physics - The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellowgreen colour and about 760 nm for red colour.

The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellowgreen colour and about 760 nm for red colour.

79 viewed last edited 2 years ago

physics dual nature of radiation and matter ncrt physics photoelectric effect and wave theory of light photon energy equation einstein's photoelectric equation: energy quantum of radiation the photoelectric effect kinetic energy of photoelectron photoelectric work function of a material photoelectric devices 12th grade

Anonym0us (Student, UG1 Grade)

(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum? (Take h = 6.63×10–34 J s and 1 eV = 1.6×10 –19J.)

(b) From which of the photosensitive materials with work functions listed in Table 11.1 and using the results of (i), (ii) and (iii) of (a), can you build a photoelectric device that operates with visible light?

Qalaxia Master Bot (last edited 2 years ago)

I found an answer from simple.wikipedia.org

Photoelectric effect - Simple English Wikipedia, the free encyclopedia

The photoelectric effect has helped physicists understand the quantum nature of light ... Not every electromagnetic wave will cause the photoelectric effect, only radiation of a ... The work function is a property of the metal and is not affected by the ... the cutoff frequency, then the emitted electron will have some kinetic energy.

For more information, see Photoelectric effect - Simple English Wikipedia, the free encyclopedia

Krishna (last edited 2 years ago) (Expert, Educator @Qalaxia)

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.