6 . 626 × 10 - 34 Js × 3 × 10 8 ms - 1 412 . 5 × 10 - 9 m = 4 . 82 × 10 - 19 J E = 4 . 82 × 10 - 19 1 . 6 × 10 - 19 eV = 3 . 01 eV From the table given in the question, the given energy is greater than the work function of sodium and potassium, but it is less than the work function for calcium and molybdenum. Thus, for the given incident light, the photoelectric emission takes places for Na and K.

Why is wave theory of electromagnetic radiation is not able to explain photo electric effect? How does photon picture resolve this problem?

The explanation based on the wave theory of electromagnetic radiation does not agree with the experimental observations ( instantaneous nature, maximum kinetic energy of the emitted photoelectron is independent of intensity, the existence of threshold frequency ) on the photoelectric effect. The photon picture resolves this problem by saying that light, in interaction with matter behaves as if it is made of quanta or packets of energy, each of energy hν . This picture enables us to get a correct explanation of all the observed experimental features of the photoelectric effect.

Write two main observations of photoelectric effect experiment which could only be explained by Einstein's photoelectric equation.

Einstein's photoelectric equation is h ν = ϕ + E k The main observations of the photoelectric effect experiment are 1 . There exist a threshold frequency below which no photoelectron is elected. 2 . The kinetic energy of electron depends linearly on the frequency and is independent of the intensity of radiation.

Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

Thus, keeping the value of frequency and intensity of light constant, the value of photoelectric current increases with the increase in applied voltage. For a given frequency, greater is the intensity of light, greater will be the photoelectric current. The upper curve, I 2 represents radiation having higher intensity while the lower curve, I 1 represents radiation having lower intensity.