Question

Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ = 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?

(Hint: Relate VBA to the stopping potentials Vs,A and Vs.B )

Answer 1

The unknown wavelength (λ) of light illuminating the right side of the metal plate is also 500 nm.

Step-by-step explanation:

To find the unknown wavelength (λ) of light illuminating the right side of the metal plate, we can use the photoelectric effect equation and relate the stopping potentials at electrodes A and B to the voltage across AB.

The photoelectric effect equation is given by:

eV_s = h * (c / λ)

Where:

e is the elementary charge (approximately 1.602 x 10^-19 C).

V_s is the stopping potential.

h is Planck's constant (approximately 6.626 x 10^-34 J·s).

c is the speed of light in vacuum (approximately 2.998 x 10^8 m/s).

λ is the wavelength of the incident light.

We have two stopping potentials for electrodes A and B:

V_s,A = e * VAB

V_s,B = e * VAB

Since the stopping potential V_s is the same for both electrodes A and B, we can set their respective photoelectric effect equations equal to each other:

h * (c / λ_A) = h * (c / λ_B)

where λ_A = 500 nm (given wavelength for the left side) and λ_B is the unknown wavelength for the right side.

Now, we can solve for λ_B:

c / λ_B = c / λ_A

λ_B = λ_A * (c / c)

λ_B = 500 nm * (2.998 x 10^8 m/s / 2.998 x 10^8 m/s)

λ_B = 500 nm

Therefore, the unknown wavelength (λ) of light illuminating the right side of the metal plate is also 500 nm.