Question

When a photon of energy $4.25eV$ strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy ${{T}_{A}}eV$ and de-Broglie wavelength ${{\lambda }_{A}}$ .The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy $4.70eV$ is ${{T}_{B}}=\left( {{T}_{A}}-1.5 \right)eV$. If the de Broglie n wavelength of these photoelectrons is ${{\lambda }_{B}}=2{{\lambda }_{A}}$, then
A. Work function of A is $2.25eV$
B. Work function of B is $4.20eV$
C. ${{T}_{A}}=2.00eV$
D. ${{T}_{B}}=2.75eV$

Answer 1

According to De-Broglie;
$\lambda =\dfrac{h}{p}$
where,
$\lambda$ is the de- Broglie’s wavelength
h is the Planck's constant
p is the momentum imparted by the wave.

Work function of a metal $({{\phi }_{A}})$ is given by subtracting the kinetic energy of the photo electrons $(K)$ ejected from the metal surface from the total energy $(E)$ carried by the radiation incident or striking the metal surface.
i.e; $(\phi )=E-K$

Complete answer:
We can obtain the following information from the given question.
Total energy of the photon striking the metal plate A,${{E}_{A}}$: $4.25eV$
Kinetic Energy of the ejected photoelectrons from metal plate A: ${{T}_{A}}eV$
De-Broglie’s wavelength of the photoelectrons ejected from metal plate A: ${{\lambda }_{A}}$
Total energy of the photon striking the metal plate B, ${{E}_{B}}$: $4.70eV$
Kinetic Energy of the ejected photoelectrons from metal plate B: ${{T}_{B}}eV$
De-Broglie’s wavelength of the photoelectrons ejected from metal plate B: ${{\lambda }_{B}}$
According to De-Broglie;
$\lambda =\dfrac{h}{p}$
where h is the Planck's constant
$\Rightarrow$ $p=\dfrac{h}{\lambda }$
Also,
Kinetic energy, K = $\dfrac{{{p}^{2}}}{2m}$
where m is the mass of the ejected photoelectron.
Therefore,
$K={{(\dfrac{h}{\lambda })}^{2}}\times \dfrac{1}{2m}$
On taking the ratio of kinetic energy of photoelectrons ejected from metal plate A $({{K}_{A}})$to the kinetic energy of the photoelectrons ejected from metal plate B$({{K}_{B}})$, we obtain:
$\dfrac{{{K}_{A}}}{{{K}_{B}}}=(\dfrac{{{h}^{2}}}{2m\lambda _{A}^{2}})\times (\dfrac{2m\lambda _{B}^{2}}{{{h}^{2}}})$
$\dfrac{{{K}_{A}}}{{{K}_{B}}}=(\dfrac{\lambda _{B}^{2}}{\lambda _{A}^{2}})$
\Rightarrow $$$\dfrac{{{T}_{A}}}{{{T}_{B}}}=(\dfrac{\lambda _{B}^{2}}{\lambda _{A}^{2}})$$ It is given in the question that there is a relation between {{T}{A}}$and${{T}{B}}$as well as between${{\lambda }{A}}and $${{\lambda }_{B}}$$. i.e; {{T}{B}}=({{T}{A}}-1.5)eV And $${{\lambda }_{B}}=2{{\lambda }_{A}}$$ Therefore, $$\dfrac{{{T}_{A}}}{{{T}_{A}}-1.5}=(\dfrac{{{2}^{2}}\lambda _{A}^{2}}{\lambda _{A}^{2}})$$ $${{T}_{A}}=4({{T}_{A}}-1.5)$$ After rearranging the terms: $$3{{T}_{A}}=6.0$$ $${{T}_{A}}=2.00eV$$ And since, {{T}{B}}=({{T}{A}}-1.5)eV$${{T}{B}}=(2.00-1.5)eV$${{T}{B}}=0.50eV$From the photoelectric effect: Work function of the metal pate A,$({{\phi }{A}})$=${{E}{A}}-{{T}{A}}$${{\phi }{A}}=(4.25-2.00)eV$${{\phi }{A}}=2.25eV$

Similarly, work function of the metal pate B,
$({{\phi }_{B}})$=${{E}_{B}}-{{T}_{B}}$
${{\phi }_{A}}=(4.70-0.50)eV$
${{\phi }_{A}}=4.20eV$

Hence,
Work function of A is $2.25eV$
Work function of B is $4.20eV$
${{T}_{A}}=2.00eV$
${{T}_{B}}=0.50eV$

Thus, the options A), B) and C) are correct and only the option D) ${{T}_{B}}=2.75eV$ is incorrect

Note:
It is to be noted that the work function of a metal plate is one of its fundamental electronic properties and solely depends on the nature of the metal.
It tells us about the minimum energy requirement for a metal to eject a single photo electron when an external radiation strikes its surface.
Also, it is to be noted that throughout the process of photoelectric emission, the total energy remains conserved as the difference in energy between the striking radiation and in the ejected photoelectrons is utilized for the purpose of ejecting the electrons from the metal surface and does not just disappear.