Light of wavelength 521 nm is incident on a target metal that has a work function of 1.6 eV.What stopping potential is required for this combination in a phototube? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

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Quizplus · Accepted Answer

To find the stopping potential, we use the photoelectric equation: $K_{max} = hf - \phi$, where $K_{max}$ is the maximum kinetic energy of the ejected electrons, $h$ is Planck's constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the metal. The stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$, where $e$ is the elementary charge (1.6 x 10^-19 C).First, convert the wavelength of the light to frequency using $c = \lambda f$, where $c$ is the speed of light and $\lambda$ is the wavelength. Given $\lambda = 521 nm = 521 \times 10^{-9} m$, $c = 3.00 \times 10^8 m/s$, and $h = 6.63 \times 10^{-34} J \cdot s$, we find:$$f = \frac{c}{\lambda} = \frac{3.00 \times 10^8 m/s}{521 \times 10^{-9} m} = 5.75 \times 10^{14} Hz$$Then, calculate the energy of the photons using $E = hf$:$$E = (6.63 \times 10^{-34} J \cdot s)(5.75 \times 10^{14} Hz) = 3.81 \times 10^{-19} J$$Convert the work function from eV to Joules:$\phi = 1.6 eV = 1.6 \times 1.60 \times 10^{-19} J = 2.56 \times 10^{-19} J$Now, calculate $K_{max}$:$$K_{max} = E - \phi = 3.81 \times 10^{-19} J - 2.56 \times 10^{-19} J = 1.25 \times 10^{-19} J$$Finally, find the stopping potential $V_s$:$$V_s = \frac{K_{max}}{e} = \frac{1.25 \times 10^{-19} J}{1.60 \times 10^{-19} C} = 0.78 V$$Rounding to the nearest option gives 0.8 V.

Light of Wavelength 521 Nm Is Incident on a Target