Question

When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have a maximum kinetic energy of T_A eV and de Broglie wavelength λ_A. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is T_B=(T_A−1.5)eV. If the de Broglie wavelength of these photoelectrons is λ_B=2λ_A then

• A. the work function of A is 2.25 eV
• B. the work function of B is 4.20 eV
• C. T_A = 2.00 eV
• D. T_B = 2.75 eV

Answer 1

Answer: (A), (B), (C)

T_A = 2.00 eV

Step 1: De Broglie wavelength

1. When studying quantum mechanics, the de Broglie wavelength is a key idea.
2. De Broglie wavelength is the wavelength that is connected to an item in relation to its momentum and mass.
3. Typically, a particle's force is inversely proportional to its de Broglie wavelength.

Step 2: Given data

1. The ejected photoelectrons have a de Broglie wavelength of A and maximal kinetic energy ${\mathrm{T}}_{\mathrm{A}}\mathrm{eV}$.
2. Photons with an energy of $4.70\mathrm{eV}$ can release photoelectrons with maximum kinetic energy ${\mathrm{T}}_{\mathrm{B}}=({\mathrm{T}}_{\mathrm{A}}-1.5)\mathrm{eV}$ from another metal B.
3. ${\mathrm{\lambda }}_{\mathrm{B}}=2{\mathrm{\lambda }}_{\mathrm{A}}$

Step 3: Formula used

De Broglie wavelength, $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{P}}$​ and kinetic energy,$$\begin{gathered} \mathrm{T}=\frac{{\mathrm{P}}^2}{2\mathrm{m}} \\ =\frac{{\mathrm{h}}^2}{2\mathrm{m}{{\mathrm{\lambda }}^2}_{\mathrm{A}}} \end{gathered}$$

Thus, $\frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{B}}}=\frac{{{\mathrm{P}}^2}_{\mathrm{A}}}{{{\mathrm{P}}^2}_{\mathrm{B}}}=\frac{{{\mathrm{\lambda }}^2}_{\mathrm{B}}}{{{\mathrm{\lambda }}^2}_{\mathrm{A}}}=4\mathrm{as}{\mathrm{\lambda }}_{\mathrm{B}}=2{\mathrm{\lambda }}_{\mathrm{A}}$

$$\begin{gathered} {\mathrm{T}}_{\mathrm{B}}=({\mathrm{T}}_{\mathrm{A}}-1.5) \\ =4{\mathrm{T}}_{\mathrm{B}}-1.5 \\ {\mathrm{T}}_{\mathrm{B}}=0.5\mathrm{eV} \end{gathered}$$

$$\begin{gathered} {\mathrm{T}}_{\mathrm{A}}=4\times 0.5 \\ =2.00\mathrm{eV} \end{gathered}$$

From photo-electric effect:

The work function of A$$\begin{gathered} =4.25-{\mathrm{T}}_{\mathrm{A}} \\ =4.25-2.00 \\ =2.25\mathrm{eV} \end{gathered}$$

The work function of B$$\begin{gathered} =4.70-{\mathrm{T}}_{\mathrm{B}} \\ =4.70-0.5 \\ =4.20\mathrm{eV} \end{gathered}$$

The explanation for the correct options A, B, and C:

From the above analysis, it is found that

In the case of option A, the work function of A$=2.25\mathrm{eV}$

In the case of option B, the work function of B$=4.20\mathrm{eV}$

In the case of option C, ${\mathrm{T}}_{\mathrm{A}}=2.00\mathrm{eV}$

The explanation of the incorrect option D:

$T_B=0.5eV$ but in the option, it is given to $2.75\mathrm{eV}$ which is incorrect.

Therefore the correct options are A, B, and C.