11

Q1. Ultraviolet light is incident on two photosensitive materials having work functions W₁ and W₂ (W₁ > W₂). In which case will kinetic energy of the emitted electron be greater ? Why ?

Solution

From Einstein"s photoelectric equation  h = W + E_K E_K = h - W Clearly, smaller the work function, greater the K.E. As W₁ > W2, K.E for metal of work function W₂ will be greater.

Q2. Show that de Broglie hypothesis of matter waves is in agreement with Bohr's theory.

Solution

 According to Bohr's theory, the angular momentum of electron in a stationary orbit is equal to integral multiple of h/ i.e.                         Bohr could give no reason for this quantum condition. However, de Broglie gave the reason for the same. Since electrons move in circular paths, the electron wave must be a circular standing wave that closes on itself .In other words; the circumference of the circular orbit contains a whole number of wavelengths.                        

Q3. Plot a graph showing the variation of photoelectric current with anode potential for two light beams of same wavelength but different intensity.

Solution

The graph is shown in figure.

Q4. The de Broglie wavelengths, associated with a proton and a neutron, are found to be equal. Which of the two has a higher value for kinetic energy ?

Solution

De - Broglie wavelength is given by, Where, E is kinetic energy. Given,  Now, mass of proton is slightly less than the mass of neutron. Hence, kinetic energy of proton is slightly higher than that of neutron.  

Q5. Why does photoelectric emission not take place if the frequency of incident radiation is less than threshold value?

Solution

According to Einstein's photoelectric equation,                            Where, m= mass of the electron                  If the frequency of incident radiation is less than the threshold value (), the K.E. of emitted electron is nagative i.e. photoelectric emission will not take place, no matter how large is the intensity of the incident radiation.         

Q6. Is a moving particle equivalent to a single wave?

Solution

 No. A moving material particle is equivalent to a group of waves called wave packet.

Q7. The work function of cesium is 1.8 eV. Light of 5000 is incident on it. Calculate (i) Threshold wavelength (ii) Maximum K.E of emitted electrons (iii) Maximum velocity of emitted photoelectric electrons. Take h=6.63x10^-34J; c=3X10⁸m/s ; m=9X10^-31Kg

Solution

Q8. Are matter waves electromagnetic in nature?

Solution

No, they are a new kind of waves proposed to locate the position of a moving particle. For this reason, they are sometimes called pilot waves.

Q9. How do matter waves differ from a light wave as regards the velocity of the particle and the wave?

Solution

In case of matter waves, the wave velocity is different from particle velocity.  But in case of light (i.e. electromagnetic wave) the particle (photon) velocity is the same as wave velocity.

Q10. Why is it difficult to remove a free electron from copper than sodium? Which has higher threshold wavelength?

Solution

The work function of copper is 4.5 eV while that for sodium is 2 eV. Therefore, radiation of more energy is required to remove a free electron from copper as compared to sodium. Since threshold radiation is inversely proportional to the work function, its value will be more for sodium.  

Take Another Test