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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
Questions With Solutions

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Question : 8
Total: 37
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut off voltage for the photoelectric emission.
Solution:  
According to Einstein’s relation, hu = hu0+
1
2
mvmax2
Maximum kinetic energy of emitted electron,
1
2
mvmax2 = h (v - v0)
= 6.63 × 10−34 (8.2 × 1014 - 3.3 × 1014)
= 32.49 × 10−20 Jouke = 2 eV
∴ Cut off potential for emitted electron will be 2 volt.
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