In a photoelectric effect experiment the cathode metal is exposed to light of wavelength 600 ~nm . The maximum kinetic energy of the ejected electron doubles when light of wavelength 400 ~nm is used. The work function of the cathode metal is approximately. [Use h=6.63 10⁻³⁴ ~J - s , c=3 10^8 ~m / s ]

Step-by-step solution

Wavelength of light, ₁=600 ~nm If the work-function is and the kinetic energy of electron is E_K= 1 2 m v^2 , then we can write, Also, when the wavelength, ₂=400 ~nm , Kinetic energy of electron is 2 E_K=m v^2 Multiplying Eq. (i) by 2 , we get By subtracting Eq (iii) from Eq. (ii), we get aligned & h c ₂ - 2 h c ₁ =m v^2+ -m v^2-2 & =h c ( 2 ₁ - 1 ₂ ) &=6.626 10⁻³⁴ 3 10^8 ( 2 600 - 1 400 ) 10^9 &=6.626 10⁻³⁴ 3 10^8 ( 1 300 - 1 400 ) 10^9 & =1.02 eV aligned