Answer:
The correct answer is option B.
Explanation:
[tex]\phi =\frac{h}{\nu ^o}=\frac{hc}{\lambda }[/tex]
[tex]\phi [/tex] = work function = energy of photon
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\nu ^0[/tex] = frequency of the metal
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] =longest wavelength of the radiation
[tex]\phi = 3.68\times 10^{-19} J[/tex]
Now put all the given values in the above formula, we get the energy of the photons.
[tex]\lambda =\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{3.68\times 10^{-19} J}[/tex]
[tex]\lambda =5.4016\times 10^{-7}m=540.16 nm[/tex]
[tex]1 m = 10^{9} nm[/tex]
540.16 nm is the wavelength of radiation.
As we know that higher the energy lower will be the wavelength.
Here, wavelength of 550 nm will have lower energy than 540.16 nm but wavelength of 500 nm will have higher energy than 540.16 nm. So, wavelength of 500 nm will be efficient to excite photo-electrons.