A monochromatic source of light operating at 200W emits $ 4 \times {10^{20}} $ photons per second. Find the wavelength of the light (in $ \times {10^{ - 7}}m $ )

Hint: Using Planck’s energy-frequency relation we have to find the energy of a single photon in terms of wavelength. Multiplying it with the total number of photons emitted per second gives us the total energy emitted per second. Since we are given the photons emitted per second, the energy will be equal to power.

Formula used
 $ {E_P} = \dfrac{{h \times c}}{\lambda } $ , where $ {E_p} $ the energy of a single photon is, $ \lambda $ is the wavelength of light in meters , $ c $ is the speed of light in free space and $ h $ is the Planck’s constant.
 $ P = \dfrac{E}{t} $ , $ E $ denotes power, $ E $ is represented by energy and $ t $ denotes time.
 $ {E_{Total}} = n \times {E_p} $ $ {E_{Total}} $ represents the total energy, $ n $ denotes the number of photons and $ {E_p} $ is the energy of photon.

Complete step-by-step answer:
We know from Planck’s energy-frequency relation that $ {E_P} = h\nu $ where $ {E_p} $ is the energy of photon (also known as photon energy), $ h $ is the Planck’s constant whose value is $ 6.626 \times {10^{ - 34}}J.s $ in SI units and $ \nu $ is the frequency of the wave. Also note that the frequency of a wave can be expressed in terms of its wavelength and speed. This is given by the equation $ \nu = \dfrac{c}{\lambda } $ .Here $ c $ is the speed of light ( $ 3 \times {10^8}m/s $ ) and $ \lambda $ is the wavelength. Substituting this equation in Planck’s energy-frequency relation we get
 $ {E_P} = \dfrac{{h \times c}}{\lambda } $ .
 $ {E_P} = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } $
Since $ t = 1s $ , according to the equation $ P = \dfrac{E}{t} $ we get $ P = E $ .Given $ P = 200W $ .
 $ P = E $
Using equation $ {E_{Total}} = n \times {E_p} $ , we get
 $ P = {E_{Total}} = n \times {E_p} $
Here the number of photons per second is given. $ n = 4 \times {10^{20}} $ .
 $ 200 = 4 \times {10^{20}} \times {E_p} $
Substituting $ {E_P} = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } $ .
 $ 200 = 4 \times {10^{20}} \times \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } $
 $ \lambda = \dfrac{{4 \times {{10}^{20}} \times 6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{200}} $
 $ \lambda = 4 \times {10^{ - 7}}m $

Note: When light of a certain minimum frequency hits the metal surface the electrons will be just ejected. This minimum frequency required is called threshold frequency $ {\nu _0} $ . .The value $ h{\nu _0} $ is called the work function. This work function varies according to each metal surface. The light falling on the metal surface therefore should be greater than the threshold frequency.
If electromagnetic radiation of frequency $ \nu $ which is greater than the threshold frequency falls on the metal surface the additional energy will be given to the electron as kinetic energy. This can be written in equation form as $ h\nu = h{\nu _0} + KE $ . Here KE represents the kinetic energy.