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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 8 of 37
Marks: +1, -0
The threshold frequency for a certain metal is 3.3 × 101410^{14} Hz. If light of frequency 8.2 × 101410^{14} Hz is incident on the metal, predict the cut off voltage for the photoelectric emission.
Solution:  
According to Einstein’s relation, hu = hu0+12mvmax2hu_0 + \frac{1}{2} m v_{max}^2
Maximum kinetic energy of emitted electron, 12mvmax2\frac{1}{2} m v^2_{max} = h (v - v0v_0)
= 6.63 × 10−3410^{-34} (8.2 × 101410^{14} - 3.3 × 101410^{14})
= 32.49 × 10−2010^{-20} Jouke = 2 eV
∴ Cut off potential for emitted electron will be 2 volt.
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