Question

The work function for metallic caesium is 2.24 eV. Calculate the kinetic energy and speed of electrons ejected if a light source of a) 250 nm and b) 600 nm is used.

Answer 1

Answer:

a) $4.362580048\times 10^{-19}\ Joule$

b)$0.27566898\times 10^{-19}\ Joule$

Explanation:

a) When, wavelength=λ=250 nm

$\text{Work function of metallic caesium}=2.24\ eV\\=2.24\times 1.6021\times 10^{-19}\\=3.588704\times 10^{-19}\ Joule \ \text{(converting to SI units)}\\\lambda =\text {Wavelength of light}=250\ nm\\Energy\\E=\frac{hc}{\lambda}\\\text{Where h=Plancks constant}=6.62607004\times 10^{-34} m^2kg / s\\\text{c=speed of light}=3\times 10^8\ m/s\\\Rightarrow E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{250\times 10^{-9}}\\\Rightarrow E=0.07951284048\times 10^{-17} Joule$

$\text{Kinetic energy}=\text{E - Work function}\\K.E.=(7.951284048\times 10^{-19})-(3.588704\times 10^{-19})\\K.E.=4.362580048\times 10^{-19}\ Joule$

b) When, λ=600 nm

$E=\frac{hc}{\lambda}\\E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{600\times 10^{-9}}\\\Rightarrow E=3.313030502\times 10^{-19}\\\text{Kinetic energy}=\text{E - Work function}\\K.E.=(3.31303502\times 10^{-19})-(3.588704\times 10^{-19})\quad \text{Work function remains constant}\\K.E.=-0.27566898\times 10^{-19}\ Joule$