CUET Preparation Today
For a photosensitive surface, work function is $3.3 × 10^{-19} J$. Taking Planck's constant to be $6.6 × 10^{-34} Js$, the threshold frequency would be |
$6 × 10^{13} Hz$ $2 × 10^{12} Hz$ $5 × 10^{14} Hz$ $1.8 × 10^{12} Hz$ |
$5 × 10^{14} Hz$ |
The correct answer is Option (3) → $5 × 10^{14} Hz$ |
