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 energy of the photon is given by $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. The stopping potential $V$ relates to the energy of the photon by $E = eV$, where $e$ is the charge of an electron. For the new light, $E_{new} = eV_{new} = 1.60 \times 10^{-19} \times 2.30$, and solving for $\lambda_{new}$ gives a wavelength of approximately 320 nm.

A Stopping Potential of 0