91影视

The given data is listed below.

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

The stopping potential,$V_1=1.85V$

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

The stopping potential,$V_2=0.820V$

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

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

Here, the Planck鈥檚 constant in the above equation is denoted by h, E is the energy, c is the speed of light, and $位$is the wavelength.

The kinetic energy is given by-

$K_{\max }=E_{photon}-蠒$ 鈥�.. (1)

Here, $蠒$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,

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

Now, Planck鈥檚 constant can be obtained as below:

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

Substitute the below values in the above equation.

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

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

The work function $蠒$is given by-

$蠒=\frac{V_2{位}_2-V_1{位}_1}{{位}_1-{位}_2}$

Substitute the below values in the above equation.

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

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.

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

Substitute known values in the above equation.

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

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