Question

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?

Short answer

1. 382 nm.

2. 1. 82 eV.

Step-by-step solution

Identification of the given data

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,$K_{m2}=1.43V$

Significance of the kinetic energy of a particle

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.

To determine value of new wavelength

(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.

To determine the work function for the surface(b)

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.