Question

a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7,calculate the maximum kinetic energy

Answer 1

Complete question:

a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7 m,calculate the maximum kinetic energy of the emitted photons.

Answer:

the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J

Explanation:

Given;

work function, Ф = 1.8 eV

wavelength of the light, λ = 3 x 10⁻⁷ m

The maximum kinetic energy of the emitted photons is calculated from photoelectric equation.

$E = K.E_{max} + \phi\\\\KE_{max} = E- \phi\\\\where;\\\\E \ is \ the \ energy \ of \ the \ incident \ light\\\\E = hf = h \frac{c}{\lambda} \\\\where;\\\\c \ is \ speed \ of \ light = 3 \times 10^8 \ m/s\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E = \frac{ (6.626 \times 10^{-34})\times ( 3 \times 10^8)}{3\times 10^{-7}} \\\\E = 6.626 \times 10^{-19} \ J$

$K.E_{max} = E - \phi\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ (1.8 \times 1.602 \times 10^{-19} \ J)\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ 2.884 \times 10^{-19} \ J\\\\K.E_{max} =3.742 \times 10^{-19} \ J$

Therefore, the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J