MCQ
The threshold frequency of a metal with work function $6.63\ eV$ is :
- A$16 \times 10^{15} Hz$
- B$16 \times 10^{12} Hz$
- C$1.6 \times 10^{12} Hz$
- D$1.6 \times 10^{15} Hz$
✓
Answer
$\phi_0=h v_0$
$6.63 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} v_0$
$v_0=\frac{1.6 \times 10^{-19}}{10^{-34}}$
$v_0=1.6 \times 10^{15} Hz$
$6.63 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} v_0$
$v_0=\frac{1.6 \times 10^{-19}}{10^{-34}}$
$v_0=1.6 \times 10^{15} Hz$
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