Question

When photon of energy $4.0\mathrm{eV}$ strikes the surface of a metal $\mathrm{A}$, the ejected photoelectrons have maximum kinetic energy ${\mathrm{T}}_{\mathrm{A}}\mathrm{eV}$ end de-Broglie wavelength ${\mathrm{\lambda }}_{\mathrm{A}}$. The maximum kinetic energy of photoelectrons liberated from another metal $\mathrm{B}$ by photon of energy $4.50\mathrm{eV}$ is ${\mathrm{T}}_{\mathrm{B}}=({\mathrm{T}}_{\mathrm{A}}-1.5)\mathrm{eV}$. If the de-Broglie wavelength of these photoelectrons ${\mathrm{\lambda }}_{\mathrm{B}}=2{\mathrm{\lambda }}_{\mathrm{A}}$, then the work function of metal $\mathrm{B}$ is

• A. $3\mathrm{eV}$
• B. $1.5\mathrm{eV}$
• C. $2\mathrm{eV}$
• D. $4\mathrm{eV}$

Answer 1

The correct option is D

$4\mathrm{eV}$

Work function: The work function of a photo metal is the minimum energy required to just liberate an electron from the metal surface.

• During the collision of the photons with the electrons, the electron absorbs the energy of the photon completely.
• The relation between kinetic energy (T) and its wavelength ($\lambda$) are related as:

Explanation of the correct answer:

The correct answer is option D:

Using de Broglie's equation,

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{{\mathrm{m}}^2{\mathrm{v}}^2}=\frac{\mathrm{h}}{\sqrt{2\mathrm{mT}}} \\ \mathrm{Given}\mathrm{that}{\mathrm{\lambda }}_{\mathrm{B}}=2{\mathrm{\lambda }}_{\mathrm{A}}\mathrm{and}{\mathrm{T}}_{\mathrm{B}}={\mathrm{T}}_{\mathrm{A}}-1.5 \end{gathered}$$

Now let us find out ${\mathrm{T}}_{\mathrm{B}}$

$$\begin{gathered} \Rightarrow {\mathrm{\lambda }}_{\mathrm{B}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{T}}_{\mathrm{B}}\mathrm{m}}}=2.\frac{\mathrm{h}}{\sqrt{2{\mathrm{T}}_{\mathrm{A}}\mathrm{m}}} \\ \Rightarrow \frac{\mathrm{h}}{\sqrt{2{\mathrm{T}}_{\mathrm{B}}\mathrm{m}}}=2.\frac{\mathrm{h}}{\sqrt{2{\mathrm{T}}_{\mathrm{A}}\mathrm{m}}} \\ \Rightarrow {\mathrm{T}}_{\mathrm{A}}=4{\mathrm{T}}_{\mathrm{B}} \\ \mathrm{Now},{\mathrm{T}}_{\mathrm{B}}=4{\mathrm{T}}_{\mathrm{B}}-1.5 \\ \Rightarrow {\mathrm{T}}_{\mathrm{B}}=0.5\mathrm{eV} \end{gathered}$$

Now for Metal B,

$\mathrm{Incident}\mathrm{energy}\left(\mathrm{h\nu }\right)=\mathrm{K}.\mathrm{E}\mathrm{of}\mathrm{electron}+\mathrm{\varphi }$ (Work Function)

$$\begin{gathered} 4.5=0.5+\mathrm{\varphi } \\ \Rightarrow \mathrm{\varphi }=4\mathrm{eV} \end{gathered}$$