Formula of energy if photon E = hf

Where, h - Planck constant = 6.63* 10^{-34} J and f - frequency of waves

E = \frac{hc}{\lambda} \because f = \frac{\text{ speed of light (c)}}{\text{ Wavelength } \lambda} .

Given that

Violet colour wavelength \lambda _v = 390 nm = 390 * 10^{-9} m

Yellow-green colour wavelength \lambda _{yg} = 550 * 10^{-9} m

Red colour wavelength \lambda _{r} = 760 nm = 760 * 10^{-9} m

(a)

(i) Incident photon energy for violet light

E = \frac{hc}{\lambda}

E = \frac{6.63 * 10^{-34} * 3*10^8}{\lambda}

E = \frac{1.989 * 10^{-25} Jm}{\lambda}

Wavelength, \lambda _v = 390 nm

E = \frac{1.989 * 10^{-25} Jm}{390 * 10^{-9} m}

E = 5.10 * 10^{-19} J

Converting joules to electron volts

E = \frac{ 5.10 * 10^{-19}}{1.6*10^{-19}} J

E = 3.19 eV

(ii) Incident photon energy for yellow-green light

E = \frac{1.989 * 10^{-25} Jm}{550 * 10^{-9} m}

E = 3.62 * 10^{-19} J

E = 2.26 eV

(iii) Incident photon energy for red light

E = \frac{1.989 * 10^{-25} Jm}{ 760 * 10^{-9} m}

E = 2.62 * 10^{-19} J

E = 1.64 eV

(b)

For a photoelectric device to operate, we need incident light energy E to be equal to or greater than the work function (\phi) of the material.

E = 3.19 eV is greater than work function of photosensitive materials Na \phi = 2.75 , K \phi = 2.74 and Cs \phi = 2.14

Hence, the photoelectric device will operate with violet light photosensitive materials Na, K and Cs

E = 2.26 eV > work function of photosensitive materials Cs ( \phi = 2.14 )

Hence, the photoelectric device will operate with yellow-green light photosensitive material Cs

E = 1.64 eV < work function of photosensitive materials

Thus, the photoelectric device will not operate with red light.