The stopping potential for electrons emitted from a surface illuminated by light of wavelength 491 nm is 0.710V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. (a) What is this new wavelength (b) What is the work function for the surface?

The given data is listed below as-

The first wavelength of light is ${\lambda }_1$

The second wavelength of light is ${\lambda }_2$

The maximum kinetic energy of electrons ejected by light with the first wavelength is, $K_{m1}=0.710V$

The maximum kinetic energy of electrons ejected by light with the second wavelength is,$K_{m2}=1.43V$

The photoelectric equation which is used to find the wavelength is given by-

$\frac{\mathrm{hc}}{\mathrm{\lambda }}=\varphi +{\mathrm{K}}_{\mathrm{m}}$

Here, the work function depends on material and condition of the surface and not on the wavelength of incident light.

(a)

The photoelectric equation is given by-

$\frac{hc}{\lambda }=\varphi +K_m$

Therefore, for first wavelength

$\frac{hc}{{\lambda }_1}=\varphi +K_{m1}$ ....(1)

Similarly, for the second wavelength

$\frac{hc}{{\lambda }_2}=\varphi +K_{m2}$ ....(2)

From equation (1)

${\varphi }_1=\frac{hc}{{\lambda }_1}-K_{m1}$

Substitute the above value of$\varphi$in equation (2)

$\frac{hc}{{\lambda }_2}=\frac{hc}{{\lambda }_1}-K_{m1}+K_{m2}$

Now, the solution for${\lambda }_2$is.

Thus, the new value of wavelength is 382 nm.

From the first equation $\varphi$is given by-

${\varphi }_1=\frac{hc}{{\lambda }_1}-K_{m1}$

Thus, the work function for the surface is 1.82 eV.