Light of wavelength 450 nm is incident on a target metal that has a work function of 1.80 eV.What stopping potential is required for this combination in a phototube? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19J,and 1 nm = 10^-9 m)

Question

Quizplus · Accepted Answer

The stopping potential can be calculated by:
$V_s = \frac{hc}{\lambda e} - \phi$
where $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of light, $e$ is the charge of an electron, and $\phi$ is the work function of the metal.

Plugging in the given values, we get:
$V_s = \frac{(6.63 	imes 10^{-34} J \cdot s)(3.00 	imes 10^8 m/s)}{(450 	imes 10^{-9} m)(1.60 	imes 10^{-19} C)} - 1.80 eV$

Simplifying this, we get:
$V_s = 0.96 V$

Therefore, the best choice is (B).

Light of Wavelength 450 Nm Is Incident on a Target