When a certain photosensitive surface is illuminated with monochromatic light of wavelength , the maximum velocity of the emitted photoelectrons is v . When the same surface is illuminated by light of wavelength 3 4 , the maximum velocity of the photoelectrons is observed to be 2v . The threshold wavelength for this surface is:

Step-by-step solution

Using Einstein's photoelectric equation, the maximum kinetic energy is K_ max = hc - hc ₀ , where ₀ is the threshold wavelength. For the first case, when the wavelength is and maximum velocity is v : 1 2 mv^2 = hc - hc ₀ For the second case, when the wavelength is 3 4 and maximum velocity is 2v : 1 2 m(2v)^2 = hc ( 3 4 ) - hc ₀ 4 ( 1 2 mv^2 ) = 4hc 3 - hc ₀ Substituting the value of 1 2 mv^2 from the first equation into the second equation: 4 ( hc - hc ₀ ) = 4hc 3 - hc ₀ Dividing the entire equation by hc : 4 - 4 ₀