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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 9 of 37
Marks: +1, -0
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Solution:  
Let us calculate the energy associated with photons incident
E = hcλ\frac{hc}{\lambda} = 6.63×10−34×3×108330×10−9\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{330 \times 10^{-9}} = 6.027 × 10−1910^{-19} J
E = 6.027×10−191.6×10−19\frac{6.027 \times 10^{-19}}{1.6 \times 10^{-19}} eV = 3.77 eV
Since, energy of incident photons i.e., 3.77 eV is less than work function, hence no emission will take place.
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