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