# A) Calculate the wavelength (in nm) of light with energy 1.63*10^-20 J per photon? b)For light of wavelength 410 nm, calculate the number of photons per joule?

## c)Determine the binding energy (in $e V$) of a metal if the kinetic energy possessed by an ejected electron (using one of the photons with wavelength $410 n m$) is $3.33 \cdot {10}^{-} 19 J$?

Jul 19, 2018

a. $\lambda = 1.22 \times {10}^{4} \textcolor{w h i t e}{l} n m$
b. E("photon") = 4.85 xx 10^(-19) color(white)(l) "J"
c. $\phi = 1.52 \times {10}^{- 19} \textcolor{w h i t e}{l} \text{J}$

#### Explanation:

a)
By the "Planck-Einstein Relation"

$E = h \cdot f = h \cdot \frac{c}{\lambda}$

where $\lambda$ the wavelength photon in question, $E$ its energy, $h$ the Planck's constant , and $c$ the speed of light.

$c = 3.00 \times {10}^{8} \textcolor{w h i t e}{l} m \cdot {s}^{- 1} = 3.00 \times {10}^{17} \textcolor{w h i t e}{l} \textcolor{p u r p \le}{n m} \cdot {s}^{- 1}$

$\lambda = h \cdot \frac{c}{E}$
$\textcolor{w h i t e}{\lambda} = \left(6.63 \times {10}^{- 34} \textcolor{w h i t e}{l} \text{J" * s * 3.00 xx 10^(17) color(white)(l) color(purple)(nm) * s^(-1)) /(1.63 xx 10^(-20) color(white)(l) "J}\right)$
$\textcolor{w h i t e}{\lambda} = 1.22 \times {10}^{4} \textcolor{w h i t e}{l} \textcolor{p u r p \le}{n m}$

b)
$E = h \cdot \frac{c}{\lambda}$
color(white)(E)= (6.63 xx 10^(-34) color(white)(l) "J" * s * 3.00 xx 10^(17) color(white)(l) color(purple)(nm) * s^(-1))/(410 color(white)(l) color(purple)(nm)
$\textcolor{w h i t e}{E} = 4.85 \times {10}^{- 19} \textcolor{w h i t e}{l} \text{J}$

c)
The $410 \textcolor{w h i t e}{l} n m$ photons are absorbed upon incidence with all their energy converted to the mechanical energy of the electrons emitted from the metal surface.

Imagine sending a cart uphill a plateau on a frictionless slope with a sufficiently-large initial velocity. The cart would gradually decelerate as it moves upslope, and continue moving at a constant velocity (lower than the initial velocity at the bottom of the slope) after reaching the flat top. The difference between the cart's kinetic energy at the bottom of the hill and that at the top of the hill would equal the potential energy it has overcome in this process.

So is the case with bombarding metals with high-energy photons. The kinetic energy of each electron (the cart) leaving the metal surface (the pit) is equal to the difference between the energy per incident photon (the kinetic energy at the bottom of the slope) and the work function $\phi$. The work function is the minimum "binding energy" that needs to be supplied to overcome the metal's attraction on the electron and remove the electron from the electrostatic potential well.

In other words,

"KE"("electron") = E("incident photon") - phi("metal")

Therefore, with reference to the value of $E \left(\text{incident photon}\right)$ deduced in b),

$\phi \left(\text{metal") = E("incident photon") - "KE"("electron}\right)$
color(white)(phi("metal")) = 4.85 xx 10^(-19) color(white)(l) "J" - 3.33 xx 10^(-19) color(white)(l) "J"
color(white)(phi("metal")) = 1.52 xx 10^(-19) color(white)(l) "J"