## Photoelectric effect post-quiz 17-19 [ENDORSED]

Ritika Saranath 3I
Posts: 24
Joined: Fri Sep 25, 2015 3:00 am

### Photoelectric effect post-quiz 17-19

Hi!

Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1.
A. What is the kinetic energy of the ejected electron?
B. How much energy is required to remove an electron from one sodium atom?
C. What is the frequency of the incident light on the sodium metal surface?

Since we are given the work function of sodium in the unit "kJ.mol-1", can we simply re-write 150.6 kJ.mol-1 as 150600 J, and use this value to solve parts B and C? Or since the unit also includes "mol-1", do we have to do more dimensional analysis?

Silvia Huang 3D
Posts: 30
Joined: Fri Sep 25, 2015 3:00 am

### Re: Photoelectric effect post-quiz 17-19

a) $E_{k}=\frac{1}{2}mv^{2}=\frac{1}{2}(9.10956*10^{-31} kg)(6.61*10^{5} \frac{m}{s})^{2}= 1.99*10^{-19} J$

b) work energy is the energy required to remove one electron, however, this is given in moles and we want it for one sodium atom, so we divide by Avogadro's number to energy get per atom
$150.6 \frac{kJ}{mol}=\frac{150600}{6.02*10^{23}}= 2.50*10^{-19}J$

c) the energy of the incident line = work energy + kinetic energy
$E=1.99*10^{-19}J+2.50*10^{-19}J=4.49*10^{-19}J$

$\nu =\frac{E}{h}=\frac{(4.49*10^{-19}J)}{6.626*10^{-34}J*s}=6.78*10^{14} Hz$

hope this help :)

Helen Shi 1J
Posts: 78
Joined: Sat Jul 22, 2017 3:00 am

### Re: Photoelectric effect post-quiz 17-19

How did you get 2.5x10^-19 as the work energy?

Rachel N 1I
Posts: 48
Joined: Thu Jul 27, 2017 3:00 am

### Re: Photoelectric effect post-quiz 17-19

@Helen
In order to get the work energy, 150.6 kJ/mol must be converted to Joules (1000 kJ= 1J) and then multiplied to Avogadro's number, 6.022 x 10^23, as a conversion factor to cancel out the mol in the denominator.

RichardValdez1L
Posts: 59
Joined: Fri Apr 06, 2018 11:05 am

### Re: Photoelectric effect post-quiz 17-19

What is an incident line and why for the second part of c. we use the E=hv equation?

Emily 1E
Posts: 27
Joined: Fri Apr 06, 2018 11:03 am

### Re: Photoelectric effect post-quiz 17-19  [ENDORSED]

Incident light means the same thing as the photon used to eject the electron in this context, and E = hv is used to convert the energy of the incident light to the frequncy

805097738
Posts: 180
Joined: Wed Sep 18, 2019 12:20 am

### Re: Photoelectric effect post-quiz 17-19

[quote="Silvia Huang 3D"]a) $E_{k}=\frac{1}{2}mv^{2}=\frac{1}{2}(9.10956*10^{-31} kg)(6.61*10^{5} \frac{m}{s})^{2}= 1.99*10^{-19} J$

how did you find m and what is m?