A horizontal cesium plate ($\Phi=1.9 \mathrm{eV}$) is moved vertically downward at a constant speed $v$ in a room full of radiation of wavelength $250 \mathrm{~nm}$ and above. What should be the minimum value of $v$ so that the vertically upward component of velocity is nonpositive for each photoelectron?

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When $\lambda = 250 , \text{nm}$:

The energy of the photon is calculated as:

$$ \frac{hc}{\lambda} = \frac{1240}{250} = 4.96 , \text{eV} $$

Therefore, the kinetic energy (K.E.) of the ejected photoelectron is given by:

$$ \text{K.E.} = \frac{hc}{\lambda} - W = 4.96 , \text{eV} - 1.9 , \text{eV} = 3.06 , \text{eV} $$

To ensure the velocity is non-positive for each photoelectron, the minimum value of the velocity $v$ of the plate should be equal to the velocity $v_{e}$ of the photoelectron.

Thus, the velocity of the photoelectron $v_{e}$ can be derived from its kinetic energy:

$$ v_{e} = \sqrt{\frac{2 , \text{K.E.}}{m}} = \sqrt{\frac{2 \times 3.06 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} = 1.04 \times 10^{6} , \text{m/s} $$

Conclusion: The minimum speed $v$ at which the cesium plate should be moved vertically downward to ensure the vertically upward component of velocity is non-positive for each photoelectron is $1.04 \times 10^{6} , \text{m/s}$.

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