The work function of a certain material is 4.9 eV.What is the photoelectric cut-off wavelength for this material? A)1.3 * 10^-7 m B)2.5 * 10^-7 m C)3.3 * 10^-7 m D)3.9 * 10^-7 m E)4.4 * 10^-7 m

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Quizplus · Accepted Answer

The photoelectric cut-off wavelength can be calculated using the equation $\lambda = \frac{hc}{\phi}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2\text{kg/s}$), $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$), and $\phi$ is the work function in joules. First, convert the work function from eV to joules ($1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$), so $4.9 \, \text{eV} = 4.9 \times 1.602 \times 10^{-19} \, \text{J}$. Then, substitute the values into the equation to find $\lambda$. The correct calculation leads to a wavelength around $2.5 \times 10^{-7} \, \text{m}$ or $250 \, \text{nm}$, which corresponds to choice B.

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