Answer
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Hint
The radiation of the source having the wavelength less than that of the given threshold wavelength will show a photoelectric effect. The wavelength of UV radiation is less than that of the infrared radiation. The wavelength of the sodium light lies between that of UV and IR radiation.
$ E = h\upsilon - h{\upsilon _0} $
Where $ E $ is the kinetic energy of the emitted particles (photons), $ h $ is the Planck's constant, $ \upsilon $ is the frequency and $ {\upsilon _0} $ is the threshold frequency,. $ h{\upsilon _0} $ is the work function.
Complete step by step solution
The condition for the photoelectric effect to take place is that the wavelength of the light should be less than the threshold or cut-off wavelength.
If in case the wavelength of the radiation is more than that of the threshold wavelength, then the photoelectric effect will not take place.
The wavelength of the ultraviolet radiation is less than that of the threshold wavelength i.e. $ 5,200A^\circ $ .
So, the photoelectric effect will take place and the photoelectrons will get emitted as $ {\lambda _{UV}} < 4000\,A^\circ $ .
The wavelength of the infrared radiation is more than that of the threshold wavelength $ 5,200A^\circ $ .
So, the photoelectric effect will not take place and the photoelectrons will not get emitted as $ {\lambda _{IR}} > 8000\,A^\circ $ .
$ \therefore $ The photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a 1-watt ultraviolet lamp.
Thus, the option (C) is correct.
Note
The other method to solve this problem is that the photoelectrons will be emitted when the energy of the incident light will be greater than that of the threshold energy. As the energy and the wavelength are inversely proportional to each other, the photoelectric effect will take place for the radiation with maximum energy.
The radiation of the source having the wavelength less than that of the given threshold wavelength will show a photoelectric effect. The wavelength of UV radiation is less than that of the infrared radiation. The wavelength of the sodium light lies between that of UV and IR radiation.
$ E = h\upsilon - h{\upsilon _0} $
Where $ E $ is the kinetic energy of the emitted particles (photons), $ h $ is the Planck's constant, $ \upsilon $ is the frequency and $ {\upsilon _0} $ is the threshold frequency,. $ h{\upsilon _0} $ is the work function.
Complete step by step solution
The condition for the photoelectric effect to take place is that the wavelength of the light should be less than the threshold or cut-off wavelength.
If in case the wavelength of the radiation is more than that of the threshold wavelength, then the photoelectric effect will not take place.
The wavelength of the ultraviolet radiation is less than that of the threshold wavelength i.e. $ 5,200A^\circ $ .
So, the photoelectric effect will take place and the photoelectrons will get emitted as $ {\lambda _{UV}} < 4000\,A^\circ $ .
The wavelength of the infrared radiation is more than that of the threshold wavelength $ 5,200A^\circ $ .
So, the photoelectric effect will not take place and the photoelectrons will not get emitted as $ {\lambda _{IR}} > 8000\,A^\circ $ .
$ \therefore $ The photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a 1-watt ultraviolet lamp.
Thus, the option (C) is correct.
Note
The other method to solve this problem is that the photoelectrons will be emitted when the energy of the incident light will be greater than that of the threshold energy. As the energy and the wavelength are inversely proportional to each other, the photoelectric effect will take place for the radiation with maximum energy.
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