In a photoelectric experiment using a sodium surface, you find a stopping potential of 1.85 V for a wavelength of $\mathbf{300}\mathit{n}\mathit{m}$ and a stopping potential of 0.820 V for a wavelength of 400 nm. From these data find (a) a value for the Planck constant, (b) the work function for sodium, and (c) the cutoff wavelength ${\mathit{\lambda }}_{\mathbf{0}}$ for sodium?

The given data is listed below.

The wavelength of light, ${\lambda }_1=300nm$

The stopping potential,$V_1=1.85V$

The wavelength of light, ${\lambda }_2=400nm$

The stopping potential,$V_2=0.820V$

The product of energy multiplied by time; a quantity called action is known as Planck’s constant. Therefore, Planck’s s constant is given by-

$$\begin{gathered} E=\frac{hc}{\lambda } \\ h=\frac{E\lambda }{c} \end{gathered}$$

Here, the Planck’s constant in the above equation is denoted by h, E is the energy, c is the speed of light, and $\lambda$is the wavelength.

The kinetic energy is given by-

$K_{\max }=E_{photon}-\varphi$ ….. (1)

Here, $\varphi$is the work function of Sodium, $K_{\max }$is the maximum kinetic energy, and $E_{photon}$is the energy of photon.

Now the stopping potentials are,

$\begin{array}{l}e{V}_{01}=\frac{hc}{{\lambda }_1}-\varphi \\ e{V}_{02}=\frac{hc}{{\lambda }_2}-\varphi \end{array}$

Now, Planck’s constant can be obtained as below:

$h=\frac{e\left(V_1-V_2\right)}{c\left({\lambda }_1^{-1}-{\lambda }_2^{-1}\right)}$

Substitute the below values in the above equation.

$$\begin{gathered} V_1=1.85eV \\ {V}_2=0.820eV \\ {\lambda }_1=300nm \\ {\lambda }_2=400nm \end{gathered}$$

Thus, the value for the Planck constant is$4.12\times {10}^{-15}eV.s$

The work function $\varphi$is given by-

$\varphi =\frac{V_2{\lambda }_2-V_1{\lambda }_1}{{\lambda }_1-{\lambda }_2}$

Substitute the below values in the above equation.

$$\begin{gathered} V_1=1.85eV \\ {V}_2=0.820eV \\ {\lambda }_1=300nm \\ {\lambda }_2=400nm \end{gathered}$$

Hence, the work function for sodium is 2.27 eV.

For finding cut-off wavelength, take the kinetic energy zero. Therefore, $K_m=0$.

Rewrite equation (1) as below.

$\begin{array}{l}{K}_{\max }=E_{photon}-\varphi \\ 0=E_{photon}-\varphi \\ \frac{hc}{\lambda }=\varphi \\ \lambda =\frac{hc}{\varphi }\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}\lambda =\frac{\left(1240eV·nm\right)}{\left(2.27eV\right)}\\ =545nm\end{array}$

Hence, the cut-off wavelength for Sodium is 545 nm.