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(C) The threshold frequency of the metal is given as $f_0 = 15 	imes 10^{14} \, Hz$. The frequency of the incident light with wavelength $\lambda = 6

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will not be emitted.

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(C) The threshold frequency of the metal is given as $f_0 = 15 	imes 10^{14} \, Hz$. The frequency of the incident light with wavelength $\lambda = 6000 \, 	ext{Å} = 6000 	imes 10^{-10} \, m$ is calculated as: $f = \frac{c}{\lambda} = \frac{3 	imes 10^8 \, m/s}{6000 	imes 10^{-10} \, m} = \frac{3 	imes 10^8}{6 	imes 10^{-7}} \, Hz = 0.5 	imes 10^{15} \, Hz = 5 	imes 10^{14} \, Hz$. Since the frequency of the incident light $(f = 5 	imes 10^{14} \, Hz)$ is less than the threshold frequency $(f_0 = 15 	imes 10^{14} \, Hz)$, the energy of the incident photons is insufficient to overcome the work function of the metal. Therefore, no photoelectric emission takes place.

Threshold frequency for a metal is $15 \times 10^{14} \, Hz$. The light of wavelength $6000 \, \text{Å}$ falls on the metal surface. Then photoelectrons [velocity of light in air, $c = 3 \times 10^8 \, m/s$]:

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