Question
If work function of caesium metal is 2.14 eV , then calculate its threshold frequency.
✓
Answer
Work function
$\phi=h v_0$
$2.14 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} v_0$
$v_0=\frac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}$
$\begin{array}{l}=0.516 \times 10^{15} \\ =5.16 \times 10^{14} Hz\end{array}$
$\phi=h v_0$
$2.14 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} v_0$
$v_0=\frac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}$
$\begin{array}{l}=0.516 \times 10^{15} \\ =5.16 \times 10^{14} Hz\end{array}$
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