Question

The work function for caesium atom s 1.9 eV. Calculate the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron. (1 eV = 1.602 * 10 ^ - 19 )​

Answer 1

Answer:

Velocity is 4000m/s

KE is 50J

Explanation:

Use E=hc/lamda to calculate Energy incident

KE+W=E incident

Answer 2

Answer:

Threshold frequency of the radiation is $4.59*10^{14}$ Hz

Kinetic energy and velocity of the ejected photoelectron is $9.32*10^{-20}$ J and $4.53*10^{5}$ m/s.

Explanation:

Work function $hv_{0}$ = 1.9 eV

                             = $1.9*1.602*10^{-19}$ J

                             = $3.044*10^{-19}$ J

where $h$ = planks constant = $6.626*10^{-34}$ Js

          $v_{o}$ = threshold frequency

∴ $3.044*10^{-19}$ = $6.626*10^{-34}$*$v_{o}$

$v_{o}$ = $\frac{3.044*10^{-19}}{6.626*10^{-34}}$

   = $4.59*10^{14}$ Hz

Kinetic energy of the ejected electron = $hv-hv_{0}$

Given wavelenght ∧ = 500 nm = $500*10^{-9}$ m

$v$ = c/∧

  = $\frac{3*10^{8} }{500*10^{-9} }$

  = $6*10^{14}$ Hz

kinetic energy = $(6.626*10^{-34}*6*10^{14})-(3.044*10^{-19})$

                        = $9.32*10^{-20}$ J

kinetic energy = $\frac{1}{2}$ m $u^{2}$

u = velocity of ejected photoelectron

$9.32*10^{-20}$ = $\frac{1}{2}$ * $9.1*10^{-31}$*$u^{2}$

$u^{2}$ = $2.05*10^{11}$ m/s

$u$ = $4.53*10^{5}$ m/s