Question

Light with a wavelength of (465 A) nm is incident on a metal surface with a work function of (1.25 (0.1)(B)) eV. Find the maximum kinetic energy of the emitted photoelectrons. Give your answer in eV and with 3 significant figures.

Answer 1

Answer:

The maximum kinetic energy of photoelectrons is 3.02 eV

Explanation:

Given:

Wavelength of light $\lambda = 465 \times 10^{-9}$ m

Work function $\phi = 1.25$ eV

From the theory of photoelectric effect,

   $KE_{max} = hf - \phi$

Where $f =$ frequency of light, $\phi$ = work function its depends on nature of metal.

$hf = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8} }{465 \times 10^{-9} }$

     $= 0.0427 \times 10^{-17}$

Now we convert energy in terms of electrovolt,

$hf = \frac{0.0427 \times 10^{-17} }{1.6 \times 10^{-19} }$

$hf = 4.27$ eV

Put above value,

$KE_{max} = 4.27 -1.25$

$KE_{max} = 3.02$ eV

Therefore, the maximum kinetic energy of photoelectrons is 3.02 eV