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Threshold frequency for a metal is $15 \times 10^{14} \mathrm{~Hz}$. The light of wavelength $6000 \AA$ falls on the metal surface. Then photoelectrons
[velocity of light in air, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ ]
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[velocity of light in air, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ ]
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will not be emitted
Frequency of light of wavelength $\lambda=6000 \mathrm{~A}^0$ is
$f=\frac{c}{\lambda}=\frac{3 \times 10^8}{6000 \times 10^{-10}} \mathrm{~Hz}=5 \times 10^{14} \mathrm{~Hz}$
which is less than the given threshold frequency. Hence, no photoelectric emission takes place.
$f=\frac{c}{\lambda}=\frac{3 \times 10^8}{6000 \times 10^{-10}} \mathrm{~Hz}=5 \times 10^{14} \mathrm{~Hz}$
which is less than the given threshold frequency. Hence, no photoelectric emission takes place.
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