Ncert Exercises & Solutions

Question 11.29: The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed I m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Hint: For photoelectric emission, the energy of the incident radiation should greater than the work function of metals.

Step 1: Find the energy of incident radiation.

The energy of incident radiation is given as:

$\begin{aligned} E & = \frac{hc}{λ} \\ = \frac{6 .6 \times 10^{- 34} \times 3 \times 10^{8}}{3300 \times 10^{- 10}} = 6 \times 10^{- 19} J \\ = \frac{6 \times 10^{- 19}}{1 .6 \times 10^{- 19}} = 3 .76 eV. \end{aligned}$

It can be observed that the energy of the incident radiation is greater than the work function of Na and K only. It is less for Mo and Ni. Hence, Mo and Ni will not show photoelectric emissions.
Step 2: Identify the effect on energy and intensity.
If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase.
This does not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.

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