Question

When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to

Answer 1

0.62 eV

Answer 2

Let

$K_1 = \frac{hc}{500} - \phi$ (for 500 nm)

$K_2 = \frac{hc}{200} - \phi$ (for 200 nm)

Given that $K_2 = 3K_1$, we have:

$\frac{hc}{200} - \phi = 3(\frac{hc}{500} - \phi)$

Expanding the right side:

$\frac{hc}{200} - \phi = \frac{3hc}{500} - 3\phi$

Rearrange to group $\phi$ terms:

$\frac{hc}{200} - \frac{3hc}{500} = -3\phi + \phi = -2\phi$

Multiply both sides by $-1$:

$\frac{3hc}{500} - \frac{hc}{200} = 2\phi$

Find a common denominator for the left-hand side:

$\frac{3hc \times 2 - hc \times 5}{1000} = \frac{6hc - 5hc}{1000} = \frac{hc}{1000}$

Thus:

$2\phi = \frac{hc}{1000} \Rightarrow \phi = \frac{hc}{2000}$

Taking $hc = 1240 \text{ eV}\cdot\text{nm}$:

$\phi = \frac{1240}{2000} \text{ eV} = 0.62 \text{ eV}$