Question

Electrons in hydrogen-like atoms (Z=3) make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Calculate the work function of the metal, and the stopping potential (in V) for the photoelectrons ejected by the longer wavelength. (\[\text{Rydberg constant} = 1.094 \times {10^7}{m^{ - 1}}\])

Answer 1

Hint: In this question, we will use Bihr’s concept and photoelectric effect equation firstly we will find the energy due to transition between the orbits and then using this energy as primary energy for photoelectric effect we will solve for stopping potential and work function of the targeted metal.

Formula Used:
The photoelectric equation is given as,
$E = \phi + K.E.$
where E is the energy of a falling photon. K.E. is the maximum kinetic energy of the photoelectrons.
The energy of nth orbit of hydrogen like atom is given by,
\[{E_n} = \dfrac{{ - 2{\pi ^2}m{e^4}{Z^2}}}{{{n^2}{h^2}}} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}\]
where m=mass of electron, e=charge of electron, Z=atomic number of hydrogens like atom, h =planck's constant, c=speed of light.

Complete step by step solution:
The energy of the different energy states in the case of the hydrogen atom are given by the expression (called Bohr formula):
\[{E_n} = \dfrac{{ - 2{\pi ^2}m{e^4}{Z^2}}}{{{n^2}{h^2}}} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}} \\ \]
By putting the values in the above equation, we get;
$\Rightarrow E = \dfrac{{hc}}{\lambda } = \dfrac{{2{\pi ^2}m{e^4}{Z^2}}}{{{h^2}}}\left( {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right)$
$\Rightarrow E = - 13.6 \times {Z^2}\left( {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right) \\ $----(4)
We can easily find out the energy of the respective radiation by using equation 4.
$\Rightarrow {E_1} = 13.6 \times 9 \times \left( {\dfrac{1}{{16}} - \dfrac{1}{{25}}} \right) = 2.75\,eV \\ $
$\Rightarrow {E_2} = 13.6 \times 9 \times \left( {\dfrac{1}{9} - \dfrac{1}{{16}}} \right) = 5.95\,eV$
Here we see that $\lambda \varpropto \dfrac{1}{E}$. So, 3.95 eV is the work function for ${E_1}$. So here we have discussed stopping potential and work function.

Electrons are released as a result of the photoelectric effect when electromagnetic radiation, such as light, strikes a substance. The value of the work function, $\phi $, which is dependent on the metal, is the minimal amount of energy needed to cause photoemission of electrons from a metal surface. We can formulate as $E = \phi + e\,V_o$ where E is the energy of a falling photon. $V_o$ is the stopping potential.

The lowest negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximum kinetic energy of the electrons is equal to the stopping voltage. Let’s find out the stopping potential of the material. So,
$E_2 = \phi + e\,{V_{o2}}$
Here, ${V_{o2}}=3.95\,V$
$\Rightarrow \phi =E_2 - e\,V_o2 = 5.95 - 3.95 = 2\,eV$
From the above discussion we can work out that the function of the material is 2.15eV.
Now by using the same $E = \phi + e\,V_o$ formula we can find out the stopping potential of the radiation with higher wavelength.
$e\,{V_{o1}} = E_1 - \phi = 2.75 - 2 = 0.75\,eV$

Hence, the values of the work function of the metal, and the stopping potential are 0.75 eV and 2 eV.

Notes: The Lyman series (from orbit 1), Balmer series (from orbit 2), Paschen series, also known as the Bohr series (from orbit 3), Brackett series (from orbit 4), Pfund series (from orbit 5) and Humphreys series are the six series that make up the majority of the line spectrum of hydrogen, as we can recall (from orbit 6). Although there are series above them, they are not as frequently utilised.