Question 107: Dual Nature of Matter and Radiation

Question

The difference between threshold wavelengths for two metal surfaces $$\mathrm{A}$$ and $$\mathrm{B}$$ having work function $$\phi_{A}=9 ~\mathrm{eV}$$ and $$\phi_{B}=4 \cdot 5 ~\mathrm{eV}$$ in $$\mathrm{nm}$$ is:
$$\{$$ Given, hc $$=1242 ~\mathrm{eV} \mathrm{nm}\}$$

Accepted Answer

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:
$$\phi = \frac{hc}{\lambda_{threshold}}$$
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 $$\mathrm{A}$$:
$$\lambda_{A} = \frac{1242}{\phi_{A}} = \frac{1242}{9}$$
For metal surface $$\mathrm{B}$$:
$$\lambda_{B} = \frac{1242}{\phi_{B}} = \frac{1242}{4.5}$$
Now let's calculate the difference between the threshold wavelengths ($$\Delta\lambda$$):
$$\Delta\lambda = \lambda_{B} - \lambda_{A} = \frac{1242}{4.5} - \frac{1242}{9}$$
By calculating the difference, we get:
$$\Delta\lambda = 276 - 138 = 138 ~\mathrm{nm}$$
So, the difference between the threshold wavelengths for the two metal surfaces is $$\boxed{138}\,\mathrm{nm}$$.

Answer

264

Answer

138

Answer

540

Answer

276

Dual Nature of Matter and Radiation

The difference between threshold wavelengths for two metal surfaces $\mathrm{A} and \mathrm{B} having work function \phi_{A}=9 ~\mathrm{eV} and \phi_{B}=4 \cdot 5 ~\mathrm{eV} in \mathrm{nm} is: \{ Given, hc =1242 ~\mathrm{eV} \mathrm{nm}\}

Choose the correct answer: