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

a)
`4.362580048* 10^(-19)\ Joule`

b)
`0.27566898* 10^(-19)\ Joule`

Step-by-step explanation:

a) When, wavelength=λ=250 nm

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

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

b) When, λ=600 nm

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