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

The wavelength is
`\lambda = 1029 nm`

Step-by-step 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 = (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 = (h * c)/(\lambda ) - W`

So

`eV_(AB) = eV_A - eV_B`

=>
`eV_(AB) = (h * c)/(\lambda_o ) - W - [(h * c)/(\lambda ) - W]`

=>
`eV_(AB) = (h * c)/(\lambda_o ) - (h * c)/(\lambda )`

=>
`(h * c)/(\lambda ) = (h * c)/(\lambda_o ) -eV_(AB)`

=>
`(1)/(\lambda ) =(1)/(\lambda_o ) - ( eV_(AB))/(hc)`

Where e is the charge on an electron with the value
`e = 1.60 *10^(-19) \ C`

=>
`(1)/(\lambda ) = (1)/(500 *10^(-9) ) - (1.60 *10^(-19) * 1.2775)/(6.626 *10^(-34) * 3.0 *10^(8))`

=>
`(1)/(\lambda ) = 9.717*10^(5) m^(-1)`

=>
`\lambda = 1.029 *10^(-6) \ m`

=>
`\lambda = 1029 nm`