Question

Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the right side by light of unknown λ. Two electrodes A and B provide the stopping potential for the ejected electrons. If the voltage across AB is VAB=1.2775 V, what is the unknown λ?

Answer 1

Answer:

The  wavelength is  $\lambda = 1029 nm$

Explanation:

From the question we are told that

    The  wavelength of the left light is  $\lambda_o = 500 nm = 500 *10^{-9} \ m$

      The  voltage across A  and  B is  $V_{AB } = 1.2775 \ V$

Let the stopping potential  at A  be $V_A$ and the electric potential at B  be  $V_B$

The voltage across A and B is mathematically represented as

      $V_{AB} = V_A - V_B$

Now  According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side  in terms of electron volt is mathematically represented as

        $eV_A = \frac{h * c}{\lambda_o } - W$

Where  W is the work function of the metal

             h is the Planck constant with values  $h = 6.626 *10^{-34} \ J \cdot s$

             c  is the speed of light with value $c = 3.0 *10^{8} \ m/s$

And  the stopping potential at B for the ejected electron from the right side  in terms of electron volt is mathematically represented as

          $eV_B = \frac{h * c}{\lambda } - W$

So  

      $eV_{AB} = eV_A - eV_B$

=>    $eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W]$

=>   $eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }$

=>   $\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}$

=>  $\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}$

Where e is the charge on an electron with the value  $e = 1.60 *10^{-19} \ C$

=>   $\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}$      

=>  $\frac{1}{\lambda } = 9.717*10^{5} m^{-1}$  

=>   $\lambda = 1.029 *10^{-6} \ m$

=>   $\lambda = 1029 nm$