Answer to Question #295173 in Mechanics | Relativity for min

Question

Ultraviolet light of wavelength 350 nm and intensity 1.00 W/m2

is directed at a potassium

surface. (a) Find the maximum KE of the photoelectrons. (b) If 0.50 percent of the incident

photons produce photoelectrons, how many are emitted per second if the potassium surface

has an area of 1.5 cm2

?

Answer 1

Answer

a) wavelength $\lambda=$ 350 nm

So energy $E=\frac{hc}{\lambda}\\=\frac{6.62\times10^{-34}\times3\times10^8}{350\times10^{-9}}\\=0.57\times10^{-18}J$

So no. Of photons

$n=\frac{1}{0.57\times10^{-18}}\\=1.75\times10^{18}per$ sec

b)Now 0.5℅ of its

=$\frac{1.75\times10^{18}}{2\times100}\\=0.87\times10^{16}$ Per sec

C) Now photoelectrons in area

$=0.87\times10^{16}\times0.00015\\=13.5\times10^{11}$