Photoelectric effect: A stopping potential of 0.50 V is required when a phototube is illuminated with monochromatic light of wavelength 590 nm. Monochromatic light of a different wavelength is now shown on the tube, and the stopping potential is measured to be 2.30 V. What is the wavelength of this new light? (c = 3.00 × 10⁸ m/s, e = -1.60 × 10^-19 C, h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J) A) 320 nm B) 300 nm C) 340 nm D) 360 nm E) 410 nm

Question

Quizplus · Accepted Answer

The photoelectric effect can be described by the equation $E_{photon} - \phi = eV$, where $E_{photon} = \frac{hc}{\lambda}$, $\phi$ is the work function, $e$ is the charge of an electron, and $V$ is the stopping potential. Since the work function ($\phi$) of the material does not change with the wavelength of the incident light, we can set up two equations for the two different wavelengths and stopping potentials and solve for the unknown wavelength.For the first wavelength (590 nm) and stopping potential (0.50 V):$$E_{photon1} - \phi = eV_1$$$$\frac{hc}{\lambda_1} - \phi = eV_1$$For the second wavelength ($\lambda_2$) and stopping potential (2.30 V):$$E_{photon2} - \phi = eV_2$$$$\frac{hc}{\lambda_2} - \phi = eV_2$$Subtracting these two equations to eliminate $\phi$, we get:$$\frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = e(V_2 - V_1)$$Plugging in the given values ($h = 6.626 \times 10^{-34} J \cdot s$, $c = 3.00 \times 10^8 m/s$, $e = 1.60 \times 10^{-19} C$, $\lambda_1 = 590 nm = 590 \times 10^{-9} m$, $V_1 = 0.50 V$, $V_2 = 2.30 V$), we can solve for $\lambda_2$:$$\frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^8}{\lambda_2} - \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^8}{590 \times 10^{-9}} = 1.60 \times 10^{-19} \cdot (2.30 - 0.50)$$Solving this equation for $\lambda_2$ yields a value close to 320 nm, which corresponds to choice A.

Photoelectric Effect: a Stopping Potential of 0