Question

The work function for two metals are 9eV and 4.5eV. Find the approx. difference between their threshold wavelength. (use hc = 1240 eV - nm)

• A. 138 nm
• B. 130 nm
• C. 112 nm
• D. 145 nm

Answer 1

The Correct Option is A

To solve the problem, we need to find the difference between the threshold wavelengths of two metals with given work functions. The work function, \(\phi\), is the minimum energy required to eject an electron from the surface of a metal. It is related to the threshold wavelength, \(\lambda\), by the formula:

\(\phi = \frac{hc}{\lambda}\)

where:

• \(h\) is Planck's constant,
• \(c\) is the speed of light,
• \(\lambda\) is the threshold wavelength.

The constant \(hc\) is provided as 1240 eV⋅nm.

Let's calculate the threshold wavelengths for each metal and find the difference.

For Metal 1 (Work function = 9 eV):

\(\lambda_1 = \frac{hc}{\phi_1} = \frac{1240 \text{ eV⋅nm}}{9 \text{ eV}} = 137.78 \text{ nm}\)

For Metal 2 (Work function = 4.5 eV):

\(\lambda_2 = \frac{hc}{\phi_2} = \frac{1240 \text{ eV⋅nm}}{4.5 \text{ eV}} = 275.56 \text{ nm}\)

Calculating the Difference:

\(\Delta \lambda = \lambda_2 - \lambda_1 = 275.56 \text{ nm} - 137.78 \text{ nm} = 137.78 \text{ nm}\)

Thus, the approximate difference between their threshold wavelengths is 138 nm.

The correct answer is therefore: 138 nm.