Ncert Exercises & Solutions

Question 11.9: The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Hint: Recall the condition of photoelectric emission.
Step 1: Find the energy of the incident photon.

The energy of the incident photon is given as:

$\begin{array}{l} E = \frac{h c}{λ} \\ = \frac{6.626 \times 10^{- 4} \times 3 \times 10^{8}}{330 \times 10^{- 9}} = 6.0 \times 10^{- 19} J \\ = \frac{6.0 \times 10^{- 17}}{1.6 \times 10^{- 19}} = 3.76 e V \end{array}$

Step 2: Compare the energy with work function.
Work function of the metal, $\phi_{0}$ = 4.2 eV
It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.

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