QJEEM23Ph68ce: JEE Main 2023 Physics - Class 12 JEEM23Ph68ce Solution: The difference between threshold wavelengths for two metal…

Threshold wavelength ( $\lambda_{threshold}$ ) is the maximum wavelength of light required to remove an electron from a metal surface, i.e., to overcome the work function ( $\phi$ ). The relationship between work function and threshold wavelength is given by:

Where:

• $\phi$ is the work function in electron-volts (eV)
• $h$ is the Planck's constant
• $c$ is the speed of light
• $\lambda_{threshold}$ is the threshold wavelength

In this problem, we are given the product $hc = 1242 ~\mathrm{eV}\,\mathrm{nm}$ , and the work functions $\phi_{A} = 9 ~\mathrm{eV}$ and $\phi_{B} = 4.5 ~\mathrm{eV}$ . Let's calculate the threshold wavelengths for metal surfaces $\mathrm{A}$ and $\mathrm{B}$ :

For metal surface $\mathrm{A}$ :

For metal surface $\mathrm{B}$ :

Now let's calculate the difference between the threshold wavelengths ( $\Delta\lambda$ ):

By calculating the difference, we get:

So, the difference between the threshold wavelengths for the two metal surfaces is $\boxed{138}\,\mathrm{nm}$ .