The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emitting photo electrons with a kinetic energy that is given by the difference betweent he photon enerzy and the work function of the metal historically, this effect was very confusing. for most wave phenon tand ight was wel known to behave as a wave phenomenon·ore nerease the increasing the amplitude of the wave. in the case of light, this means increase- there are simple more electrons emitted increasing the ententisty. however, with increased intensity, the kinetic energy of the photoelections emited does not albert einstein (1879-1955) offered a solution to this perplecing problem in hi 190s einen die erzeugung und verwandlung des lichtes betrefflenden heuristischen gesichtspunkr. einstein proposed the following relationship: kinetic- hw-e work function energy of an emitted photoelectron, hw is the energy of a photon of frequency v and ework function is the wrok function for the metal used in the ion is the energy required to remove an electron from the surface of the metal experiment. the work functi references einstein, albert (1905). the work function of cesium is 2.1 ev. what is this energy in? hint: 1 ev-1,602 x 10-19) annalen der physik 17(6 132-148 (1905) 34x 1019) 21x 1019 64x 1019 43x 1019 19, 0 19, 0


The ionization energy increases
The answer for this is A.
Ionization energy increases.
I believe that the answer is B.

your answer to your question is B

I think answer of this question is D

The ionisation energy increases.


This is because the force of attraction between the electrons and the positive nucleus will increase

The energy of the electron in third energy level.

Explanation: The energy of the electrons depend on the value of n, i.e the value of principal quantum number which tells the shell of the electron.


E= energy

Z= atomic number

n= principal quantum number

Thus E_3 represents the energy of the electron which is present in n=3.

3.4\cdot 10^{-19} J


In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

1 eV = 1.6 \cdot 10^{-19} J

In our problem, the work function of cesium is

E=2.1 eV

so, we can convert it into Joules by using the following proportion:

1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J

Do you know the answer?

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