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(D) The threshold frequency is given as $
u_{0} = 15 	imes 10^{14} \,Hz$.The threshold wavelength $\lambda_{0}$ is calculated as $\lambda_{0} = \fra

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photoelectrons will not be emitted.

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(D) The threshold frequency is given as $
u_{0} = 15 	imes 10^{14} \,Hz$.The threshold wavelength $\lambda_{0}$ is calculated as $\lambda_{0} = \frac{c}{
u_{0}} = \frac{3 	imes 10^{8}}{15 	imes 10^{14}} = 0.2 	imes 10^{-6} \,m = 2000 	ext{ Å}$.The wavelength of incident light is $\lambda = 6000 	ext{ Å}$.For photoelectric emission to occur, the incident wavelength must be less than or equal to the threshold wavelength $(\lambda \leq \lambda_{0})$.Since $\lambda = 6000 	ext{ Å} > \lambda_{0} = 2000 	ext{ Å}$, the energy of the incident photons is less than the work function of the metal.Therefore, photoelectrons will not be emitted.

Threshold frequency for a metal is $15 \times 10^{14} \,Hz$. The light of wavelength $6000 \text{ Å}$ falls on the metal surface. Which one of the following statements is correct? [velocity of light, $c = 3 \times 10^{8} \,m/s$]

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